Displaying 1-10 of 213 results found.
page
1
2
3
4
5
6
7
8
9
10
... 22
Triangle of unsigned Stirling numbers of the first kind (see A048994), read by rows, T(n,k) for 0 <= k <= n.
+20
120
1, 0, 1, 0, 1, 1, 0, 2, 3, 1, 0, 6, 11, 6, 1, 0, 24, 50, 35, 10, 1, 0, 120, 274, 225, 85, 15, 1, 0, 720, 1764, 1624, 735, 175, 21, 1, 0, 5040, 13068, 13132, 6769, 1960, 322, 28, 1, 0, 40320, 109584, 118124, 67284, 22449, 4536, 546, 36, 1, 0, 362880, 1026576, 1172700
COMMENTS
Another name: Triangle of signless Stirling numbers of the first kind.
Triangle T(n,k), 0<=k<=n, read by rows given by [0,1,1,2,2,3,3,4,4,5,5,...] DELTA [1,0,1,0,1,0,1,0,1,...] where DELTA is the operator defined in A084938. Exponential Riordan array [1/(1-x), log(1/(1-x))]. - Ralf Stephan, Feb 07 2014 Also the Bell transform of the factorial numbers ( A000142). For the definition of the Bell transform see A264428 and for cross-references A265606. - Peter Luschny, Dec 31 2015 This is the lower triagonal Sheffer matrix of the associated or Jabotinsky type |S1| = (1, -log(1-x)) (see the W. Lang link under A006232 for the notation and references). This implies the e.g.f.s given below. |S1| is the transition matrix from the monomial basis {x^n} to the rising factorial basis {risefac(x,n)}, n >= 0. - Wolfdieter Lang, Feb 21 2017 T(n, k), for n >= k >= 1, is also the total volume of the n-k dimensional cell (polytope) built from the n-k orthogonal vectors of pairwise different lengths chosen from the set {1, 2, ..., n-1}. See the elementary symmetric function formula for T(n, k) and an example below. - Wolfdieter Lang, May 28 2017 The compositional inverse w.r.t. x of y = y(t;x) = x*(1 - t(-log(1-x)/x)) = x + t*log(1-x) is x = x(t;y) = ED(y,t) := Sum_{d>=0} D(d,t)*y^(d+1)/(d+1)!, the e.g.f. of the o.g.f.s D(d,t) = Sum_{m>=0} T(d+m, m)*t^m of the diagonal sequences of the present triangle. See the P. Bala link for a proof (there d = n-1, n >= 1, is the label for the diagonals).
This inversion gives D(d,t) = P(d, t)/(1-t)^(2*d+1), with the numerator polynomials P(d, t) = Sum_{m=0..d} A288874(d, m)*t^m. See an example below. See also the P. Bala formula in A112007. (End) For n > 0, T(n,k) is the number of permutations of the integers from 1 to n which have k visible digits when viewed from a specific end, in the sense that a higher value hides a lower one in a subsequent position. - Ian Duff, Jul 12 2019
REFERENCES
Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, pages 31, 187, 441, 996.
R. L. Graham, D. E. Knuth, and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 2nd. ed., Table 259, p. 259.
Steve Roman, The Umbral Calculus, Dover Publications, New York (1984), pp. 149-150
FORMULA
T(n,k) = T(n-1,k-1)+(n-1)*T(n-1,k), n,k>=1; T(n,0)=T(0,k); T(0,0)=1.
Sum_{k=0..n} T(n,k)*x^(n-k) = A000012(n), A000142(n), A001147(n), A007559(n), A007696(n), A008548(n), A008542(n), A045754(n), A045755(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8 respectively. - Philippe Deléham, Nov 13 2007 Expand 1/(1-t)^x = Sum_{n>=0}p(x,n)*t^n/n!; then the coefficients of the p(x,n) produce the triangle. - Roger L. Bagula, Apr 18 2008 Sum_{k=0..n} T(n,k)*2^k*x^(n-k) = A000142(n+1), A000165(n), A008544(n), A001813(n), A047055(n), A047657(n), A084947(n), A084948(n), A084949(n) for x = 1, 2, 3, 4, 5, 6, 7, 8, 9 respectively. - Philippe Deléham, Sep 18 2008 a(n) = Sum_{k=0..n} T(n,k)*3^k*x^(n-k) = A001710(n+2), A001147(n+1), A032031(n), A008545(n), A047056(n), A011781(n), A144739(n), A144756(n), A144758(n) for x=1,2,3,4,5,6,7,8,9,respectively. - Philippe Deléham, Sep 20 2008 Sum_{k=0..n} T(n,k)*4^k*x^(n-k) = A001715(n+3), A002866(n+1), A007559(n+1), A047053(n), A008546(n), A049308(n), A144827(n), A144828(n), A144829(n) for x=1,2,3,4,5,6,7,8,9 respectively. - Philippe Deléham, Sep 21 2008 Sum_{k=0..n} x^k*T(n,k) = x*(1+x)*(2+x)*...*(n-1+x), n>=1. - Philippe Deléham, Oct 17 2008 E.g.f. k-th column: (-log(1 - x))^k, k >= 0.
E.g.f. triangle (see the Apr 18 2008 Baluga comment): exp(-x*log(1-z)).
E.g.f. a-sequence: x/(1 - exp(-x)). See A164555/ A027642. The e.g.f. for the z-sequence is 0. (End) The row polynomials R(n, x) = Sum_{k=0..n} T(n, k)*x^k, for n >= 0, are R(n, x) = risefac(x,n-1) := Product_{j=0..n-1} x+j, with the empty product for n=0 put to 1. See the Feb 21 2017 comment above. This implies:
T(n, k) = sigma^{(n-1)}_(n-k), for n >= k >= 1, with the elementary symmetric functions sigma^{(n-1))_m of degree m in the n-1 symbols 1, 2, ..., n-1, with binomial(n-1, m) terms. See an example below.(End)
Boas-Buck type recurrence for column sequence k: T(n, k) = (n!*k/(n - k)) * Sum_{p=k..n-1} beta(n-1-p)*T(p, k)/p!, for n > k >= 0, with input T(k, k) = 1, and beta(k) = A002208(k+1)/ A002209(k+1). See a comment and references in A286718. - Wolfdieter Lang, Aug 11 2017 T(n,k) = Sum_{j=k..n} j^(j-k)*binomial(j-1, k-1)* A354795(n,j) for n > 0. - Mélika Tebni, Mar 02 2023 n-th row polynomial: n!*Sum_{k = 0..2*n} (-1)^k*binomial(-x, k)*binomial(-x, 2*n-k) = n!*Sum_{k = 0..2*n} (-1)^k*binomial(1-x, k)*binomial(-x, 2*n-k). - Peter Bala, Mar 31 2024
EXAMPLE
Triangle T(n,k) begins:
1;
0, 1;
0, 1, 1;
0, 2, 3, 1;
0, 6, 11, 6, 1;
0, 24, 50, 35, 10, 1;
0, 120, 274, 225, 85, 15, 1;
0, 720, 1764, 1624, 735, 175, 21, 1;
0, 5040, 13068, 13132, 6769, 1960, 322, 28, 1;
---------------------------------------------------
Production matrix is
0, 1
0, 1, 1
0, 1, 2, 1
0, 1, 3, 3, 1
0, 1, 4, 6, 4, 1
0, 1, 5, 10, 10, 5, 1
0, 1, 6, 15, 20, 15, 6, 1
0, 1, 7, 21, 35, 35, 21, 7, 1
...
Three term recurrence: 50 = T(5, 2) = 1*6 + (5-1)*11 = 50.
Recurrence from the Sheffer a-sequence [1, 1/2, 1/6, 0, ...]: 50 = T(5, 2) = (5/2)*(binomial(1, 1)*1*6 + binomial(2, 1)*(1/2)*11 + binomial(3, 1)*(1/6)*6 + 0) = 50. The vanishing z-sequence produces the k=0 column from T(0, 0) = 1. (End)
Elementary symmetric function T(4, 2) = sigma^{(3)}_2 = 1*2 + 1*3 + 2*3 = 11. Here the cells (polytopes) are 3 rectangles with total area 11. - Wolfdieter Lang, May 28 2017 O.g.f.s of diagonals: d=2 (third diagonal) [0, 6, 50, ...] has D(2,t) = P(2, t)/(1-t)^5, with P(2, t) = 2 + t, the n = 2 row of A288874. - Wolfdieter Lang, Jul 20 2017 Boas-Buck recurrence for column k = 2 and n= 5: T(5, 2) = (5!*2/3)*((3/8)*T(2,2)/2! + (5/12)*T(3,2)/3! + (1/2)*T(4,2)/4!) = (5!*2/3)*((3/16 + (5/12)*3/3! + (1/2)*11/4!) = 50. The beta sequence begins: {1/2, 5/12, 3/8, ...}. - Wolfdieter Lang, Aug 11 2017
MAPLE
a132393_row := proc(n) local k; seq(coeff(expand(pochhammer (x, n)), x, k), k=0..n) end: # Peter Luschny, Nov 28 2010
MATHEMATICA
p[t_] = 1/(1 - t)^x; Table[ ExpandAll[(n!)SeriesCoefficient[ Series[p[t], {t, 0, 30}], n]], {n, 0, 10}]; a = Table[(n!)* CoefficientList[SeriesCoefficient[ Series[p[t], {t, 0, 30}], n], x], {n, 0, 10}]; Flatten[a] (* Roger L. Bagula, Apr 18 2008 *) Flatten[Table[Abs[StirlingS1[n, i]], {n, 0, 10}, {i, 0, n}]] (* Harvey P. Dale, Feb 04 2014 *)
PROG
(Maxima) create_list(abs(stirling1(n, k)), n, 0, 12, k, 0, n); /* Emanuele Munarini, Mar 11 2011 */ (Haskell)
a132393 n k = a132393_tabl !! n !! k
a132393_row n = a132393_tabl !! n
a132393_tabl = map (map abs) a048994_tabl
Numerators of upper right triangle of a(i,j) = Integral_{x=i..i+1} Sum_{k=0..j} A048994(j,k)*x^k.
+20
7
1, 1, 3, -1, 5, 23, 1, -1, 9, 55, -19, 11, -19, 251, 1901, 27, -11, 11, -27, 475, 4277, -863, 271, -191, 271, -863, 19087, 198721, 1375, -351, 191, -191, 351, -1375, 36799, 434241, -33953, 7297, -3233, 2497, -3233, 7297, -33953, 1070017, 14097247, 57281, -10625, 3969
COMMENTS
Denominators of the j-th column are A002790(j). Note that the fractions defined by division are not fully reduced to coprime numerator and denominator.
REFERENCES
P. Curtz, Intégration numérique des systèmes différentiels à conditions initiales, Centre de Calcul Scientifique de l'Armement, Arcueil, 1969.
EXAMPLE
The array a(i,j) starts with rows i>=0 and columns j>=0 as:
1 1/2 -1/6 1/4 -19/30 9/4 -863/84 1375/24 ...
1 3/2 5/6 -1/4 11/30 -11/12 271/84 -117/8 ...
1 5/2 23/6 9/4 -19/30 11/12 -191/84 191/24 ...
1 7/2 53/6 55/4 251/30 -9/4 271/84 -191/24 ...
1 9/2 95/6 161/4 1901/30 475/12 -863/84 117/8 ...
1 11/2 149/6 351/4 6731/30 4277/12 19087/84 -1375/24 ...
The sequence lists the numerators of the j-th column from row 0 down to row j.
The fractions of the j=5 column, 9/4, -11/12, 11/12, -9/4, 475/12, 4277/12, are listed with a common denominator A002790(5)=12 as 27, -11, 11, -27, 475, 4277.
MATHEMATICA
a[i_, j_] := Sum[((1+i)^(k+1)-i^(k+1))*StirlingS1[j, k]/(k+1), {k, 0, j}]; col[j_] := Total[Table[a[i, j], {i, 0, j} ]*x^Range[0, j]] // Together // Numerator // CoefficientList[#, x]&; Table[col[j], {j, 0, 9}] // Flatten (* Jean-François Alcover, Jan 10 2016 *)
Triangle formed by reading triangle of Stirling numbers of the first kind ( A048994) mod 2.
+20
4
1, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1
REFERENCES
Brand, Neal; Das, Sajal; Jacob, Tom. The number of nonzero entries in recursively defined tables modulo primes. Proceedings of the Twenty-first Southeastern Conference on Combinatorics, Graph Theory, and Computing (Boca Raton, FL, 1990). Congr. Numer. 78 (1990), 47--59. MR1140469 (92h:05004). - From N. J. A. Sloane, Jun 03 2012
FORMULA
T(n, k) = A087755(n, k) = A048994(n, k) mod 2 = A047999([n/2], k-[(n+1)/2]) = T(n-2, k-2) XOR T(n-2, k-1) with T(0, 0) = T(1, 1) = 1 and T(1, 0) = 0; T(2n, k) = T(2n-1, k-1) XOR T(2n-1, k); T(2n+1, k) = T(2n, k-1). - Henry Bottomley, Dec 01 2003
EXAMPLE
Triangle begins:
1,
0, 1,
0, 1, 1,
0, 0, 1, 1,
0, 0, 1, 0, 1,
0, 0, 0, 1, 0, 1,
0, 0, 0, 1, 1, 1, 1,
0, 0, 0, 0, 1, 1, 1, 1,
0, 0, 0, 0, 1, 0, 0, 0, 1,
0, 0, 0, 0, 0, 1, 0, 0, 0, 1,
0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1,
0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1,
0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1,
...
a(n) = sum_{k=0,...,[n/2]} s(n-k,k)^2, s = A048994, Stirling numbers of the first kind.
+20
0
1, 0, 1, 1, 5, 45, 698, 16936, 594702, 28564021, 1799119626, 143863537330, 14236046853139, 1707698666758297, 244136528082097062, 41008147506862052681, 7995945735393199219626, 1791074412870104676689001, 456745672286592586379503743
LINKS
Edyta Hetmaniok, Barbara Smoleń, Roman Wituła, The Stirling triangles, Proceedings of the Symposium for Young Scientists in Technology, Engineering and Mathematics (SYSTEM 2017), Kaunas, Lithuania, April 28, 2017, p. 35-41.
a(n) = sum_{k=0,...,[n/2]} |s(n-k,k)|^3, s = A048994, Stirling numbers of the first kind.
+20
0
1, 0, 1, 1, 9, 243, 15156, 1853216, 393861700, 133524487369, 67784261131182, 49102947079265422, 48868239988727255585, 64803779202807835851565, 111657015638972745549794074, 244745390650212498564219429909, 670332605628298040569504378787338
LINKS
Edyta Hetmaniok, Barbara Smoleń, Roman Wituła, The Stirling triangles, Proceedings of the Symposium for Young Scientists in Technology, Engineering and Mathematics (SYSTEM 2017), Kaunas, Lithuania, April 28, 2017, p. 35-41.
MATHEMATICA
Abs[Table[Sum[StirlingS1[n-k, k]^3, {k, 0, Floor[n/2]}], {n, 0, 20}]] (* Harvey P. Dale, Apr 03 2021 *)
Triangular numbers: a(n) = binomial(n+1,2) = n*(n+1)/2 = 0 + 1 + 2 + ... + n.
(Formerly M2535 N1002)
+10
4676
0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136, 153, 171, 190, 210, 231, 253, 276, 300, 325, 351, 378, 406, 435, 465, 496, 528, 561, 595, 630, 666, 703, 741, 780, 820, 861, 903, 946, 990, 1035, 1081, 1128, 1176, 1225, 1275, 1326, 1378, 1431
COMMENTS
Also referred to as T(n) or C(n+1, 2) or binomial(n+1, 2) (preferred).
Also generalized hexagonal numbers: n*(2*n-1), n=0, +-1, +-2, +-3, ... Generalized k-gonal numbers are second k-gonal numbers and positive terms of k-gonal numbers interleaved, k >= 5. In this case k = 6. - Omar E. Pol, Sep 13 2011 and Aug 04 2012 Number of edges in complete graph of order n+1, K_{n+1}.
