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A038754
a(2n) = 3^n, a(2n+1) = 2*3^n.
95
1, 2, 3, 6, 9, 18, 27, 54, 81, 162, 243, 486, 729, 1458, 2187, 4374, 6561, 13122, 19683, 39366, 59049, 118098, 177147, 354294, 531441, 1062882, 1594323, 3188646, 4782969, 9565938, 14348907, 28697814, 43046721, 86093442, 129140163, 258280326, 387420489
OFFSET
0,2
COMMENTS
In general, for the recurrence a(n) = a(n-1)*a(n-2)/a(n-3), all terms are integers iff a(0) divides a(2) and first three terms are positive integers, since a(2n+k) = a(k)*(a(2)/a(0))^n for all nonnegative integers n and k.
Equals eigensequence of triangle A070909; (1, 1, 2, 3, 6, 9, 18, ...) shifts to the left with multiplication by triangle A070909. - Gary W. Adamson, May 15 2010
The a(n) represent all paths of length (n+1), n >= 0, starting at the initial node on the path graph P_5, see the second Maple program. - Johannes W. Meijer, May 29 2010
a(n) is the difference between numbers of multiple of 3 evil (A001969) and odious (A000069) numbers in interval [0, 2^(n+1)). - Vladimir Shevelev, May 16 2012
A "half-geometric progression": to obtain a term (beginning with the third one) we multiply the before previous one by 3. - Vladimir Shevelev, May 21 2012
Pisano periods: 1, 2, 1, 4, 8, 2, 12, 4, 1, 8, 10, 4, 6, 12, 8, 8, 32, 2, 36, 8, ... . - R. J. Mathar, Aug 10 2012
Numbers k such that the k-th cyclotomic polynomial has a root mod 3. - Eric M. Schmidt, Jul 31 2013
Range of row n of the circular Pascal array of order 6. - Shaun V. Ault, Jun 05 2014
LINKS
Indranil Ghosh, Table of n, a(n) for n = 0..1500 (first 401 terms from T. D. Noe)
S. V. Ault and C. Kicey, Counting paths in corridors using circular Pascal arrays, Discrete Mathematics, Vol. 332 (2014), pp. 45-54.
Paul Barry, Three Études on a sequence transformation pipeline, arXiv:1803.06408 [math.CO], 2018.
Nachum Dershowitz, Between Broadway and the Hudson, arXiv:2006.06516 [math.CO], 2020.
Richard L. Ollerton and Anthony G. Shannon, Some properties of generalized Pascal squares and triangles, Fib. Q., 36 (1998), 98-109. See pages 106-7.
M. B. Wells, Elements of Combinatorial Computing, Pergamon, Oxford, 1971. [Annotated scanned copy of pages 237-240]
FORMULA
a(n) = a(n-1)*a(n-2)/a(n-3) with a(0)=1, a(1)=2, a(2)=3.
a(2*n) = (3/2)*a(2*n-1) = 3^n, a(2*n+1) = 2*a(2*n) = 2*3^n.
From Benoit Cloitre, Apr 27 2003: (Start)
a(1)=1, a(n)= 2*a(n-1) if a(n-1) is odd, or a(n)= (3/2)*a(n-1) if a(n-1) is even.
a(n) = (1/6)*(5-(-1)^n)*3^floor(n/2).
a(2*n) = a(2*n-1) + a(2*n-2) + a(2*n-3).
a(2*n+1) = a(2*n) + a(2*n-1). (End)
G.f.: (1+2*x)/(1-3*x^2). - Paul Barry, Aug 25 2003
From Reinhard Zumkeller, Sep 11 2003: (Start)
a(n) = (1 + n mod 2) * 3^floor(n/2).
a(n) = A087503(n) - A087503(n-1). (End)
a(n) = sqrt(3)*(2+sqrt(3))*(sqrt(3))^n/6 - sqrt(3)*(2-sqrt(3))*(-sqrt(3))^n/6. - Paul Barry, Sep 16 2003
From Reinhard Zumkeller, May 26 2008: (Start)
a(n) = A140740(n+2,2).