Number of legal ways to insert a pair of parentheses in a string of n letters. E.g., there are 6 ways for three letters: (a)bc, (ab)c, (abc), a(b)c, a(bc), ab(c). Proof: there are C(n+2,2) ways to choose where the parentheses might go, but n + 1 of them are illegal because the parentheses are adjacent. Cf. A002415. For n >= 1, a(n) is also the genus of a nonsingular curve of degree n+2, such as the Fermat curve x^(n+2) + y^(n+2) = 1. - Ahmed Fares (ahmedfares(AT)my_deja.com), Feb 21 2001
From Harnack's theorem (1876), the number of branches of a nonsingular curve of order n is bounded by a(n-1)+1, and the bound can be achieved. See also A152947. - Benoit Cloitre, Aug 29 2002. Corrected by Robert McLachlan, Aug 19 2024 Number of tiles in the set of double-n dominoes. - Scott A. Brown, Sep 24 2002 Number of ways a chain of n non-identical links can be broken up. This is based on a similar problem in the field of proteomics: the number of ways a peptide of n amino acid residues can be broken up in a mass spectrometer. In general, each amino acid has a different mass, so AB and BC would have different masses. - James A. Raymond, Apr 08 2003 Triangular numbers - odd numbers = shifted triangular numbers; 1, 3, 6, 10, 15, 21, ... - 1, 3, 5, 7, 9, 11, ... = 0, 0, 1, 3, 6, 10, ... - Xavier Acloque, Oct 31 2003 [Corrected by Derek Orr, May 05 2015] Centered polygonal numbers are the result of [number of sides * A000217 + 1]. E.g., centered pentagonal numbers (1,6,16,31,...) = 5 * (0,1,3,6,...) + 1. Centered heptagonal numbers (1,8,22,43,...) = 7 * (0,1,3,6,...) + 1. - Xavier Acloque, Oct 31 2003 Maximum number of lines formed by the intersection of n+1 planes. - Ron R. King, Mar 29 2004 Number of permutations of [n] which avoid the pattern 132 and have exactly 1 descent. - Mike Zabrocki, Aug 26 2004 Number of ternary words of length n-1 with subwords (0,1), (0,2) and (1,2) not allowed. - Olivier Gérard, Aug 28 2012 Number of ways two different numbers can be selected from the set {0,1,2,...,n} without repetition, or, number of ways two different numbers can be selected from the set {1,2,...,n} with repetition.
Conjecturally, 1, 6, 120 are the only numbers that are both triangular and factorial. - Christopher M. Tomaszewski (cmt1288(AT)comcast.net), Mar 30 2005
Binomial transform is {0, 1, 5, 18, 56, 160, 432, ...}, A001793 with one leading zero. - Philippe Deléham, Aug 02 2005 Each pair of neighboring terms adds to a perfect square. - Zak Seidov, Mar 21 2006 Number of transpositions in the symmetric group of n+1 letters, i.e., the number of permutations that leave all but two elements fixed. - Geoffrey Critzer, Jun 23 2006 With rho(n):=exp(i*2*Pi/n) (an n-th root of 1) one has, for n >= 1, rho(n)^a(n) = (-1)^(n+1). Just use the triviality a(2*k+1) == 0 (mod (2*k+1)) and a(2*k) == k (mod (2*k)).
a(n) is the number of terms in the expansion of (a_1 + a_2 + a_3)^(n-1). - Sergio Falcon, Feb 12 2007 a(n+1) is the number of terms in the complete homogeneous symmetric polynomial of degree n in 2 variables. - Richard Barnes, Sep 06 2017 Equal to the rank (minimal cardinality of a generating set) of the semigroup PT_n\S_n, where PT_n and S_n denote the partial transformation semigroup and symmetric group on [n]. - James East, May 03 2007 a(n) gives the total number of triangles found when cevians are drawn from a single vertex on a triangle to the side opposite that vertex, where n = the number of cevians drawn+1. For instance, with 1 cevian drawn, n = 1+1 = 2 and a(n)= 2*(2+1)/2 = 3 so there is a total of 3 triangles in the figure. If 2 cevians are drawn from one point to the opposite side, then n = 1+2 = 3 and a(n) = 3*(3+1)/2 = 6 so there is a total of 6 triangles in the figure. - Noah Priluck (npriluck(AT)gmail.com), Apr 30 2007
For n >= 1, a(n) is the number of ways in which n-1 can be written as a sum of three nonnegative integers if representations differing in the order of the terms are considered to be different. In other words, for n >= 1, a(n) is the number of nonnegative integral solutions of the equation x + y + z = n-1. - Amarnath Murthy, Apr 22 2001 (edited by Robert A. Beeler) a(n) is the number of levels with energy n + 3/2 (in units of h*f0, with Planck's constant h and the oscillator frequency f0) of the three-dimensional isotropic harmonic quantum oscillator. See the comment by A. Murthy above: n = n1 + n2 + n3 with positive integers and ordered. Proof from the o.g.f. See the A. Messiah reference. - Wolfdieter Lang, Jun 29 2007 Numbers m >= 0 such that round(sqrt(2m+1)) - round(sqrt(2m)) = 1.
Numbers m >= 0 such that ceiling(2*sqrt(2m+1)) - 1 = 1 + floor(2*sqrt(2m)).
Numbers m >= 0 such that fract(sqrt(2m+1)) > 1/2 and fract(sqrt(2m)) < 1/2, where fract(x) is the fractional part of x (i.e., x - floor(x), x >= 0). (End)
If Y and Z are 3-blocks of an n-set X, then, for n >= 6, a(n-1) is the number of (n-2)-subsets of X intersecting both Y and Z. - Milan Janjic, Nov 09 2007 a(n) is also a perfect number A000396 if n is a Mersenne prime A000668, assuming there are no odd perfect numbers. - Omar E. Pol, Sep 05 2008 The number of matches played in a round robin tournament: n*(n-1)/2 gives the number of matches needed for n players. Everyone plays against everyone else exactly once. - Georg Wrede (georg(AT)iki.fi), Dec 18 2008
-a(n+1) = E(2)*binomial(n+2,2) (n >= 0) where E(n) are the Euler numbers in the enumeration A122045. Viewed this way, a(n) is the special case k=2 in the sequence of diagonals in the triangle A153641. - Peter Luschny, Jan 06 2009 Equivalent to the first differences of successive tetrahedral numbers. See A000292. - Jeremy Cahill (jcahill(AT)inbox.com), Apr 15 2009 The general formula for alternating sums of powers is in terms of the Swiss-Knife polynomials P(n,x) A153641 2^(-n-1)(P(n,1)-(-1)^k P(n,2k+1)). Thus a(k) = |2^(-3)(P(2,1)-(-1)^k P(2,2k+1))|. - Peter Luschny, Jul 12 2009 a(n) is the smallest number > a(n-1) such that gcd(n,a(n)) = gcd(n,a(n-1)). If n is odd this gcd is n; if n is even it is n/2. - Franklin T. Adams-Watters, Aug 06 2009 The numbers along the right edge of Floyd's triangle are 1, 3, 6, 10, 15, .... - Paul Muljadi, Jan 25 2010 More generally, a(2k+1) == j*(2j-1) (mod 2k+2j+1) and
a(2k) == [-k + 2j*(j-1)] (mod 2k+2j).
Column sums of:
1 3 5 7 9 ...
1 3 5 ...
1 ...
...............
---------------
1 3 6 10 15 ...
Sum_{n>=1} 1/a(n)^2 = 4*Pi^2/3-12 = 12 less than the volume of a sphere with radius Pi^(1/3).
(End)
1/a(n+1), n >= 0, has e.g.f. -2*(1+x-exp(x))/x^2, and o.g.f. 2*(x+(1-x)*log(1-x))/x^2 (see the Stephen Crowley formula line). -1/(2*a(n+1)) is the z-sequence for the Sheffer triangle of the coefficients of the Bernoulli polynomials A196838/ A196839. - Wolfdieter Lang, Oct 26 2011 (End)
Plot the three points (0,0), (a(n), a(n+1)), (a(n+1), a(n+2)) to form a triangle. The area will be a(n+1)/2. - J. M. Bergot, May 04 2012 The sum of four consecutive triangular numbers, beginning with a(n)=n*(n+1)/2, minus 2 is 2*(n+2)^2. a(n)*a(n+2)/2 = a(a(n+1)-1). - J. M. Bergot, May 17 2012 (a(n)*a(n+3) - a(n+1)*a(n+2))*(a(n+1)*a(n+4) - a(n+2)*a(n+3))/8 = a((n^2+5*n+4)/2). - J. M. Bergot, May 18 2012 a(n)*a(n+1) + a(n+2)*a(n+3) + 3 = a(n^2 + 4*n + 6). - J. M. Bergot, May 22 2012 In general, a(n)*a(n+1) + a(n+k)*a(n+k+1) + a(k-1)*a(k) = a(n^2 + (k+2)*n + k*(k+1)). - Charlie Marion, Sep 11 2012 a(n)*a(n+3) + a(n+1)*a(n+2) = a(n^2 + 4*n + 2). - J. M. Bergot, May 22 2012 In general, a(n)*a(n+k) + a(n+1)*a(n+k-1) = a(n^2 + (k+1)*n + k-1). - Charlie Marion, Sep 11 2012 a(n)*a(n+2) + a(n+1)*a(n+3) = a(n^2 + 4*n + 3). - J. M. Bergot, May 22 2012 Three points (a(n),a(n+1)), (a(n+1),a(n)) and (a(n+2),a(n+3)) form a triangle with area 4*a(n+1). - J. M. Bergot, May 23 2012 a(n) + a(n+k) = (n+k)^2 - (k^2 + (2n-1)*k -2n)/2. For k=1 we obtain a(n) + a(n+1) = (n+1)^2 (see below). - Charlie Marion, Oct 02 2012 In n-space we can define a(n-1) nontrivial orthogonal projections. For example, in 3-space there are a(2)=3 (namely point onto line, point onto plane, line onto plane). - Douglas Latimer, Dec 17 2012 For n >= 1, a(n) is equal to the rank (minimal cardinality of a generating set) and idempotent rank (minimal cardinality of an idempotent generating set) of the semigroup P_n\S_n, where P_n and S_n denote the partition monoid and symmetric group on [n].
For n >= 3, a(n-1) is equal to the rank and idempotent rank of the semigroup T_n\S_n, where T_n and S_n denote the full transformation semigroup and symmetric group on [n].
(End)
For n >= 3, a(n) is equal to the rank and idempotent rank of the semigroup PT_n\S_n, where PT_n and S_n denote the partial transformation semigroup and symmetric group on [n]. - James East, Jan 15 2013 The formula, a(n)*a(n+4k+2)/2 + a(k) = a(a(n+2k+1) - (k^2+(k+1)^2)), is a generalization of the formula a(n)*a(n+2)/2 = a(a(n+1)-1) in Bergot's comment dated May 17 2012. - Charlie Marion, Mar 28 2013 For odd m = 2k+1, we have the recurrence a(m*n + k) = m^2*a(n) + a(k). Corollary: If number T is in the sequence then so is 9*T+1. - Lekraj Beedassy, May 29 2013 Euler, in Section 87 of the Opera Postuma, shows that whenever T is a triangular number then 9*T + 1, 25*T + 3, 49*T + 6 and 81*T + 10 are also triangular numbers. In general, if T is a triangular number then (2*k + 1)^2*T + k*(k + 1)/2 is also a triangular number. - Peter Bala, Jan 05 2015 Using 1/b and 1/(b+2) will give a Pythagorean triangle with sides 2*b + 2, b^2 + 2*b, and b^2 + 2*b + 2. Set b=n-1 to give a triangle with sides of lengths 2*n,n^2-1, and n^2 + 1. One-fourth the perimeter = a(n) for n > 1. - J. M. Bergot, Jul 24 2013 a(n) = A028896(n)/6, where A028896(n) = s(n) - s(n-1) are the first differences of s(n) = n^3 + 3*n^2 + 2*n - 8. s(n) can be interpreted as the sum of the 12 edge lengths plus the sum of the 6 face areas plus the volume of an n X (n-1) X (n-2) rectangular prism. - J. M. Bergot, Aug 13 2013 Number of positive roots in the root system of type A_n (for n > 0). - Tom Edgar, Nov 05 2013 A formula for the r-th successive summation of k, for k = 1 to n, is binomial(n+r,r+1) [H. W. Gould]. - Gary Detlefs, Jan 02 2014 For n >= 3, a(n-2) is the number of permutations of 1,2,...,n with the distribution of up (1) - down (0) elements 0...011 (n-3 zeros), or, the same, a(n-2) is up-down coefficient {n,3} (see comment in A060351). - Vladimir Shevelev, Feb 14 2014 a(n) is the dimension of the vector space of symmetric n X n matrices. - Derek Orr, Mar 29 2014 Non-vanishing subdiagonal of A132440^2/2, aside from the initial zero. First subdiagonal of unsigned A238363. Cf. A130534 for relations to colored forests, disposition of flags on flagpoles, and colorings of the vertices of complete graphs. - Tom Copeland, Apr 05 2014 The number of Sidon subsets of {1,...,n+1} of size 2. - Carl Najafi, Apr 27 2014 Number of factors in the definition of the Vandermonde determinant V(x_1,x_2,...,x_n) = Product_{1 <= i < k <= n} x_i - x_k. - Tom Copeland, Apr 27 2014 Number of weak compositions of n into three parts. - Robert A. Beeler, May 20 2014 Suppose a bag contains a(n) red marbles and a(n+1) blue marbles, where a(n), a(n+1) are consecutive triangular numbers. Then, for n > 0, the probability of choosing two marbles at random and getting two red or two blue is 1/2. In general, for k > 2, let b(0) = 0, b(1) = 1 and, for n > 1, b(n) = (k-1)*b(n-1) - b(n-2) + 1. Suppose, for n > 0, a bag contains b(n) red marbles and b(n+1) blue marbles. Then the probability of choosing two marbles at random and getting two red or two blue is (k-1)/(k+1). See also A027941, A061278, A089817, A053142, A092521. - Charlie Marion, Nov 03 2014 Let O(n) be the oblong number n(n+1) = A002378 and S(n) the square number n^2 = A000290(n). Then a(4n) = O(3n) - O(n), a(4n+1) = S(3n+1) - S(n), a(4n+2) = S(3n+2) - S(n+1) and a(4n+3) = O(3n+2) - O(n). - Charlie Marion, Feb 21 2015 Consider the partition of the natural numbers into parts from the set S=(1,2,3,...,n). The length (order) of the signature of the resulting sequence is given by the triangular numbers. E.g., for n=10, the signature length is 55. - David Neil McGrath, May 05 2015 a(n) counts the partitions of (n-1) unlabeled objects into three (3) parts (labeled a,b,c), e.g., a(5)=15 for (n-1)=4. These are (aaaa),(bbbb),(cccc),(aaab),(aaac),(aabb),(aacc),(aabc),(abbc),(abcc),(abbb),(accc ),(bbcc),(bccc),(bbbc). - David Neil McGrath, May 21 2015Conjecture: the sequence is the genus/deficiency of the sinusoidal spirals of index n which are algebraic curves. The value 0 corresponds to the case of the Bernoulli Lemniscate n=2. So the formula conjectured is (n-1)(n-2)/2. - Wolfgang Tintemann, Aug 02 2015 Conjecture: Let m be any positive integer. Then, for each n = 1,2,3,... the set {Sum_{k=s..t} 1/k^m: 1 <= s <= t <= n} has cardinality a(n) = n*(n+1)/2; in other words, all the sums Sum_{k=s..t} 1/k^m with 1 <= s <= t are pairwise distinct. (I have checked this conjecture via a computer and found no counterexample.) - Zhi-Wei Sun, Sep 09 2015 The Pisano period lengths of reading the sequence modulo m seem to be A022998(m). - R. J. Mathar, Nov 29 2015 For n >= 1, a(n) is the number of compositions of n+4 into n parts avoiding the part 2. - Milan Janjic, Jan 07 2016 In this sequence only 3 is prime. - Fabian Kopp, Jan 09 2016 Suppose you are playing Bulgarian Solitaire (see A242424 and Chamberland's and Gardner's books) and, for n > 0, you are starting with a single pile of a(n) cards. Then the number of operations needed to reach the fixed state {n, n-1,...,1} is a(n-1). For example, {6}->{5,1}->{4,2}->{3,2,1}. - Charlie Marion, Jan 14 2016 Every perfect cube is the difference of the squares of two consecutive triangular numbers. 1^2-0^2 = 1^3, 3^2-1^2 = 2^3, 6^2-3^2 = 3^3. - Miquel Cerda, Jun 26 2016 For n > 1, a(n) = tau_n(k*) where tau_n(k) is the number of ordered n-factorizations of k and k* is the square of a prime. For example, tau_3(4) = tau_3(9) = tau_3(25) = tau_3(49) = 6 (see A007425) since the number of divisors of 4, 9, 25, and 49's divisors is 6, and a(3) = 6. - Melvin Peralta, Aug 29 2016 In an (n+1)-dimensional hypercube, number of two-dimensional faces congruent with a vertex (see also A001788). - Stanislav Sykora, Oct 23 2016 Generalizations of the familiar formulas, a(n) + a(n+1) = (n+1)^2 (Feb 19 2004) and a(n)^2 + a(n+1)^2 = a((n+1)^2) (Nov 22 2006), follow: a(n) + a(n+2k-1) + 4a(k-1) = (n+k)^2 + 6a(k-1) and a(n)^2 + a(n+2k-1)^2 + (4a(k-1))^2 + 3a(k-1) = a((n+k)^2 + 6a(k-1)). - Charlie Marion, Nov 27 2016 a(n) is also the greatest possible number of diagonals in a polyhedron with n+4 vertices. - Vladimir Letsko, Dec 19 2016 For n > 0, 2^5 * (binomial(n+1,2))^2 represents the first integer in a sum of 2*(2*n + 1)^2 consecutive integers that equals (2*n + 1)^6. - Patrick J. McNab, Dec 25 2016 Does not satisfy Benford's law (cf. Ross, 2012). - N. J. A. Sloane, Feb 12 2017 Number of ordered triples (a,b,c) of positive integers not larger than n such that a+b+c = 2n+1. - Aviel Livay, Feb 13 2017 Number of inequivalent tetrahedral face colorings using at most n colors so that no color appears only once. - David Nacin, Feb 22 2017 Also the Wiener index of the complete graph K_{n+1}. - Eric W. Weisstein, Sep 07 2017 Number of intersections between the Bernstein polynomials of degree n. - Eric Desbiaux, Apr 01 2018 a(n) is the area of a triangle with vertices at (1,1), (n+1,n+2), and ((n+1)^2, (n+2)^2). - Art Baker, Dec 06 2018 For n > 0, a(n) is the smallest k > 0 such that n divides numerator of (1/a(1) + 1/a(2) + ... + 1/a(n-1) + 1/k). It should be noted that 1/1 + 1/3 + 1/6 + ... + 2/(n(n+1)) = 2n/(n+1). - Thomas Ordowski, Aug 04 2019 Upper bound of the number of lines in an n-homogeneous supersolvable line arrangement (see Theorem 1.1 in Dimca). - Stefano Spezia, Oct 04 2019 For n > 0, a(n+1) is the number of lattice points on a triangular grid with side length n. - Wesley Ivan Hurt, Aug 12 2020 Maximum number of distinct nonempty substrings of a string of length n.
Maximum cardinality of the sumset A+A, where A is a set of n numbers. (End)
a(n) is the number of parking functions of size n avoiding the patterns 123, 132, and 312. - Lara Pudwell, Apr 10 2023 Suppose two rows, each consisting of n evenly spaced dots, are drawn in parallel. Suppose we bijectively draw lines between the dots of the two rows. For n >= 1, a(n - 1) is the maximal possible number of intersections between the lines. Equivalently, the maximal number of inversions in a permutation of [n]. - Sela Fried, Apr 18 2023 The following equation complements the generalization in Bala's Comment (Jan 05 2015). (2k + 1)^2*a(n) + a(k) = a((2k + 1)*n + k). - Charlie Marion, Aug 28 2023 a(n) + a(n+k) + a(k-1) + (k-1)*n = (n+k)^2. For k = 1, we have a(n) + a(n+1) = (n+1)^2. - Charlie Marion, Nov 17 2023
REFERENCES
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 828.
C. Alsina and R. B. Nelson, Charming Proofs: A Journey into Elegant Mathematics, MAA, 2010. See Chapter 1.
T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 2.
A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 189.
A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 109ff.
Marc Chamberland, Single Digits: In Praise of Small Numbers, Chapter 3, The Number Three, p. 72, Princeton University Press, 2015.
L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 155.
J. M. De Koninck and A. Mercier, 1001 Problèmes en Théorie Classique des Nombres, Problème 309 pp 46-196, Ellipses, Paris, 2004
E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 6.
L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 2, p. 1.
Martin Gardner, Colossal Book of Mathematics, Chapter 34, Bulgarian Solitaire and Other Seemingly Endless Tasks, pp. 455-467, W. W. Norton & Company, 2001.
James Gleick, The Information: A History, A Theory, A Flood, Pantheon, 2011. [On page 82 mentions a table of the first 19999 triangular numbers published by E. de Joncort in 1762.]
Cay S. Horstmann, Scala for the Impatient. Upper Saddle River, New Jersey: Addison-Wesley (2012): 171.
Labos E.: On the number of RGB-colors we can distinguish. Partition Spectra. Lecture at 7th Hungarian Conference on Biometry and Biomathematics. Budapest. Jul 06 2005.
A. Messiah, Quantum Mechanics, Vol.1, North Holland, Amsterdam, 1965, p. 457.
J. C. P. Miller, editor, Table of Binomial Coefficients. Royal Society Mathematical Tables, Vol. 3, Cambridge Univ. Press, 1954.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
T. Trotter, Some Identities for the Triangular Numbers, Journal of Recreational Mathematics, Spring 1973, 6(2).
D. Wells, The Penguin Dictionary of Curious and Interesting Numbers, pp. 91-93 Penguin Books 1987.
LINKS
C. Hamberg, Triangular Numbers Are Everywhere, llinois Mathematics and Science Academy, IMSA Math Journal: a Resource Notebook for High School Mathematics (1992), pp. 7-10. Ed Pegg, Jr., Sequence Pictures, Math Games column, Dec 08 2003 [Cached copy, with permission (pdf only)] Michel Waldschmidt, Continued fractions, École de recherche CIMPA-Oujda, Théorie des Nombres et ses Applications, 18 - 29 mai 2015: Oujda (Maroc). Eric Weisstein's World of Mathematics, Absolute Value, Binomial Coefficient, Composition, Distance, Golomb Ruler, Line Line Picking, Polygonal Number, Triangular Number, Trinomial Coefficient, and Wiener Index
FORMULA
E.g.f.: exp(x)*(x+x^2/2).
a(n) = a(-1-n).
a(n+1) = ((n+2)/n)*a(n), Sum_{n>=1} 1/a(n) = 2. - Jon Perry, Jul 13 2003 For n > 0, a(n) = A001109(n) - Sum_{k=0..n-1} (2*k+1)* A001652(n-1-k); e.g., 10 = 204 - (1*119 + 3*20 + 5*3 + 7*0). - Charlie Marion, Jul 18 2003 With interpolated zeros, this is n*(n+2)*(1+(-1)^n)/16. - Benoit Cloitre, Aug 19 2003 a(n+1) is the determinant of the n X n symmetric Pascal matrix M_(i, j) = binomial(i+j+1, i). - Benoit Cloitre, Aug 19 2003 a(n) = ((n+1)^3 - n^3 - 1)/6. - Xavier Acloque, Oct 24 2003
a(n) = a(n-1) + (1 + sqrt(1 + 8*a(n-1)))/2. This recursive relation is inverted when taking the negative branch of the square root, i.e., a(n) is transformed into a(n-1) rather than a(n+1). - Carl R. White, Nov 04 2003 a(n) = a(n-2) + 2*n - 1. - Paul Barry, Jul 17 2004 a(n) = sqrt(sqrt(Sum_{i=1..n} Sum_{j=1..n} (i*j)^3)) = (Sum_{i=1..n} Sum_{j=1..n} Sum_{k=1..n} (i*j*k)^3)^(1/6). - Alexander Adamchuk, Oct 26 2004 a(n) == 1 (mod n+2) if n is odd and a(n) == n/2+2 (mod n+2) if n is even. - Jon Perry, Dec 16 2004 a(0) = 0, a(1) = 1, a(n) = 2*a(n-1) - a(n-2) + 1. - Miklos Kristof, Mar 09 2005 a(n) = floor((2*n+1)^2/8). - Paul Barry, May 29 2006 a(n)^2 + a(n+1)^2 = a((n+1)^2) [R B Nelsen, Math Mag 70 (2) (1997), p. 130]. - R. J. Mathar, Nov 22 2006 a(n)*a(n+k)+a(n+1)*a(n+1+k) = a((n+1)*(n+1+k)). Generalizes previous formula dated Nov 22 2006 [and comments by J. M. Bergot dated May 22 2012]. - Charlie Marion, Feb 04 2011 (sqrt(8*a(n)+1)-1)/2 = n. - David W. Cantrell (DWCantrell(AT)sigmaxi.net), Feb 26 2007
A general formula for polygonal numbers is P(k,n) = (k-2)*(n-1)n/2 + n = n + (k-2)* A000217(n-1), for n >= 1, k >= 3. - Omar E. Pol, Apr 28 2008 and Mar 31 2013 If we define f(n,i,a) = Sum_{j=0..k-1} (binomial(n,k)*Stirling1(n-k,i)*Product_{j=0..k-1} (-a-j)), then a(n) = -f(n,n-1,1), for n >= 1. - Milan Janjic, Dec 20 2008 An exponential generating function for the inverse of this sequence is given by Sum_{m>=0} ((Pochhammer(1, m)*Pochhammer(1, m))*x^m/(Pochhammer(3, m)*factorial(m))) = ((2-2*x)*log(1-x)+2*x)/x^2, the n-th derivative of which has a closed form which must be evaluated by taking the limit as x->0. A000217(n+1) = (lim_{x->0} d^n/dx^n (((2-2*x)*log(1-x)+2*x)/x^2))^-1 = (lim_{x->0} (2*Gamma(n)*(-1/x)^n*(n*(x/(-1+x))^n*(-x+1+n)*LerchPhi(x/(-1+x), 1, n) + (-1+x)*(n+1)*(x/(-1+x))^n + n*(log(1-x)+log(-1/(-1+x)))*(-x+1+n))/x^2))^-1. - Stephen Crowley, Jun 28 2009 With offset 1, a(n) = floor(n^3/(n+1))/2. - Gary Detlefs, Feb 14 2010 a(n) = 4*a(floor(n/2)) + (-1)^(n+1)*floor((n+1)/2). - Bruno Berselli, May 23 2010 a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3); a(0)=0, a(1)=1. - Mark Dols, Aug 20 2010 a(n) + 2*a(n-1) + a(n-2) = n^2 + (n-1)^2; and
a(n) + 3*a(n-1) + 3*a(n-2) + a(n-3) = n^2 + 2*(n-1)^2 + (n-2)^2.
In general, for n >= m > 2, Sum_{k=0..m} binomial(m,m-k)*a(n-k) = Sum_{k=0..m-1} binomial(m-1,m-1-k)*(n-k)^2.
a(n) - 2*a(n-1) + a(n-2) = 1, a(n) - 3*a(n-1) + 3*a(n-2) - a(n-3) = 0 and a(n) - 4*a(n-1) + 6*a(n-2) - 4*(a-3) + a(n-4) = 0.
In general, for n >= m > 2, Sum_{k=0..m} (-1)^k*binomial(m,m-k)*a(n-k) = 0.
(End)
For n > 0, a(n) = 1/(Integral_{x=0..Pi/2} 4*(sin(x))^(2*n-1)*(cos(x))^3). - Francesco Daddi, Aug 02 2011 a(n) = -s(n+1,n), where s(n,k) are the Stirling numbers of the first kind, A048994. - Mircea Merca, May 03 2012 a(n)*a(n+1) = a(Sum_{m=1..n} A005408(m))/2, for n >= 1. For example, if n=8, then a(8)*a(9) = a(80)/2 = 1620. - Ivan N. Ianakiev, May 27 2012 G.f.: x * (1 + 3x + 6x^2 + ...) = x * Product_{j>=0} (1+x^(2^j))^3 = x * A(x) * A(x^2) * A(x^4) * ..., where A(x) = (1 + 3x + 3x^2 + x^3). - Gary W. Adamson, Jun 26 2012 G.f.: G(0) where G(k) = 1 + (2*k+3)*x/(2*k+1 - x*(k+2)*(2*k+1)/(x*(k+2) + (k+1)/G(k+1))); (continued fraction, 3rd kind, 3-step). - Sergei N. Gladkovskii, Nov 23 2012 G.f.: x + 3*x^2/(Q(0)-3*x) where Q(k) = 1 + k*(x+1) + 3*x - x*(k+1)*(k+4)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Mar 14 2013 a(n) + a(n+1) + ... + a(n+8) + 6*n = a(3*n+15). - Charlie Marion, Mar 18 2013 a(n) + a(n+1) + ... + a(n+20) + 2*n^2 + 57*n = a(5*n+55). - Charlie Marion, Mar 18 2013 In general, a(k*n) = (2*k-1)*a(n) + a((k-1)*n-1). - Charlie Marion, Apr 20 2015 Also, a(k*n) = a(k)*a(n) + a(k-1)*a(n-1). - Robert Israel, Apr 20 2015 a(n+1) = det(binomial(i+2,j+1), 1 <= i,j <= n). - Mircea Merca, Apr 06 2013 a(n) = floor(n/2) + ceiling(n^2/2) = n - floor(n/2) + floor(n^2/2). - Wesley Ivan Hurt, Jun 15 2013 Sum_{n>=1} a(n)/n! = 3*exp(1)/2 by the e.g.f. Also see A067764 regarding ratios calculated this way for binomial coefficients in general. - Richard R. Forberg, Jul 15 2013 Sum_{n>=1} (-1)^(n+1)/a(n) = 4*log(2) - 2 = 0.7725887... . - Richard R. Forberg, Aug 11 2014 Let O(n) be the oblong number n(n+1) = A002378(n) and S(n) the square number n^2 = A000290(n). Then a(n) + a(n+2k) = O(n+k) + S(k) and a(n) + a(n+2k+1) = S(n+k+1) + O(k). - Charlie Marion, Jul 16 2015 A generalization of the Nov 22 2006 formula, a(n)^2 + a(n+1)^2 = a((n+1)^2), follows. Let T(k,n) = a(n) + k. Then for all k, T(k,n)^2 + T(k,n+1)^2 = T(k,(n+1)^2 + 2*k) - 2*k. - Charlie Marion, Dec 10 2015 a(n)^2 + a(n+1)^2 = a(a(n) + a(n+1)). Deducible from N. J. A. Sloane's a(n) + a(n+1) = (n+1)^2 and R. B. Nelson's a(n)^2 + a(n+1)^2 = a((n+1)^2). - Ben Paul Thurston, Dec 28 2015 a(n)^2 + a(n+3)^2 + 19 = a(n^2 + 4*n + 10). - Charlie Marion, Nov 23 2016 G.f.: x/(1-x)^3 = (x * r(x) * r(x^3) * r(x^9) * r(x^27) * ...), where r(x) = (1 + x + x^2)^3 = (1 + 3*x + 6*x^2 + 7*x^3 + 6*x^4 + 3*x^5 + x^6). - Gary W. Adamson, Dec 03 2016 a(n) = sum of the elements of inverse of matrix Q(n), where Q(n) has elements q_i,j = 1/(1-4*(i-j)^2). So if e = appropriately sized vector consisting of 1's, then a(n) = e'.Q(n)^-1.e. - Michael Yukish, Mar 20 2017 a(n) = Sum_{k=1..n} ((2*k-1)!!*(2*n-2*k-1)!!)/((2*k-2)!!*(2*n-2*k)!!). - Michael Yukish, Mar 20 2017 Sum_{i=0..k-1} a(n+i) = (3*k*n^2 + 3*n*k^2 + k^3 - k)/6. - Christopher Hohl, Feb 23 2019 a(n) == 0 (mod n) iff n is odd (see De Koninck reference). - Bernard Schott, Jan 10 2020 8*a(k)*a(n) + ((a(k)-1)*n + a(k))^2 = ((a(k)+1)*n + a(k))^2. This formula reduces to the well-known formula, 8*a(n) + 1 = (2*n+1)^2, when k = 1. - Charlie Marion, Jul 23 2020 a(k)*a(n) = Sum_{i = 0..k-1} (-1)^i*a((k-i)*(n-i)). - Charlie Marion, Dec 04 2020 Product_{n>=1} (1 + 1/a(n)) = cosh(sqrt(7)*Pi/2)/(2*Pi).