a(n+1) = a(n) + a(n - n mod 2). (End)
If p(i) = Fibonacci(i-3) and if A is the Hessenberg matrix of order n defined by A(i,j) = p(j-i+1), (i<=j), A(i,j)=-1, (i=j+1), and A(i,j)=0 otherwise. Then, for n>=1, a(n-1) = (-1)^n det A. - Milan Janjic, May 08 2010
a(n) = A182751(n) for n >= 2. - Jaroslav Krizek, Nov 27 2010
a(n) = Sum_{i=0..2^(n+1), i==0 (mod 3)} (-1)^A000120(i). - Vladimir Shevelev, May 16 2012
a(0)=1, a(1)=2, for n>=3, a(n)=3*a(n-2). - Vladimir Shevelev, May 21 2012
Sum_(n>=0) 1/a(n) = 9/4. - Alexander R. Povolotsky, Aug 24 2012
a(n) = sqrt(3*a(n-1)^2 + (-3)^(n-1)). - Richard R. Forberg, Sep 04 2013
a(n) = 2^((1-(-1)^n)/2)*3^((2*n-1+(-1)^n)/4). - Luce ETIENNE, Aug 11 2014
From Reinhard Zumkeller, Oct 19 2015: (Start)
a(2*n) = A000244(n), a(2*n+1) = A008776(n).
For n > 0: a(n+1) = a(n) + if a(n) odd then min{a(n), a(n-1)} else max{a(n), a(n-1)}, see also A128588. (End)
E.g.f.: (7*cosh(sqrt(3)*x) + 4*sqrt(3)*sinh(sqrt(3)*x) - 4)/3. - Stefano Spezia, Feb 17 2022
Sum_{n>=0} (-1)^n/a(n) = 3/4. - Amiram Eldar, Dec 02 2022
EXAMPLE
In the interval [0,2^5) we have 11 multiples of 3 numbers, from which 10 are evil and only one (21) is odious. Thus a(4) = 10 - 1 = 9. - Vladimir Shevelev, May 16 2012
MAPLE
a[0]:=0:a[1]:=1:for n from 2 to 50 do a[n]:=3*a[n-2]+2 od: seq(a[n]+1, n=0..34); # Zerinvary Lajos, Mar 20 2008
with(GraphTheory): P:=5: G:=PathGraph(P): A:= AdjacencyMatrix(G): nmax:=35; for n from 1 to nmax do B(n):=A^n; a(n):=add(B(n)[1, k], k=1..P) od: seq(a(n), n=1..nmax); # Johannes W. Meijer, May 29 2010
MATHEMATICA
LinearRecurrence[{0, 3}, {1, 2}, 40] (* Harvey P. Dale, Jan 26 2014 *)
CoefficientList[Series[(1+2x)/(1-3x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Aug 18 2016 *)
Module[{nn=20, c}, c=3^Range[0, nn]; Riffle[c, 2c]] (* Harvey P. Dale, Aug 21 2021 *)
PROG
(PARI) a(n)=(1/6)*(5-(-1)^n)*3^floor(n/2)
(PARI) a(n)=3^(n>>1)<<bittest(n, 0)
(Haskell)
import Data.List (transpose)
a038754 n = a038754_list !! n
a038754_list = concat $ transpose [a000244_list, a008776_list]
-- Reinhard Zumkeller, Oct 19 2015
(Magma) [n le 2 select n else 3*Self(n-2): n in [1..40]]; // Vincenzo Librandi, Aug 18 2016
(SageMath) [2^(n%2)*3^((n-(n%2))/2) for n in range(61)] # G. C. Greubel, Oct 10 2022
KEYWORD
easy,nice,nonn
AUTHOR
Henry Bottomley, May 03 2000
STATUS
approved