Product_{n>=2} (1 - 1/a(n)) = 1/3. (End)
a(n) = Sum_{k=1..2*n-1} (-1)^(k+1)*a(k)*a(2*n-k). For example, for n = 4, 1*28 - 3*21 + 6*15 - 10*10 + 15*6 - 21*3 + 28*1 = 10. - Charlie Marion, Mar 23 2022 2*a(n) = A000384(n) - n^2 + 2*n. In general, if P(k,n) = the n-th k-gonal number, then (j+1)*a(n) = P(5 + j, n) - n^2 + (j+1)*n. More generally, (j+1)*P(k,n) = P(2*k + (k-2)*(j-1),n) - n^2 + (j+1)*n. - Charlie Marion, Mar 14 2023 a(n) = (1/6)* Sum_{k = 0..3*n} (-1)^(n+k+1) * k*(k + 1) * binomial(3*n+k, 2*k). - Peter Bala, Nov 03 2024
EXAMPLE
G.f.: x + 3*x^2 + 6*x^3 + 10*x^4 + 15*x^5 + 21*x^6 + 28*x^7 + 36*x^8 + 45*x^9 + ...
When n=3, a(3) = 4*3/2 = 6.
Example(a(4)=10): ABCD where A, B, C and D are different links in a chain or different amino acids in a peptide possible fragments: A, B, C, D, AB, ABC, ABCD, BC, BCD, CD = 10.
a(2): hollyhock leaves on the Tokugawa Mon, a(4): points in Pythagorean tetractys, a(5): object balls in eight-ball billiards. - Bradley Klee, Aug 24 2015 The a(1) = 1 through a(5) = 15 ordered triples of positive integers summing to n + 2 [Beeler, McGrath above] are the following. These compositions are ranked by A014311. (111) (112) (113) (114) (115)
(121) (122) (123) (124)
(211) (131) (132) (133)
(212) (141) (142)
(221) (213) (151)
(311) (222) (214)
(231) (223)
(312) (232)
(321) (241)
(411) (313)
(322)
(331)
(412)
(421)
(511)
The unordered strict case is A001399(n-6), with Heinz numbers A007304. (End)
MAPLE
istriangular:=proc(n) local t1; t1:=floor(sqrt(2*n)); if n = t1*(t1+1)/2 then return true else return false; end if; end proc; # N. J. A. Sloane, May 25 2008 ZL := [S, {S=Prod(B, B, B), B=Set(Z, 1 <= card)}, unlabeled]:
seq(combstruct[count](ZL, size=n), n=2..55); # Zerinvary Lajos, Mar 24 2007 isA000217 := proc(n)
issqr(1+8*n) ;
end proc: # R. J. Mathar, Nov 29 2015 [This is the recipe Leonhard Euler proposes in chapter VII of his "Vollständige Anleitung zur Algebra", 1765. Peter Luschny, Sep 02 2022]
MATHEMATICA
CoefficientList[Series[x / (1 - x)^3, {x, 0, 50}], x] (* Vincenzo Librandi, Jul 30 2014 *) (* For Mathematica 10.4+ *) Table[PolygonalNumber[n], {n, 0, 53}] (* Arkadiusz Wesolowski, Aug 27 2016 *) (* The following Mathematica program, courtesy of Steven J. Miller, is useful for testing if a sequence is Benford. To test a different sequence only one line needs to be changed. This strongly suggests that the triangular numbers are not Benford, since the second and third columns of the output disagree. - N. J. A. Sloane, Feb 12 2017 *) fd[x_] := Floor[10^Mod[Log[10, x], 1]]
benfordtest[num_] := Module[{},
For[d = 1, d <= 9, d++, digit[d] = 0];
For[n = 1, n <= num, n++,
{
d = fd[n(n+1)/2];
If[d != 0, digit[d] = digit[d] + 1];
}];
For[d = 1, d <= 9, d++, digit[d] = 1.0 digit[d]/num];
For[d = 1, d <= 9, d++,
Print[d, " ", 100.0 digit[d], " ", 100.0 Log[10, (d + 1)/d]]];
];
benfordtest[20000]
Table[Length[Join@@Permutations/@IntegerPartitions[n, {3}]], {n, 0, 15}] (* Gus Wiseman, Oct 28 2020 *)
PROG
(PARI) A000217(n) = n * (n + 1) / 2; (PARI) list(lim)=my(v=List(), n, t); while((t=n*n++/2)<=lim, listput(v, t)); Vec(v) \\ Charles R Greathouse IV, Jun 18 2021 (Haskell)
a000217 n = a000217_list !! n
(Scala) (1 to 53).scanLeft(0)(_ + _) // Horstmann (2012), p. 171
(Python) for n in range(0, 60): print(n*(n+1)/2, end=', ') # Stefano Spezia, Dec 06 2018 (Python) # Intended to compute the initial segment of the sequence, not
# isolated terms. If in the iteration the line "x, y = x + y + 1, y + 1"
# is replaced by "x, y = x + y + k, y + k" then the figurate numbers are obtained,
# for k = 0 (natural A001477), k = 1 (triangular), k = 2 (squares), k = 3 (pentagonal), k = 4 (hexagonal), k = 5 (heptagonal), k = 6 (octagonal), etc. def aList():
x, y = 1, 1
yield 0
while True:
yield x
x, y = x + y + 1, y + 1
CROSSREFS
Cf. A000096, A000124, A000292, A000330, A000396, A000668, A001082, A001788, A002024, A002378, A002415, A003056 (inverse function), A004526, A006011, A007318, A008953, A008954, A010054 (characteristic function), A028347, A036666, A046092, A051942, A055998, A055999, A056000, A056115, A056119, A056121, A056126, A062717, A087475, A101859, A109613, A143320, A210569, A245031, A245300, A060544, A016754. Cf. A002817 (doubly triangular numbers), A075528 (solutions of a(n)=a(m)/2). Cf. A104712 (first column, starting with a(1)). Some generalized k-gonal numbers are A001318 (k=5), this sequence (k=6), A085787 (k=7), etc. A011782 counts compositions of any length.
A337461 counts pairwise coprime triples, with unordered version A307719.
The positive integers. Also called the natural numbers, the whole numbers or the counting numbers, but these terms are ambiguous.
(Formerly M0472 N0173)
+10
2113
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77
COMMENTS
For some authors, the terms "natural numbers" and "counting numbers" include 0, i.e., refer to the nonnegative integers A001477; the term "whole numbers" frequently also designates the whole set of (signed) integers A001057. a(n) is smallest positive integer which is consistent with sequence being monotonically increasing and satisfying a(a(n)) = n (cf. A007378). Inverse Euler transform of A000219. The rectangular array having A000027 as antidiagonals is the dispersion of the complement of the triangular numbers, A000217 (which triangularly form column 1 of this array). The array is also the transpose of A038722. - Clark Kimberling, Apr 05 2003 For nonzero x, define f(n) = floor(nx) - floor(n/x). Then f= A000027 if and only if x=tau or x=-tau. - Clark Kimberling, Jan 09 2005 If the offset were changed to 0, we would have the following pattern: a(n)=binomial(n,0) + binomial(n,1) for the present sequence (number of regions in 1-space defined by n points), A000124 (number of regions in 2-space defined by n straight lines), A000125 (number of regions in 3-space defined by n planes), A000127 (number of regions in 4-space defined by n hyperplanes), A006261, A008859, A008860, A008861, A008862 and A008863, where the last six sequences are interpreted analogously and in each "... by n ..." clause an offset of 0 has been assumed, resulting in a(0)=1 for all of them, which corresponds to the case of not cutting with a hyperplane at all and therefore having one region. - Peter C. Heinig (algorithms(AT)gmx.de), Oct 19 2006 Define a number of points on a straight line to be in general arrangement when no two points coincide. Then these are the numbers of regions defined by n points in general arrangement on a straight line, when an offset of 0 is assumed. For instance, a(0)=1, since using no point at all leaves one region. The sequence satisfies the recursion a(n) = a(n-1) + 1. This has the following geometrical interpretation: Suppose there are already n-1 points in general arrangement, thus defining the maximal number of regions on a straight line obtainable by n-1 points, and now one more point is added in general arrangement. Then it will coincide with no other point and act as a dividing wall thereby creating one new region in addition to the a(n-1)=(n-1)+1=n regions already there, hence a(n)=a(n-1)+1. Cf. the comments on A000124 for an analogous interpretation. - Peter C. Heinig (algorithms(AT)gmx.de), Oct 19 2006 The sequence a(n)=n (for n=1,2,3) and a(n)=n+1 (for n=4,5,...) gives to the rank (minimal cardinality of a generating set) for the semigroup I_n\S_n, where I_n and S_n denote the symmetric inverse semigroup and symmetric group on [n]. - James East, May 03 2007 The sequence a(n)=n (for n=1,2), a(n)=n+1 (for n=3) and a(n)=n+2 (for n=4,5,...) gives the rank (minimal cardinality of a generating set) for the semigroup PT_n\T_n, where PT_n and T_n denote the partial transformation semigroup and transformation semigroup on [n]. - James East, May 03 2007 "God made the integers; all else is the work of man." This famous quotation is a translation of "Die ganzen Zahlen hat der liebe Gott gemacht, alles andere ist Menschenwerk," spoken by Leopold Kronecker in a lecture at the Berliner Naturforscher-Versammlung in 1886. Possibly the first publication of the statement is in Heinrich Weber's "Leopold Kronecker," Jahresberichte D.M.V. 2 (1893) 5-31. - Clark Kimberling, Jul 07 2007 Writing A000027 as N, perhaps the simplest one-to-one correspondence between N X N and N is this: f(m,n) = ((m+n)^2 - m - 3n + 2)/2. Its inverse is given by I(k)=(g,h), where g = k - J(J-1)/2, h = J + 1 - g, J = floor((1 + sqrt(8k - 7))/2). Thus I(1)=(1,1), I(2)=(1,2), I(3)=(2,1) and so on; the mapping I fills the first-quadrant lattice by successive antidiagonals. - Clark Kimberling, Sep 11 2008 a(n) is also the mean of the first n odd integers. - Ian Kent, Dec 23 2008 Equals INVERTi transform of A001906, the even-indexed Fibonacci numbers starting (1, 3, 8, 21, 55, ...). - Gary W. Adamson, Jun 05 2009 These are also the 2-rough numbers: positive integers that have no prime factors less than 2. - Michael B. Porter, Oct 08 2009 Totally multiplicative sequence with a(p) = p for prime p. Totally multiplicative sequence with a(p) = a(p-1) + 1 for prime p. - Jaroslav Krizek, Oct 18 2009 Triangle T(k,j) of natural numbers, read by rows, with T(k,j) = binomial(k,2) + j = (k^2-k)/2 + j where 1 <= j <= k. In other words, a(n) = n = binomial(k,2) + j where k is the largest integer such that binomial(k,2) < n and j = n - binomial(k,2). For example, T(4,1)=7, T(4,2)=8, T(4,3)=9, and T(4,4)=10. Note that T(n,n)= A000217(n), the n-th triangular number. - Dennis P. Walsh, Nov 19 2009 Hofstadter-Conway-like sequence (see A004001): a(n) = a(a(n-1)) + a(n-a(n-1)) with a(1) = 1, a(2) = 2. - Jaroslav Krizek, Dec 11 2009 a(n) is also the dimension of the irreducible representations of the Lie algebra sl(2). - Leonid Bedratyuk, Jan 04 2010 Generated from a(2n) = r*a(n), a(2n+1) = a(n) + a(n+1), r = 2; in an infinite set, row 2 of the array shown in A178568. - Gary W. Adamson, May 29 2010 1/n = continued fraction [n]. Let barover[n] = [n,n,n,...] = 1/k. Then k - 1/k = n. Example: [2,2,2,...] = (sqrt(2) - 1) = 1/k, with k = (sqrt(2) + 1). Then 2 = k - 1/k. - Gary W. Adamson, Jul 15 2010 Number of n-digit numbers the binary expansion of which contains one run of 1's. - Vladimir Shevelev, Jul 30 2010 Let T denote the "natural number array A000027": 1 2 4 7 ...
3 5 8 12 ...
6 9 13 18 ...
10 14 19 25 ...
T(n,k) = n+(n+k-2)*(n+k-1)/2. See A185787 for a list of sequences based on T, such as rows, columns, diagonals, and sub-arrays. (End) The Stern polynomial B(n,x) evaluated at x=2. See A125184. - T. D. Noe, Feb 28 2011 The denominator in the Maclaurin series of log(2), which is 1 - 1/2 + 1/3 - 1/4 + .... - Mohammad K. Azarian, Oct 13 2011 As a function of Bernoulli numbers B_n (cf. A027641: (1, -1/2, 1/6, 0, -1/30, 0, 1/42, ...)): let V = a variant of B_n changing the (-1/2) to (1/2). Then triangle A074909 (the beheaded Pascal's triangle) * [1, 1/2, 1/6, 0, -1/30, ...] = the vector [1, 2, 3, 4, 5, ...]. - Gary W. Adamson, Mar 05 2012 Number of partitions of 2n+1 into exactly two parts. - Wesley Ivan Hurt, Jul 15 2013 Integers n dividing u(n) = 2u(n-1) - u(n-2); u(0)=0, u(1)=1 (Lucas sequence A001477). - Thomas M. Bridge, Nov 03 2013 For this sequence, the generalized continued fraction a(1)+a(1)/(a(2)+a(2)/(a(3)+a(3)/(a(4)+...))), evaluates to 1/(e-2) = A194807. - Stanislav Sykora, Jan 20 2014 a(n) is the number of permutations of length n simultaneously avoiding 213, 231 and 321 in the classical sense which are breadth-first search reading words of increasing unary-binary trees. For more details, see the entry for permutations avoiding 231 at A245898. - Manda Riehl, Aug 05 2014 a(n) is also the number of permutations simultaneously avoiding 213, 231 and 321 in the classical sense which can be realized as labels on an increasing strict binary tree with 2n-1 nodes. See A245904 for more information on increasing strict binary trees. - Manda Riehl, Aug 07 2014 a(n) = least k such that 2*Pi - Sum_{h=1..k} 1/(h^2 - h + 3/16) < 1/n. - Clark Kimberling, Sep 28 2014 a(n) = least k such that Pi^2/6 - Sum_{h=1..k} 1/h^2 < 1/n. - Clark Kimberling, Oct 02 2014 Determinants of the spiral knots S(2,k,(1)). a(k) = det(S(2,k,(1))). These knots are also the torus knots T(2,k). - Ryan Stees, Dec 15 2014 As a function, the restriction of the identity map on the nonnegative integers {0,1,2,3...}, A001477, to the positive integers {1,2,3,...}. - M. F. Hasler, Jan 18 2015 See also A131685(k) = smallest positive number m such that c(i) = m (i^1 + 1) (i^2 + 2) ... (i^k+ k) / k! takes integral values for all i>=0: For k=1, A131685(k)=1, which implies that this is a well defined integer sequence. - Alexander R. Povolotsky, Apr 24 2015 a(n) is the number of compositions of n+2 into n parts avoiding the part 2. - Milan Janjic, Jan 07 2016 Does not satisfy Benford's law [Berger-Hill, 2017] - N. J. A. Sloane, Feb 07 2017 For n >= 3, a(n)=n is the least area that can be obtained for an irregular octagon drawn in a square of n units side, whose sides are parallel to the axes, with 4 vertices that coincide with the 4 vertices of the square, and the 4 remaining vertices having integer coordinates. See Affaire de Logique link. - Michel Marcus, Apr 28 2018 a(n+1) is the order of rowmotion on a poset defined by a disjoint union of chains of length n. - Nick Mayers, Jun 08 2018 Number of 1's in n-th generation of 1-D Cellular Automata using Rules 50, 58, 114, 122, 178, 186, 206, 220, 238, 242, 250 or 252 in the Wolfram numbering scheme, started with a single 1. - Frank Hollstein, Mar 25 2019 (1, 2, 3, 4, 5, ...) is the fourth INVERT transform of (1, -2, 3, -4, 5, ...). - Gary W. Adamson, Jul 15 2019
REFERENCES
T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 1.
T. M. Apostol, Modular Functions and Dirichlet Series in Number Theory, Springer-Verlag, 1990, page 25.
W. Fulton and J. Harris, Representation theory: a first course, (1991), page 149. [From Leonid Bedratyuk, Jan 04 2010] I. S. Gradstein and I. M. Ryshik, Tables of series, products, and integrals, Volume 1, Verlag Harri Deutsch, 1981.
R. E. Schwartz, You Can Count on Monsters: The First 100 numbers and Their Characters, A. K. Peters and MAA, 2010.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
LINKS
N. Brothers, S. Evans, L. Taalman, L. Van Wyk, D. Witczak, and C. Yarnall, Spiral knots, Missouri J. of Math. Sci., 22 (2010). M. DeLong, M. Russell, and J. Schrock, Colorability and determinants of T(m,n,r,s) twisted torus knots for n equiv. +/-1(mod m), Involve, Vol. 8 (2015), No. 3, 361-384. Eric Weisstein's World of Mathematics, Natural Number, Positive Integer, Counting Number Composition, Davenport-Schinzel Sequence, Idempotent Number, N, Smarandache Ceil Function, Whole Number, Engel Expansion, and Trinomial Coefficient
FORMULA
Another g.f.: Sum_{n>0} phi(n)*x^n/(1-x^n) (Apostol).
When seen as an array: T(k, n) = n+1 + (k+n)*(k+n+1)/2. Main diagonal is 2n*(n+1)+1 ( A001844), antidiagonal sums are n*(n^2+1)/2 ( A006003). - Ralf Stephan, Oct 17 2004 G.f.: x/(1-x)^2. E.g.f.: x*exp(x). a(n)=n. a(-n)=-a(n).
Series reversion of g.f. A(x) is x*C(-x)^2 where C(x) is the g.f. of A000108. - Michael Somos, Sep 04 2006 G.f. A(x) satisfies 0 = f(A(x), A(x^2)) where f(u, v) = u^2 - v - 4*u*v. - Michael Somos, Oct 03 2006 a(n) = 2*a(n-1)-a(n-2); a(1)=1, a(2)=2. a(n) = 1+a(n-1). - Philippe Deléham, Nov 03 2008 G.f.: x * Product_{j>=0} (1+x^(2^j))^2 = x * (1+2*x+x^2) * (1+2*x^2+x^4) * (1+2*x^4+x^8) * ... = x + 2x^2 + 3x^3 + ... . - Gary W. Adamson, Jun 26 2012 a(n) = det(binomial(i+1,j), 1 <= i,j <= n). - Mircea Merca, Apr 06 2013 E.g.f.: x*E(0), where E(k) = 1 + 1/(x - x^3/(x^2 + (k+1)/E(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 03 2013 a(n) = Product_{k=1..n-1} 2*sin(Pi*k/n), n > 1.
a(n) = Product_{k=1..n-1} (2*sin(Pi*k/(2*n)))^2, n > 1.
These identities are used in the calculation of products of ratios of lengths of certain lines in a regular n-gon. For the first identity see the Gradstein-Ryshik reference, p. 62, 1.392 1., bringing the first factor there to the left hand side and taking the limit x -> 0 (L'Hôpital). The second line follows from the first one. Thanks to Seppo Mustonen who led me to consider n-gon lengths products. (End) a(n) = Sum_{j=0..k} (-1)^(j-1)*j*binomial(n,j)*binomial(n-1+k-j,k-j), k>=0. - Mircea Merca, Jan 25 2014 a(n) = Sum_{k=1..n^2+2*n} 1/(sqrt(k)+sqrt(k+1)). - Pierre CAMI, Apr 25 2014 a(n) = floor(1/sin(1/n)) = floor(cot(1/(n+1))) = ceiling(cot(1/n)). - Clark Kimberling, Oct 08 2014 a(n) = 1/(1/(n+1) + 1/(n+1)^2 + 1/(n+1)^3 + ...). - Pierre CAMI, Jan 22 2015 a(n) = Sum_{m=0..n-1} Stirling1(n-1,m)*Bell(m+1), for n >= 1. This corresponds to Bell(m+1) = Sum_{k=0..m} Stirling2(m, k)*(k+1), for m >= 0, from the fact that Stirling2*Stirling1 = identity matrix. See A048993, A048994 and A000110. - Wolfdieter Lang, Feb 03 2015 a(n) = Sum_{k=1..2n-1}(-1)^(k+1)*k*(2n-k). In addition, surprisingly, a(n) = Sum_{k=1..2n-1}(-1)^(k+1)*k^2*(2n-k)^2. - Charlie Marion, Jan 05 2016 G.f.: x/(1-x)^2 = (x * r(x) *r(x^3) * r(x^9) * r(x^27) * ...), where r(x) = (1 + x + x^2)^2 = (1 + 2x + 3x^2 + 2x^3 + x^4). - Gary W. Adamson, Jan 11 2017 For positive integer m, a(n) = (1/m)* Sum_{k = 1..2*m*n-1} (-1)^(k+1) * k * (2*m*n - k) = (1/m) * Sum_{k = 1..2*m*n-1} (-1)^(k+1) * k^2 * (2*m*n - k)^2 (the case m = 1 is given above).
a(n) = Sum_{k = 0..3*n} (-1)^(n+k+1) * k * binomial(3*n+k, 2*k). (End)
PROG
(Magma) [ n : n in [1..100]];
(PARI) {a(n) = n};
(R) 1:100
(Shell) seq 1 100
(Haskell)
a000027 = id
(Maxima) makelist(n, n, 1, 30); /* Martin Ettl, Nov 07 2012 */ (Python)
(Julia) print([n for n in 1:280]) # Paul Muljadi, Apr 09 2024
CROSSREFS
Cf. A026081 = integers in reverse alphabetical order in U.S. English, A107322 = English name for number and its reverse have the same number of letters, A119796 = zero through ten in alphabetical order of English reverse spelling, A005589, etc. Cf. A185787 (includes a list of sequences based on the natural number array A000027). Cf. A038722 (mirrored when seen as triangle), A056011 (boustrophedon).
Triangle of Stirling numbers of the second kind, S2(n,k), n >= 1, 1 <= k <= n.
+10
638
1, 1, 1, 1, 3, 1, 1, 7, 6, 1, 1, 15, 25, 10, 1, 1, 31, 90, 65, 15, 1, 1, 63, 301, 350, 140, 21, 1, 1, 127, 966, 1701, 1050, 266, 28, 1, 1, 255, 3025, 7770, 6951, 2646, 462, 36, 1, 1, 511, 9330, 34105, 42525, 22827, 5880, 750, 45, 1, 1, 1023, 28501, 145750, 246730, 179487, 63987, 11880, 1155, 55, 1
COMMENTS
Also known as Stirling set numbers and written {n, k}.
S2(n,k) counts partitions of an n-set into k nonempty subsets.
With regard to the preceding comment: For arbitrary (including non-disjoint) covers of an n-set by k nonempty subsets see A055154. - Manfred Boergens, May 20 2024 Triangle S2(n,k), 1 <= k <= n, read by rows, given by [0, 1, 0, 2, 0, 3, 0, 4, 0, 5, 0, 6, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, ...] where DELTA is Deléham's operator defined in A084938. Number of partitions of {1, ..., n+1} into k+1 nonempty subsets of nonconsecutive integers, including the partition 1|2|...|n+1 if n=k. E.g., S2(3,2)=3 since the number of partitions of {1,2,3,4} into three subsets of nonconsecutive integers is 3, i.e., 13|2|4, 14|2|3, 1|24|3. - Augustine O. Munagi, Mar 20 2005 Draw n cards (with replacement) from a deck of k cards. Let prob(n,k) be the probability that each card was drawn at least once. Then prob(n,k) = S2(n,k)*k!/k^n (see A090582). - Rainer Rosenthal, Oct 22 2005 Define f_1(x), f_2(x), ..., such that f_1(x)=e^x and for n = 2, 3, ..., f_{n+1}(x) = (d/dx)(x*f_n(x)). Then f_n(x) = e^x*Sum_{k=1..n} S2(n,k)*x^(k-1). - Milan Janjic, May 30 2008 For tables of restricted Stirling numbers of the second kind see A143494 - A143496. S2(n,k) gives the number of 'patterns' of words of length n using k distinct symbols - see [Cooper & Kennedy] for an exact definition of the term 'pattern'. As an example, the words AADCBB and XXEGTT, both of length 6, have the same pattern of letters. The five patterns of words of length 3 are AAA, AAB, ABA, BAA and ABC giving row 3 of this table as (1,3,1).
Equivalently, S2(n,k) gives the number of sequences of positive integers (N_1,...,N_n) of length n, with k distinct entries, such that N_1 = 1 and N_(i+1) <= 1 + max{j = 1..i} N_j for i >= 1 (restricted growth functions). For example, Stirling(4,2) = 7 since the sequences of length 4 having 2 distinct entries that satisfy the conditions are (1,1,1,2), (1,1,2,1), (1,2,1,1), (1,1,2,2), (1,2,2,2), (1,2,2,1) and (1,2,1,2).
(End)
Number of combinations of subsets in the plane. - Mats Granvik, Jan 13 2009 S2(n+1,k+1) is the number of size k collections of pairwise disjoint, nonempty subsets of [n]. For example: S2(4,3)=6 because there are six such collections of subsets of [3] that have cardinality two: {(1)(23)},{(12)(3)}, {(13)(2)}, {(1)(2)}, {(1)(3)}, {(2)(3)}. - Geoffrey Critzer, Apr 06 2009 Consider a set of A000217(n) balls of n colors in which, for each integer k = 1 to n, exactly one color appears in the set a total of k times. (Each ball has exactly one color and is indistinguishable from other balls of the same color.) a(n+1, k+1) equals the number of ways to choose 0 or more balls of each color in such a way that exactly (n-k) colors are chosen at least once, and no two colors are chosen the same positive number of times. - Matthew Vandermast, Nov 22 2010 S2(n,k) is the number of monotonic-labeled forests on n vertices with exactly k rooted trees, each of height one or less. See link "Counting forests with Stirling and Bell numbers" below. - Dennis P. Walsh, Nov 16 2011 If D is the operator d/dx, and E the operator xd/dx, Stirling numbers are given by: E^n = Sum_{k=1..n} S2(n,k) * x^k*D^k. - Hyunwoo Jang, Dec 13 2011
The Stirling polynomials of the second kind (a.k.a. the Bell / Touchard polynomials) are the umbral compositional inverses of the falling factorials (a.k.a. the Pochhammer symbol or Stirling polynomials of the first kind), i.e., binomial(Bell(.,x),n) = x^n/n! (cf. Copeland's 2007 formulas), implying binomial(xD,n) = binomial(Bell(.,:xD:),n) = :xD:^n/n! where D = d/dx and :xD:^n = x^n*D^n. - Tom Copeland, Apr 17 2014 S2(n,k) is the number of ways to nest n matryoshkas (Russian nesting dolls) so that exactly k matryoshkas are not contained in any other matryoshka. - Carlo Sanna, Oct 17 2015 The row polynomials R(n, x) = Sum_{k=1..n} S2(n, k)*x^k appear in the numerator of the e.g.f. of n-th powers, E(n, x) = Sum_{m>=0} m^n*x^m/m!, as E(n, x) = exp(x)*x*R(n, x), for n >= 1. - Wolfdieter Lang, Apr 02 2017 With offsets 0 for n and k this is the Sheffer product matrix A007318* A048993 denoted by (exp(t), (exp(t) - 1)) with e.g.f. exp(t)*exp(x*(exp(t) - 1)). - Wolfdieter Lang, Jun 20 2017 Number of words on k+1 unlabeled letters of length n+1 with no repeated letters. - Thomas Anton, Mar 14 2019 Also coefficients of moments of Poisson distribution about the origin expressed as polynomials in lambda. [Haight] (see also A331155). - N. J. A. Sloane, Jan 14 2020 k!*S2(n,k) is the number of surjections from an n-element set to a k-element set. - Jianing Song, Jun 01 2022
REFERENCES
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 835.
A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 103ff.
B. A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5-648-00738-8.
G. Boole, Finite Differences, 5th ed. New York, NY: Chelsea, 1970.
C. A. Charalambides, Enumerative Combinatorics, Chapman & Hall/CRC, 2002, Theorem 8.11, pp. 298-299.
L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 310.
J. H. Conway and R. K. Guy, The Book of Numbers, Springer, p. 92.
F. N. David, M. G. Kendall, and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 223.
S.N. Elaydi, An Introduction to Difference Equations, 3rd ed. Springer, 2005.
H. H. Goldstine, A History of Numerical Analysis, Springer-Verlag, 1977; Section 2.7.
R. L. Graham, D. E. Knuth, and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 244.
Frank Avery Haight, Handbook of the Poisson distribution, John Wiley, 1967. See pages 6,7.
A. D. Korshunov, Asymptotic behavior of Stirling numbers of the second kind. (Russian) Metody Diskret. Analiz. No. 39 (1983), 24-41.
E. Kuz'min and A. I. Shirshov: On the number e, pp. 111-119, eq.(6), in: Kvant Selecta: Algebra and Analysis, I, ed. S. Tabachnikov, Am.Math.Soc., 1999, p. 116, eq. (11).
J. Riordan, An Introduction to Combinatorial Analysis, p. 48.
J. Stirling, The Differential Method, London, 1749; see p. 7.
LINKS
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy]. T. Copeland's Shadows of Simplicity, A Class of Differential Operators and the Stirling Numbers,2015; Generators, Inversion, and Matrix, Binomial, and Integral Transforms, 2015; The Inverse Mellin Transform, Bell Polynomials, a Generalized Dobinski Relation, and the Confluent Hypergeometric Functions, 2011; Mathemagical Forests, 2008; and Addendum to "Mathemagical Forests", 2010.
FORMULA
S2(n, k) = k*S2(n-1, k) + S2(n-1, k-1), n > 1. S2(1, k) = 0, k > 1. S2(1, 1) = 1.
E.g.f.: A(x, y) = e^(y*e^x-y). E.g.f. for m-th column: (e^x-1)^m/m!.
S2(n, k) = (1/k!) * Sum_{i=0..k} (-1)^(k-i)*binomial(k, i)*i^n.
Row sums: Bell number A000110(n) = Sum_{k=1..n} S2(n, k), n>0. S(n, k) = Sum (i_1*i_2*...*i_(n-k)) summed over all (n-k)-combinations {i_1, i_2, ..., i_k} with repetitions of the numbers {1, 2, ..., k}. Also S(n, k) = Sum (1^(r_1)*2^(r_2)*...* k^(r_k)) summed over integers r_j >= 0, for j=1..k, with Sum{j=1..k} r_j = n-k. [Charalambides]. - Wolfdieter Lang, Aug 15 2019. For asymptotics see Hsu (1948), among other sources.
Sum_{n>=0} S2(n, k)*x^n = x^k/((1-x)(1-2x)(1-3x)...(1-kx)).
Let P(n) = the number of integer partitions of n ( A000041), p(i) = the number of parts of the i-th partition of n, d(i) = the number of distinct parts of the i-th partition of n, p(j, i) = the j-th part of the i-th partition of n, m(i, j) = multiplicity of the j-th part of the i-th partition of n, and Sum_{i=1..P(n), p(i)=m} = sum running from i=1 to i=P(n) but taking only partitions with p(i)=m parts into account. Then S2(n, m) = Sum_{i=1..P(n), p(i)=m} n!/(Product_{j=1..p(i)} p(i, j)!) * 1/(Product_{j=1..d(i)} m(i, j)!). For example, S2(6, 3) = 90 because n=6 has the following partitions with m=3 parts: (114), (123), (222). Their complexions are: (114): 6!/(1!*1!*4!) * 1/(2!*1!) = 15, (123): 6!/(1!*2!*3!) * 1/(1!*1!*1!) = 60, (222): 6!/(2!*2!*2!) * 1/(3!) = 15. The sum of the complexions is 15+60+15 = 90 = S2(6, 3). - Thomas Wieder, Jun 02 2005 Sum_{k=1..n} k*S2(n,k) = B(n+1)-B(n), where B(q) are the Bell numbers ( A000110). - Emeric Deutsch, Nov 01 2006 Recurrence: S2(n+1,k) = Sum_{i=0..n} binomial(n,i)*S2(i,k-1). With the starting conditions S2(n,k) = 1 for n = 0 or k = 1 and S2(n,k) = 0 for k = 0 we have the closely related recurrence S2(n,k) = Sum_{i=k..n} binomial(n-1,i-1)*S2(i-1,k-1). - Thomas Wieder, Jan 27 2007 Representation of Stirling numbers of the second kind S2(n,k), n=1,2,..., k=1,2,...,n, as special values of hypergeometric function of type (n)F(n-1): S2(n,k)= (-1)^(k-1)*hypergeom([ -k+1,2,2,...,2],[1,1,...,1],1)/(k-1)!, i.e., having n parameters in the numerator: one equal to -k+1 and n-1 parameters all equal to 2; and having n-1 parameters in the denominator all equal to 1 and the value of the argument equal to 1. Example: S2(6,k)= seq(evalf((-1)^(k-1)*hypergeom([ -k+1,2,2,2,2,2],[1,1,1,1,1],1)/(k-1)!),k=1..6)=1,31,90,65,15,1. - Karol A. Penson, Mar 28 2007 Bell_n(x) = Sum_{j=0..n} S2(n,j) * x^j = Sum_{j=0..n} E(n,j) * Lag(n,-x, j-n) = Sum_{j=0..n} (E(n,j)/n!) * (n!*Lag(n,-x, j-n)) = Sum_{j=0..n} E(n,j) * binomial(Bell.(x)+j, n) umbrally where Bell_n(x) are the Bell / Touchard / exponential polynomials; S2(n,j), the Stirling numbers of the second kind; E(n,j), the Eulerian numbers; and Lag(n,x,m), the associated Laguerre polynomials of order m.
For x = 0, the equation gives Sum_{j=0..n} E(n,j) * binomial(j,n) = 1 for n=0 and 0 for all other n. By substituting the umbral compositional inverse of the Bell polynomials, the lower factorial n!*binomial(y,n), for x in the equation, the Worpitzky identity is obtained; y^n = Sum_{j=0..n} E(n,j) * binomial(y+j,n).
Note that E(n,j)/n! = E(n,j)/(Sum_{k=0..n}E(n,k)). Also (n!*Lag(n, -1, j-n)) is A086885 with a simple combinatorial interpretation in terms of seating arrangements, giving a combinatorial interpretation to the equation for x=1; n!*Bell_n(1) = n!*Sum_{j=0..n} S2(n,j) = Sum_{j=0..n} E(n,j) * (n!*Lag(n, -1, j-n)). (Appended Sep 16 2020) For connections to the Bernoulli numbers, extensions, proofs, and a clear presentation of the number arrays involved in the identities above, see my post Reciprocity and Umbral Witchcraft. (End)
n-th row = leftmost column of nonzero terms of A127701^(n-1). Also, (n+1)-th row of the triangle = A127701 * n-th row; deleting the zeros. Example: A127701 * [1, 3, 1, 0, 0, 0, ...] = [1, 7, 6, 1, 0, 0, 0, ...]. - Gary W. Adamson, Nov 21 2007 The row polynomials are given by D^n(e^(x*t)) evaluated at x = 0, where D is the operator (1+x)*d/dx. Cf. A147315 and A094198. See also A185422. - Peter Bala, Nov 25 2011 O.g.f. for the n-th diagonal is D^n(x), where D is the operator x/(1-x)*d/dx. - Peter Bala, Jul 02 2012 n*i!*S2(n-1,i) = Sum_{j=(i+1)..n} (-1)^(j-i+1)*j!/(j-i)*S2(n,j). - Leonid Bedratyuk, Aug 19 2012 G.f.: (1/Q(0)-1)/(x*y), where Q(k) = 1 - (y+k)*x - (k+1)*y*x^2/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Nov 09 2013 Multiply each n-th diagonal of the Pascal lower triangular matrix by x^n and designate the result as A007318(x) = P(x). With Bell(n,x)=B(n,x) defined above, D = d/dx, and :xD:^n = x^n*D^n, a Dobinski formula gives umbrally f(y)^B(.,x) = e^(-x)*e^(f(y)*x). Then f(y)^B(.,:xD:)g(x) = [f(y)^(xD)]g(x) = e^[-(1-f(y)):xD:]g(x) = g[f(y)x].
In particular, for f(y) = (1+y),
A) (1+y)^B(.,x) = e^(-x)*e^((1+y)*x) = e^(x*y) = e^[log(1+y)B(.,x)],
B) (I+dP)^B(.,x) = e^(x*dP) = P(x) = e^[x*(e^M-I)]= e^[M*B(.,x)] with dP = A132440, M = A238385-I = log(I+dP), and I = identity matrix, and C) (1+dP)^(xD) = e^(dP:xD:) = P(:xD:) = e^[(e^M-I):xD:] = e^[M*xD] with action e^(dP:xD:)g(x) = g[(I+dP)*x].
D) P(x)^m = P(m*x), which implies (Sum_{k=1..m} a_k)^j = B(j,m*x) where the sum is umbrally evaluated only after exponentiation with (a_k)^q = B(.,x)^q = B(q,x). E.g., (a1+a2+a3)^2=a1^2+a2^2+a3^2+2(a1*a2+a1*a3+a2*a3) = 3*B(2,x)+6*B(1,x)^2 = 9x^2+3x = B(2,3x).
E) P(x)^2 = P(2x) = e^[M*B(.,2x)] = A038207(x), the face vectors of the n-Dim hypercubes. (End)
As a matrix equivalent of some inversions mentioned above, A008277* A008275 = I, the identity matrix, regarded as lower triangular matrices. - Tom Copeland, Apr 26 2014 O.g.f. for the n-th diagonal of the triangle (n = 0,1,2,...): Sum_{k>=0} k^(k+n)*(x*e^(-x))^k/k!. Cf. the generating functions of the diagonals of A039755. Also cf. A112492. - Peter Bala, Jun 22 2014 Floor(1/(-1 + Sum_{n>=k} 1/S2(n,k))) = A034856(k-1), for k>=2. The fractional portion goes to zero at large k. - Richard R. Forberg, Jan 17 2015 Let x_(n), called a factorial term (Boole, 1970) or a factorial polynomial (Elaydi, 2005: p. 60), denote the falling factorial Product_{k=0..n-1} (x-k). Then, for n >= 1, x_(n) = Sum_{k=1..n} A008275(n,k) * x^k, x^n = Sum_{k=1..n} T(n,k) * x_(k), where A008275(n,k) are Stirling numbers of the first kind. For n >= 1, the row sums yield the exponential numbers (or Bell numbers): Sum_{k=1..n} T(n,k) = A000110(n), and Sum_{k=1..n} (-1)^(n+k) * T(n,k) = (-1)^n * Sum_{k=1..n} (-1)^k * T(n,k) = (-1)^n * A000587(n), where A000587 are the complementary Bell numbers. (End) O.g.f. for the m-th column: x^m/(Product_{j=1..m} 1-j*x). - Daniel Checa, Aug 25 2022 S2(n,k) ~ (k^n)/k!, for fixed k as n->oo. - Daniel Checa, Nov 08 2022
EXAMPLE
The triangle S2(n, k) begins:
\ k 1 2 3 4 5 6 7 8 9
n \ 10 11 12 13 14 15 ...
----------------------------------------------------------------------------------
1 | 1
2 | 1 1
3 | 1 3 1
4 | 1 7 6 1
5 | 1 15 25 10 1
6 | 1 31 90 65 15 1
7 | 1 63 301 350 140 21 1
8 | 1 127 966 1701 1050 266 28 1
9 | 1 255 3025 7770 6951 2646 462 36 1
10 | 1 511 9330 34105 42525 22827 5880 750 45
1
11 | 1 1023 28501 145750 246730 179487 63987 11880 1155
55 1
12 | 1 2047 86526 611501 1379400 1323652 627396 159027 22275
1705 66 1
13 | 1 4095 261625 2532530 7508501 9321312 5715424 1899612 359502
39325 2431 78 1
14 | 1 8191 788970 10391745 40075035 63436373 49329280 20912320 5135130
752752 66066 3367 91 1
15 | 1 16383 2375101 42355950 210766920 420693273 408741333 216627840 67128490
12662650 1479478 106470 4550 105 1
...
----------------------------------------------------------------------------------
x^4 = 1 x_(1) + 7 x_(2) + 6 x_(3) + 1 x_(4), where x_(k) = P(x,k) = k!*C(x,k). - Daniel Forgues, Jan 16 2016
MAPLE
seq(seq(combinat[stirling2](n, k), k=1..n), n=1..10); # Zerinvary Lajos, Jun 02 2007 stirling_2 := (n, k) -> (1/k!) * add((-1)^(k-i)*binomial(k, i)*i^n, i=0..k);
MATHEMATICA
BellMatrix[f_, len_] := With[{t = Array[f, len, 0]}, Table[BellY[n, k, t], {n, 0, len - 1}, {k, 0, len - 1}]];
rows = 12;
B = BellMatrix[1&, rows];
a[n_, n_] := 1; a[n_, 1] := 1;
a[n_, k_] := a[n, k] = a[n-1, k-1] + k a[n-1, k]; Flatten@
Table[a[n, k], {n, 1, 11}, {k, 1, n}] (* Oliver Seipel, Jun 12 2024 *) With[{m = 11},
Flatten@MapIndexed[Take[#, #2[[1]]] &,
Transpose@
Table[Range[1, m]! Coefficient[(E^x-1)^k/k! + O[x]^(m+1), x,
PROG
(PARI) for(n=1, 22, for(k=1, n, print1(stirling(n, k, 2), ", ")); print()); \\ Joerg Arndt, Apr 21 2013 (PARI) Stirling2(n, k)=sum(i=0, k, (-1)^i*binomial(k, i)*i^n)*(-1)^k/k! \\ M. F. Hasler, Mar 06 2012 (Haskell)
a008277 n k = a008277_tabl !! (n-1) !! (k-1)
a008277_row n = a008277_tabl !! (n-1)
(Maxima) create_list(stirling2(n+1, k+1), n, 0, 30, k, 0, n); /* Emanuele Munarini, Jun 01 2012 */ (J) n ((] (1 % !)) * +/@((^~ * (] (_1 ^ |.)) * (! {:)@]) i.@>:)) k NB. Stephen Makdisi, Apr 06 2016 (Magma) [[StirlingSecond(n, k): k in [1..n]]: n in [1..12]]; // G. C. Greubel, May 22 2019
CROSSREFS
Cf. A008275 (Stirling numbers of first kind), A048993 (another version of this triangle). Cf. A000217, A001296, A001297, A001298, A007318, A028246, A039810- A039813, A048994, A087107- A087111, A087127, A094262, A127701.
Double factorial of odd numbers: a(n) = (2*n-1)!! = 1*3*5*...*(2*n-1).
(Formerly M3002 N1217)
+10
624
1, 1, 3, 15, 105, 945, 10395, 135135, 2027025, 34459425, 654729075, 13749310575, 316234143225, 7905853580625, 213458046676875, 6190283353629375, 191898783962510625, 6332659870762850625, 221643095476699771875, 8200794532637891559375, 319830986772877770815625
COMMENTS
The solution to Schröder's third problem.
Number of fixed-point-free involutions in symmetric group S_{2n} (cf. A000085). a(n-2) is the number of full Steiner topologies on n points with n-2 Steiner points. [corrected by Lyle Ramshaw, Jul 20 2022] a(n) is also the number of perfect matchings in the complete graph K(2n). - Ola Veshta (olaveshta(AT)my-deja.com), Mar 25 2001
Number of ways to choose n disjoint pairs of items from 2*n items. - Ron Zeno (rzeno(AT)hotmail.com), Feb 06 2002
Number of ways to choose n-1 disjoint pairs of items from 2*n-1 items (one item remains unpaired). - Bartosz Zoltak, Oct 16 2012 For n >= 1 a(n) is the number of permutations in the symmetric group S_(2n) whose cycle decomposition is a product of n disjoint transpositions. - Ahmed Fares (ahmedfares(AT)my-deja.com), Apr 21 2001
a(n) is the number of distinct products of n+1 variables with commutative, nonassociative multiplication. - Andrew Walters (awalters3(AT)yahoo.com), Jan 17 2004. For example, a(3)=15 because the product of the four variables w, x, y and z can be constructed in exactly 15 ways, assuming commutativity but not associativity: 1. w(x(yz)) 2. w(y(xz)) 3. w(z(xy)) 4. x(w(yz)) 5. x(y(wz)) 6. x(z(wy)) 7. y(w(xz)) 8. y(x(wz)) 9. y(z(wx)) 10. z(w(xy)) 11. z(x(wy)) 12. z(y(wx)) 13. (wx)(yz) 14. (wy)(xz) 15. (wz)(xy).
a(n) = E(X^(2n)), where X is a standard normal random variable (i.e., X is normal with mean = 0, variance = 1). So for instance a(3) = E(X^6) = 15, etc. See Abramowitz and Stegun or Hoel, Port and Stone. - Jerome Coleman, Apr 06 2004
Second Eulerian transform of 1,1,1,1,1,1,... The second Eulerian transform transforms a sequence s to a sequence t by the formula t(n) = Sum_{k=0..n} E(n,k)s(k), where E(n,k) is a second-order Eulerian number ( A008517). - Ross La Haye, Feb 13 2005 Integral representation as n-th moment of a positive function on the positive axis, in Maple notation: a(n) = int(x^n*exp(-x/2)/sqrt(2*Pi*x), x=0..infinity), n=0,1... . - Karol A. Penson, Oct 10 2005 a(n) is the number of binary total partitions of n+1 (each non-singleton block must be partitioned into exactly two blocks) or, equivalently, the number of unordered full binary trees with n+1 labeled leaves (Stanley, ex 5.2.6). - Mitch Harris, Aug 01 2006 a(n) is the Pfaffian of the skew-symmetric 2n X 2n matrix whose (i,j) entry is i for i<j. - David Callan, Sep 25 2006 a(n) is the number of increasing ordered rooted trees on n+1 vertices where "increasing" means the vertices are labeled 0,1,2,...,n so that each path from the root has increasing labels. Increasing unordered rooted trees are counted by the factorial numbers A000142. - David Callan, Oct 26 2006 Number of perfect multi Skolem-type sequences of order n. - Emeric Deutsch, Nov 24 2006 a(n) = total weight of all Dyck n-paths ( A000108) when each path is weighted with the product of the heights of the terminal points of its upsteps. For example with n=3, the 5 Dyck 3-paths UUUDDD, UUDUDD, UUDDUD, UDUUDD, UDUDUD have weights 1*2*3=6, 1*2*2=4, 1*2*1=2, 1*1*2=2, 1*1*1=1 respectively and 6+4+2+2+1=15. Counting weights by height of last upstep yields A102625. - David Callan, Dec 29 2006 a(n) is the number of increasing ternary trees on n vertices. Increasing binary trees are counted by ordinary factorials ( A000142) and increasing quaternary trees by triple factorials ( A007559). - David Callan, Mar 30 2007 From Tom Copeland, Nov 13 2007, clarified in first and extended in second paragraph, Jun 12 2021: (Start) a(n) has the e.g.f. (1-2x)^(-1/2) = 1 + x + 3*x^2/2! + ..., whose reciprocal is (1-2x)^(1/2) = 1 - x - x^2/2! - 3*x^3/3! - ... = b(0) - b(1)*x - b(2)*x^2/2! - ... with b(0) = 1 and b(n+1) = -a(n) otherwise. By the formalism of A133314, Sum_{k=0..n} binomial(n,k)*b(k)*a(n-k) = 0^n where 0^0 := 1. In this sense, the sequence a(n) is essentially self-inverse. See A132382 for an extension of this result. See A094638 for interpretations. This sequence aerated has the e.g.f. e^(t^2/2) = 1 + t^2/2! + 3*t^4/4! + ... = c(0) + c(1)*t + c(2)*t^2/2! + ... and the reciprocal e^(-t^2/2); therefore, Sum_{k=0..n} cos(Pi k/2)*binomial(n,k)*c(k)*c(n-k) = 0^n; i.e., the aerated sequence is essentially self-inverse. Consequently, Sum_{k=0..n} (-1)^k*binomial(2n,2k)*a(k)*a(n-k) = 0^n. (End)
This is also the number of ways of arranging the elements of n distinct pairs, assuming the order of elements is significant but the pairs are not distinguishable, i.e., arrangements which are the same after permutations of the labels are equivalent.
If this sequence and A000680 are denoted by a(n) and b(n) respectively, then a(n) = b(n)/n! where n! = the number of ways of permuting the pair labels. For example, there are 90 ways of arranging the elements of 3 pairs [1 1], [2 2], [3 3] when the pairs are distinguishable: A = { [112233], [112323], ..., [332211] }.
By applying the 6 relabeling permutations to A, we can partition A into 90/6 = 15 subsets: B = { {[112233], [113322], [221133], [223311], [331122], [332211]}, {[112323], [113232], [221313], [223131], [331212], [332121]}, ....}
Each subset or equivalence class in B represents a unique pattern of pair relationships. For example, subset B1 above represents {3 disjoint pairs} and subset B2 represents {1 disjoint pair + 2 interleaved pairs}, with the order being significant (contrast A132101). (End) a(n) is the number of adjacent transpositions in all fixed-point-free involutions of {1,2,...,2n}. Example: a(2)=3 because in 2143=(12)(34), 3412=(13)(24), and 4321=(14)(23) we have 2 + 0 + 1 adjacent transpositions.
(End)
(1, 3, 15, 105, ...) = INVERT transform of A000698 starting (1, 2, 10, 74, ...). - Gary W. Adamson, Oct 21 2009 a(n) = (-1)^(n+1)*H(2*n,0), where H(n,x) is the probabilists' Hermite polynomial. The generating function for the probabilists' Hermite polynomials is as follows: exp(x*t-t^2/2) = Sum_{i>=0} H(i,x)*t^i/i!. - Leonid Bedratyuk, Oct 31 2009 a(n) is the number of subsets of {1,...,n^2} that contain exactly k elements from {1,...,k^2} for k=1,...,n. For example, a(3)=15 since there are 15 subsets of {1,2,...,9} that satisfy the conditions, namely, {1,2,5}, {1,2,6}, {1,2,7}, {1,2,8}, {1,2,9}, {1,3,5}, {1,3,6}, {1,3,7}, {1,3,8}, {1,3,9}, {1,4,5}, {1,4,6}, {1,4,7}, {1,4,8}, and {1,4,9}. - Dennis P. Walsh, Dec 02 2011 For n>0: a(n) is also the determinant of the symmetric n X n matrix M defined by M(i,j) = min(i,j)^2 for 1 <= i,j <= n. - Enrique Pérez Herrero, Jan 14 2013 a(n) is also the numerator of the mean value from 0 to Pi/2 of sin(x)^(2n). - Jean-François Alcover, Jun 13 2013 For n>1: a(n) is the numerator of M(n)/M(1) where the numbers M(i) have the property that M(n+1)/M(n) ~ n-1/2 (for example, large Kendell-Mann numbers, see A000140 or A181609, as n --> infinity). - Mikhail Gaichenkov, Jan 14 2014 a(n) = the number of upper-triangular matrix representations required for the symbolic representation of a first order central moment of the multivariate normal distribution of dimension 2(n-1), i.e., E[X_1*X_2...*X_(2n-2)|mu=0, Sigma]. See vignette for symmoments R package on CRAN and Phillips reference below. - Kem Phillips, Aug 10 2014 For n>1: a(n) is the number of Feynman diagrams of order 2n (number of internal vertices) for the vacuum polarization with one charged loop only, in quantum electrodynamics. - Robert Coquereaux, Sep 15 2014 Aerated with intervening zeros (1,0,1,0,3,...) = a(n) (cf. A123023), the e.g.f. is e^(t^2/2), so this is the base for the Appell sequence A099174 with e.g.f. e^(t^2/2) e^(x*t) = exp(P(.,x),t) = unsigned A066325(x,t), the probabilist's (or normalized) Hermite polynomials. P(n,x) = (a. + x)^n with (a.)^n = a_n and comprise the umbral compositional inverses for A066325(x,t) = exp(UP(.,x),t), i.e., UP(n,P(.,t)) = x^n = P(n,UP(.,t)), where UP(n,t) are the polynomials of A066325 and, e.g., (P(.,t))^n = P(n,t). - Tom Copeland, Nov 15 2014 a(n) = the number of relaxed compacted binary trees of right height at most one of size n. A relaxed compacted binary tree of size n is a directed acyclic graph consisting of a binary tree with n internal nodes, one leaf, and n pointers. It is constructed from a binary tree of size n, where the first leaf in a post-order traversal is kept and all other leaves are replaced by pointers. These links may point to any node that has already been visited by the post-order traversal. The right height is the maximal number of right-edges (or right children) on all paths from the root to any leaf after deleting all pointers. The number of unbounded relaxed compacted binary trees of size n is A082161(n). See the Genitrini et al. link. - Michael Wallner, Jun 20 2017 Also the number of distinct adjacency matrices in the n-ladder rung graph. - Eric W. Weisstein, Jul 22 2017 a(n) = the number of essentially different ways of writing a probability distribution taking n+1 values as a sum of products of binary probability distributions. See comment of Mitch Harris above. This is because each such way corresponds to a full binary tree with n+1 leaves, with the leaves labeled by the values. (This comment is due to Niko Brummer.)
Also the number of binary trees with root labeled by an (n+1)-set S, its n+1 leaves by the singleton subsets of S, and other nodes labeled by subsets T of S so that the two daughter nodes of the node labeled by T are labeled by the two parts of a 2-partition of T. This also follows from Mitch Harris' comment above, since the leaf labels determine the labels of the other vertices of the tree.
(End)
a(n) is the n-th moment of the chi-squared distribution with one degree of freedom (equivalent to Coleman's Apr 06 2004 comment). - Bryan R. Gillespie, Mar 07 2021 Let b(n) = 0 for n odd and b(2k) = a(k); i.e., let the sequence b(n) be an aerated version of this entry. After expanding the differential operator (x + D)^n and normal ordering the resulting terms, the integer coefficient of the term x^k D^m is n! b(n-k-m) / [(n-k-m)! k! m!] with 0 <= k,m <= n and (k+m) <= n. E.g., (x+D)^2 = x^2 + 2xD + D^2 + 1 with D = d/dx. The result generalizes to the raising (R) and lowering (L) operators of any Sheffer polynomial sequence by replacing x by R and D by L and follows from the disentangling relation e^{t(L+R)} = e^{t^2/2} e^{tR} e^{tL}. Consequently, these are also the coefficients of the reordered 2^n permutations of the binary symbols L and R under the condition LR = RL + 1. E.g., (L+R)^2 = LL + LR + RL + RR = LL + 2RL + RR + 1. (Cf. A344678.) - Tom Copeland, May 25 2021 Lando and Zvonkin present several scenarios in which the double factorials occur in their role of enumerating perfect matchings (pairings) and as the nonzero moments of the Gaussian e^(x^2/2).
Speyer and Sturmfels (p. 6) state that the number of facets of the abstract simplicial complex known as the tropical Grassmannian G'''(2,n), the space of phylogenetic T_n trees (see A134991), or Whitehouse complex is a shifted double factorial. These are also the unsigned coefficients of the x[2]^m terms in the partition polynomials of A134685 for compositional inversion of e.g.f.s, a refinement of A134991. a(n)*2^n = A001813(n) and A001813(n)/(n+1)! = A000108(n), the Catalan numbers, the unsigned coefficients of the x[2]^m terms in the partition polynomials A133437 for compositional inversion of o.g.f.s, a refinement of A033282, A126216, and A086810. Then the double factorials inherit a multitude of analytic and combinatoric interpretations from those of the Catalan numbers, associahedra, and the noncrossing partitions of A134264 with the Catalan numbers as unsigned-row sums. (End) Connections among the Catalan numbers A000108, the odd double factorials, values of the Riemann zeta function and its derivative for integer arguments, and series expansions of the reduced action for the simple harmonic oscillator and the arc length of the spiral of Archimedes are given in the MathOverflow post on the Riemann zeta function. - Tom Copeland, Oct 02 2021 b(n) = a(n) / (n! 2^n) = Sum_{k = 0..n} (-1)^n binomial(n,k) (-1)^k a(k) / (k! 2^k) = (1-b.)^n, umbrally; i.e., the normalized double factorial a(n) is self-inverse under the binomial transform. This can be proved by applying the Euler binomial transformation for o.g.f.s Sum_{n >= 0} (1-b_n)^n x^n = (1/(1-x)) Sum_{n >= 0} b_n (x / (x-1))^n to the o.g.f. (1-x)^{-1/2} = Sum_{n >= 0} b_n x^n. Other proofs are suggested by the discussion in Watson on pages 104-5 of transformations of the Bessel functions of the first kind with b(n) = (-1)^n binomial(-1/2,n) = binomial(n-1/2,n) = (2n)! / (n! 2^n)^2. - Tom Copeland, Dec 10 2022
REFERENCES
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, Tenth Printing, 1972, (26.2.28).
Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 317.
L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 228, #19.
Hoel, Port and Stone, Introduction to Probability Theory, Section 7.3.
F. K. Hwang, D. S. Richards and P. Winter, The Steiner Tree Problem, North-Holland, 1992, see p. 14.
C. Itzykson and J.-B. Zuber, Quantum Field Theory, McGraw-Hill, 1980, pages 466-467.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Example 5.2.6 and also p. 178.
R. Vein and P. Dale, Determinants and Their Applications in Mathematical Physics, Springer-Verlag, New York, 1999, p. 73.
G. Watson, The Theory of Bessel Functions, Cambridge Univ. Press, 1922.
LINKS
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy]. O. Bodini, M. Dien, X. Fontaine, A. Genitrini, and H. K. Hwang, Increasing Diamonds, in LATIN 2016: 12th Latin American Symposium, Ensenada, Mexico, April 11-15, 2016, Proceedings Pages pp. 207-219 2016 DOI 10.1007/978-3-662-49529-2_16; Lecture Notes in Computer Science Series Volume 9644. Peter J. Cameron, Some treelike objects Quart. J. Math. Oxford Ser. 38 (1987), 155-183. MR0891613 (89a:05009). See p. 155. Paul W. Haggard, On Legendre numbers, International Journal of Mathematics and Mathematical Sciences, vol. 8, Article ID 787189, 5 pages, 1985. See Table 1 p. 408. M. Kauers and S.-L. Ko, Problem 11545, Amer. Math. Monthly, 118 (2011), p. 84. Eric Weisstein's World of Mathematics, Erf
FORMULA
E.g.f.: 1 / sqrt(1 - 2*x).
D-finite with recurrence: a(n) = a(n-1)*(2*n-1) = (2*n)!/(n!*2^n) = A010050(n)/ A000165(n). a(n) ~ sqrt(2) * 2^n * (n/e)^n.
Rational part of numerator of Gamma(n+1/2): a(n) * sqrt(Pi) / 2^n = Gamma(n+1/2). - Yuriy Brun, Ewa Dominowska (brun(AT)mit.edu), May 12 2001
With interpolated zeros, the sequence has e.g.f. exp(x^2/2). - Paul Barry, Jun 27 2003 The Ramanujan polynomial psi(n+1, n) has value a(n). - Ralf Stephan, Apr 16 2004 Log(1 + x + 3*x^2 + 15*x^3 + 105*x^4 + 945*x^5 + 10395*x^6 + ...) = x + 5/2*x^2 + 37/3*x^3 + 353/4*x^4 + 4081/5*x^5 + 55205/6*x^6 + ..., where [1, 5, 37, 353, 4081, 55205, ...] = A004208. - Philippe Deléham, Jun 20 2006 1/3 + 2/15 + 3/105 + ... = 1/2. [Jolley eq. 216]
Sum_{j=1..n} j/a(j+1) = (1 - 1/a(n+1))/2. [Jolley eq. 216]
1/1 + 1/3 + 2/15 + 6/105 + 24/945 + ... = Pi/2. - Gary W. Adamson, Dec 21 2006 a(n) = (1/sqrt(2*Pi))*Integral_{x>=0} x^n*exp(-x/2)/sqrt(x). - Paul Barry, Jan 28 2008 G.f.: 1/(1-x-2x^2/(1-5x-12x^2/(1-9x-30x^2/(1-13x-56x^2/(1- ... (continued fraction). - Paul Barry, Sep 18 2009 a(n) = (-1)^n*subs({log(e)=1,x=0},coeff(simplify(series(e^(x*t-t^2/2),t,2*n+1)),t^(2*n))*(2*n)!). - Leonid Bedratyuk, Oct 31 2009 G.f.: 1/(1-x/(1-2x/(1-3x/(1-4x/(1-5x/(1- ...(continued fraction). - Aoife Hennessy (aoife.hennessy(AT)gmail.com), Dec 02 2009
The g.f. of a(n+1) is 1/(1-3x/(1-2x/(1-5x/(1-4x/(1-7x/(1-6x/(1-.... (continued fraction). - Paul Barry, Dec 04 2009 E.g.f.: A(x) = 1 - sqrt(1-2*x) satisfies the differential equation A'(x) - A'(x)*A(x) - 1 = 0. - Vladimir Kruchinin, Jan 17 2011 a(n) = (1/2)*Sum_{i=1..n} binomial(n+1,i)*a(i-1)*a(n-i). See link above. - Dennis P. Walsh, Dec 02 2011 a(n) = Sum_{k=0..n} (-1)^k*binomial(2*n,n+k)*Stirling_1(n+k,k) [Kauers and Ko].
a(n) = A035342(n, 1), n >= 1 (first column of triangle). a(n) = A001497(n, 0) = A001498(n, n), first column, resp. main diagonal, of Bessel triangle. a(n) = upper left term of M^n and sum of top row terms of M^(n-1), where M = a variant of the (1,2) Pascal triangle (Cf. A029635) as the following production matrix: 1, 2, 0, 0, 0, ...
1, 3, 2, 0, 0, ...
1, 4, 5, 2, 0, ...
1, 5, 9, 7, 2, ...
...
For example, a(3) = 15 is the left term in top row of M^3: (15, 46, 36, 8) and a(4) = 105 = (15 + 46 + 36 + 8).
(End)
G.f.: A(x) = 1 + x/(W(0) - x); W(k) = 1 + x + x*2*k - x*(2*k + 3)/W(k+1); (continued fraction). - Sergei N. Gladkovskii, Nov 17 2011 a(n) = Sum_{i=1..n} binomial(n,i-1)*a(i-1)*a(n-i). - Dennis P. Walsh, Dec 02 2011 a(n) = (-1)^n*Sum_{k=0..n} 2^(n-k)*s(n+1,k+1), where s(n,k) are the Stirling numbers of the first kind, A048994. - Mircea Merca, May 03 2012 a(n) = (2*n)_4! = Gauss_factorial(2*n,4) = Product_{j=1..2*n, gcd(j,4)=1} j. - Peter Luschny, Oct 01 2012 G.f.: (1 - 1/Q(0))/x where Q(k) = 1 - x*(2*k - 1)/(1 - x*(2*k + 2)/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Mar 19 2013 G.f.: 1 + x/Q(0), where Q(k) = 1 + (2*k - 1)*x - 2*x*(k + 1)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, May 01 2013 G.f.: 2/G(0), where G(k) = 1 + 1/(1 - 2*x*(2*k + 1)/(2*x*(2*k + 1) - 1 + 2*x*(2*k + 2)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 31 2013 G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x/(x + 1/(2*k + 1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 01 2013 G.f.: G(0), where G(k) = 1 + 2*x*(4*k + 1)/(4*k + 2 - 2*x*(2*k + 1)*(4*k + 3)/(x*(4*k + 3) + 2*(k + 1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 22 2013 a(n) = (2*n - 3)*a(n-2) + (2*n - 2)*a(n-1), n > 1. - Ivan N. Ianakiev, Jul 08 2013 G.f.: G(0), where G(k) = 1 - x*(k+1)/(x*(k+1) - 1/G(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Aug 04 2013 a(n) = 2*a(n-1) + (2n-3)^2*a(n-2), a(0) = a(1) = 1. - Philippe Deléham, Oct 27 2013 G.f. of reciprocals: Sum_{n>=0} x^n/a(n) = 1F1(1; 1/2; x/2), confluent hypergeometric Function. - R. J. Mathar, Jul 25 2014 0 = a(n)*(+2*a(n+1) - a(n+2)) + a(n+1)*(+a(n+1)) for all n in Z. - Michael Somos, Sep 18 2014 a(n) = (-1)^n / a(-n) = 2*a(n-1) + a(n-1)^2 / a(n-2) for all n in Z. - Michael Somos, Sep 18 2014 Recurrence equation: a(n) = (3*n - 2)*a(n-1) - (n - 1)*(2*n - 3)*a(n-2) with a(1) = 1 and a(2) = 3.
The sequence b(n) = A087547(n), beginning [1, 4, 52, 608, 12624, ... ], satisfies the same second-order recurrence equation. This leads to the generalized continued fraction expansion lim_{n -> infinity} b(n)/a(n) = Pi/2 = 1 + 1/(3 - 6/(7 - 15/(10 - ... - n*(2*n - 1)/((3*n + 1) - ... )))). (End) E.g.f of the sequence whose n-th element (n = 1,2,...) equals a(n-1) is 1-sqrt(1-2*x). - Stanislav Sykora, Jan 06 2017 Sum_{n >= 1} a(n)/(2*n-1)! = exp(1/2). - Daniel Suteu, Feb 06 2017 a(n) = (Product_{k=0..n-2} binomial(2*(n-k),2))/n!. - Stefano Spezia, Nov 13 2018 a(n) = Sum_{i=0..n-1} Sum_{j=0..n-i-1} C(n-1,i)*C(n-i-1,j)*a(i)*a(j)*a(n-i-j-1), a(0)=1, - Vladimir Kruchinin, May 06 2020 Sum_{n>=1} 1/a(n) = sqrt(e*Pi/2)*erf(1/sqrt(2)), where erf is the error function.
Sum_{n>=1} (-1)^(n+1)/a(n) = sqrt(Pi/(2*e))*erfi(1/sqrt(2)), where erfi is the imaginary error function. (End)
G.f. of reciprocals: R(x) = Sum_{n>=0} x^n/a(n) satisfies (1 + x)*R(x) = 1 + 2*x*R'(x). - Werner Schulte, Nov 04 2024
EXAMPLE
a(3) = 1*3*5 = 15.
There are a(3)=15 involutions of 6 elements without fixed points:
#: permutation transpositions
01: [ 1 0 3 2 5 4 ] (0, 1) (2, 3) (4, 5)
02: [ 1 0 4 5 2 3 ] (0, 1) (2, 4) (3, 5)
03: [ 1 0 5 4 3 2 ] (0, 1) (2, 5) (3, 4)
04: [ 2 3 0 1 5 4 ] (0, 2) (1, 3) (4, 5)
05: [ 2 4 0 5 1 3 ] (0, 2) (1, 4) (3, 5)
06: [ 2 5 0 4 3 1 ] (0, 2) (1, 5) (3, 4)
07: [ 3 2 1 0 5 4 ] (0, 3) (1, 2) (4, 5)
08: [ 3 4 5 0 1 2 ] (0, 3) (1, 4) (2, 5)
09: [ 3 5 4 0 2 1 ] (0, 3) (1, 5) (2, 4)
10: [ 4 2 1 5 0 3 ] (0, 4) (1, 2) (3, 5)
11: [ 4 3 5 1 0 2 ] (0, 4) (1, 3) (2, 5)
12: [ 4 5 3 2 0 1 ] (0, 4) (1, 5) (2, 3)
13: [ 5 2 1 4 3 0 ] (0, 5) (1, 2) (3, 4)
14: [ 5 3 4 1 2 0 ] (0, 5) (1, 3) (2, 4)
15: [ 5 4 3 2 1 0 ] (0, 5) (1, 4) (2, 3)
(End)
G.f. = 1 + x + 3*x^2 + 15*x^3 + 105*x^4 + 945*x^5 + 10395*x^6 + 135135*x^7 + ...
MAPLE
f := n->(2*n)!/(n!*2^n);
G(x):=(1-2*x)^(-1/2): f[0]:=G(x): for n from 1 to 29 do f[n]:=diff(f[n-1], x) od: x:=0: seq(f[n], n=0..19); # Zerinvary Lajos, Apr 03 2009; aligned with offset by Johannes W. Meijer, Aug 11 2009 series(hypergeom([1, 1/2], [], 2*x), x=0, 20); # Mark van Hoeij, Apr 07 2013
MATHEMATICA
a[ n_] := 2^n Gamma[n + 1/2] / Gamma[1/2]; (* Michael Somos, Sep 18 2014 *) a[ n_] := If[ n < 0, (-1)^n / a[-n], SeriesCoefficient[ Product[1 - (1 - x)^(2 k - 1), {k, n}], {x, 0, n}]]; (* Michael Somos, Jun 27 2017 *)
PROG
(PARI) {a(n) = if( n<0, (-1)^n / a(-n), (2*n)! / n! / 2^n)}; /* Michael Somos, Sep 18 2014 */ (PARI) x='x+O('x^33); Vec(serlaplace((1-2*x)^(-1/2))) \\ Joerg Arndt, Apr 24 2011 (Magma) I:=[1, 3]; [1] cat [n le 2 select I[n] else (3*n-2)*Self(n-1)-(n-1)*(2*n-3)*Self(n-2): n in [1..25] ]; // Vincenzo Librandi, Feb 19 2015 (Haskell)
a001147 n = product [1, 3 .. 2 * n - 1]
a001147_list = 1 : zipWith (*) [1, 3 ..] a001147_list
(Sage) [rising_factorial(n+1, n)/2^n for n in (0..15)] # Peter Luschny, Jun 26 2012 (Python)
from sympy import factorial2
def a(n): return factorial2(2 * n - 1)
(GAP) A001147 := function(n) local i, s, t; t := 1; i := 0; Print(t, ", "); for i in [1 .. n] do t := t*(2*i-1); Print(t, ", "); od; end; A001147(100); # Stefano Spezia, Nov 13 2018 (Maxima)
a(n):=if n=0 then 1 else sum(sum(binomial(n-1, i)*binomial(n-i-1, j)*a(i)*a(j)*a(n-i-j-1), j, 0, n-i-1), i, 0, n-1); /* Vladimir Kruchinin, May 06 2020 */
CROSSREFS
Cf. A086677; A055142 (for this sequence, |a(n+1)| + 1 is the number of distinct products which can be formed using commutative, nonassociative multiplication and a nonempty subset of n given variables). Cf. A079267, A000698, A029635, A161198, A076795, A123023, A161124, A051125, A181983, A099174, A087547, A028338 (first column). Cf. A082161 (relaxed compacted binary trees of unbounded right height). Cf. A000108, A001813, A033282, A060540, A086810, A094638, A126216, A133437, A134264, A134685, A134991, A344678.
KEYWORD
nonn,easy,nice,core,changed
EXTENSIONS
Removed erroneous comments: neither the number of n X n binary matrices A such that A^2 = 0 nor the number of simple directed graphs on n vertices with no directed path of length two are counted by this sequence (for n = 3, both are 13). - Dan Drake, Jun 02 2009
a(n) = n^n; number of labeled mappings from n points to themselves (endofunctions).
(Formerly M3619 N1469)
+10
584
1, 1, 4, 27, 256, 3125, 46656, 823543, 16777216, 387420489, 10000000000, 285311670611, 8916100448256, 302875106592253, 11112006825558016, 437893890380859375, 18446744073709551616, 827240261886336764177, 39346408075296537575424, 1978419655660313589123979
COMMENTS
Also number of labeled pointed rooted trees (or vertebrates) on n nodes.
For n >= 1 a(n) is also the number of n X n (0,1) matrices in which each row contains exactly one entry equal to 1. - Avi Peretz (njk(AT)netvision.net.il), Apr 21 2001
Also the number of labeled rooted trees on (n+1) nodes such that the root is lower than its children. Also the number of alternating labeled rooted ordered trees on (n+1) nodes such that the root is lower than its children. - Cedric Chauve (chauve(AT)lacim.uqam.ca), Mar 27 2002
With p(n) = the number of integer partitions of n, p(i) = the number of parts of the i-th partition of n, d(i) = the number of different parts of the i-th partition of n, p(j, i) = the j-th part of the i-th partition of n, m(i, j) = multiplicity of the j-th part of the i-th partition of n, one has: a(n) = Sum_{i=1..p(n)} (n!/(Product_{j=1..p(i)} p(i, j)!)) * ((n!/(n - p(i)))!/(Product_{j=1..d(i)} m(i, j)!)). - Thomas Wieder, May 18 2005 a(n) is the total number of leaves in all (n+1)^(n-1) trees on {0,1,2,...,n} rooted at 0. For example, with edges directed away from the root, the trees on {0,1,2} are {0->1,0->2},{0->1->2},{0->2->1} and contain a total of a(2)=4 leaves. - David Callan, Feb 01 2007 Limit_{n->infinity} A000169(n+1)/a(n) = exp(1). Convergence is slow, e.g., it takes n > 74 to get one decimal place correct and n > 163 to get two of them. - Alonso del Arte, Jun 20 2011 Also smallest k such that binomial(k, n) is divisible by n^(n-1), n > 0. - Michel Lagneau, Jul 29 2013 For n >= 2 a(n) is represented in base n as "one followed by n zeros". - R. J. Cano, Aug 22 2014 Number of length-n words over the alphabet of n letters. - Joerg Arndt, May 15 2015 Number of prime parking functions of length n+1. - Rui Duarte, Jul 27 2015 The probability density functions p(x, m=q, n=q, mu=1) = A000312(q)*E(x, q, q) and p(x, m=q, n=1, mu=q) = ( A000312(q)/ A000142(q-1))*x^(q-1)*E(x, q, 1), with q >= 1, lead to this sequence, see A163931, A274181 and A008276. - Johannes W. Meijer, Jun 17 2016 A signed version of this sequence apart from the first term (1, -4, -27, 256, 3125, -46656, ...), has the following property: for every prime p == 1 (mod 2n), (-1)^(n(n-1)/2)*n^n = A057077(n)*a(n) is always a 2n-th power residue modulo p. - Jianing Song, Sep 05 2018 n^n is both Sum_{i=0..n} binomial(n,i)*(n-1)^(n-i)
and Sum_{i=0..n} binomial(n,i)*(n-1)^(n-i)*i.
The former is the familiar binomial distribution of a throw of n n-sided dice, according to how many times a required side appears, 0 to n. The latter is the same but each term is multiplied by its amount. This means that if the bank pays the player 1 token for each die that has the chosen side, it is always a fair game if the player pays 1 token to enter - neither bank nor player wins on average.
Examples:
2-sided dice (2 coins): 4 = 1 + 2 + 1 = 1*0 + 2*1 + 1*2 (0 omitted from now on);
3-sided dice (3 long triangular prisms): 27 = 8 + 12 + 6 + 1 = 12*1 + 6*2 + 1*3;
4-sided dice (4 long square prisms or 4 tetrahedrons): 256 = 81 + 108 + 54 + 12 + 1 = 108*1 + 54*2 + 12*3 + 1*4;
5-sided dice (5 long pentagonal prisms): 3125 = 1024 + 1280 + 640 + 160 + 20 + 1 = 1280*1 + 640*2 + 160*3 + 20*4 + 1*5;
6-sided dice (6 cubes): 46656 = 15625 + 18750 + 9375 + 2500 + 375 + 30 + 1 = 18750*1 + 9375*2 + 2500*3 + 375*4 + 30*5 + 1*6.
(End)
For each n >= 1 there is a graph on a(n) vertices whose largest independent set has size n and whose independent set sequence is constant (specifically, for each k=1,2,...,n, the graph has n^n independent sets of size k). There is no graph of smaller order with this property (Ball et al. 2019). - David Galvin, Jun 13 2019 For n >= 2 and 1 <= k <= n, a(n)*(n + 1)/4 + a(n)*(k - 1)*(n + 1 - k)/2*n is equal to the sum over all words w = w(1)...w(n) of length n over the alphabet {1, 2, ..., n} of the following quantity: Sum_{i=1..w(k)} w(i). Inspired by Problem 12432 in the AMM (see links). - Sela Fried, Dec 10 2023
REFERENCES
F. Bergeron, G. Labelle and P. Leroux, Combinatorial Species and Tree-Like Structures, Cambridge, 1998, pp. 62, 63, 87.
L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 173, #39.
A. P. Prudnikov, Yu. A. Brychkov and O.I. Marichev, "Integrals and Series", Volume 1: "Elementary Functions", Chapter 4: "Finite Sums", New York, Gordon and Breach Science Publishers, 1986-1992, Eq. (4.2.2.37)
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
LINKS
E. Vigren (Proposer), Problem 12432, Amer. Math. Monthly 130 (2023), p. 953.
FORMULA
a(n-1) = -Sum_{i=1..n} (-1)^i*i*n^(n-1-i)*binomial(n, i). - Yong Kong (ykong(AT)curagen.com), Dec 28 2000
E.g.f.: 1/(1 + W(-x)), W(x) = principal branch of Lambert's function.
E.g.f.: 1/(1 - T), where T = T(x) is Euler's tree function (see A000169). Comment on power series with denominators a(n): Let f(x) = 1 + Sum_{n>=1} x^n/n^n. Then as x -> infinity, f(x) ~ exp(x/e)*sqrt(2*Pi*x/e). - Philippe Flajolet, Sep 11 2008 E.g.f.: 1 - exp(W(-x)) with an offset of 1 where W(x) = principal branch of Lambert's function. - Vladimir Kruchinin, Sep 15 2010 a(n) = (n-1)*a(n-1) + Sum_{i=1..n} binomial(n, i)*a(i-1)*a(n-i). - Vladimir Shevelev, Sep 30 2010 With an offset of 1, the e.g.f. is the compositional inverse ((x - 1)*log(1 - x))^(-1) = x + x^2/2! + 4*x^3/3! + 27*x^4/4! + .... - Peter Bala, Dec 09 2011 a(n) = (n-1)^(n-1)*(2*n) + Sum_{i=1..n-2} binomial(n, i)*(i^i*(n-i-1)^(n-i-1)), n > 1, a(0) = 1, a(1) = 1. - Vladimir Kruchinin, Nov 28 2014 Sum_{n>=1} 1/a(n) = 1.291285997... = A073009. Sum_{n>=1} 1/a(n)^2 = 1.063887103... = A086648. Sum_{n>=1} n!/a(n) = 1.879853862... = A094082. (End) a(n-1) = abs(p_n(2-n)), for n > 2, the single local extremum of the n-th row polynomial of A055137 with Bagula's sign convention. - Tom Copeland, Nov 15 2019 Limit_{n->oo} (a(n+1)/a(n) - a(n)/a(n-1)) = e (see Brothers/Knox link). - Harlan J. Brothers, Oct 24 2021
EXAMPLE
G.f. = 1 + x + 4*x^2 + 27*x^3 + 256*x^4 + 3125*x^5 + 46656*x^6 + 823543*x^7 + ...
MATHEMATICA
Table[Sum[StirlingS2[n, i] i! Binomial[n, i], {i, 0, n}], {n, 0, 20}] (* Geoffrey Critzer, Mar 17 2009 *) a[ n_] := If[ n < 1, Boole[n == 0], n^n]; (* Michael Somos, May 24 2014 *) a[ n_] := If[ n < 0, 0, n! SeriesCoefficient[ 1 / (1 + LambertW[-x]), {x, 0, n}]]; (* Michael Somos, May 24 2014 *) a[ n_] := If[n < 0, 0, n! SeriesCoefficient[ Nest[ 1 / (1 - x / (1 - Integrate[#, x])) &, 1 + O[x], n], {x, 0, n}]]; (* Michael Somos, May 24 2014 *) a[ n_] := If[ n < 0, 0, With[{m = n + 1}, m! SeriesCoefficient[ InverseSeries[ Series[ (x - 1) Log[1 - x], {x, 0, m}]], m]]]; (* Michael Somos, May 24 2014 *)
PROG
(PARI) {a(n) = n^n};
(PARI) is(n)=my(b, k=ispower(n, , &b)); if(k, for(e=1, valuation(k, b), if(k/b^e == e, return(1)))); n==1 \\ Charles R Greathouse IV, Jan 14 2013 (PARI) {a(n) = my(A = 1 + O(x)); if( n<0, 0, for(k=1, n, A = 1 / (1 - x / (1 - intformal( A)))); n! * polcoeff( A, n))}; /* Michael Somos, May 24 2014 */ (Haskell)
a000312 n = n ^ n
(Maxima) A000312[n]:=if n=0 then 1 else n^n$ (Python)
CROSSREFS
Cf. A000107, A000169, A000272, A001372, A007778, A007830, A008785- A008791, A019538, A048993, A008279, A085741, A062206, A212333.
Search completed in 0.197 seconds
| 
