OFFSET
1,1
COMMENTS
Liljestrom shows that n is in this sequence if and only if 4n^2+1 is composite.
Complement of A001912.
From Hermann Stamm-Wilbrandt, Sep 16 2014: (Start)
For n > 1, A047209 is a subset of this sequence [ 4*n^2+1 is divisible by 5 if n is (1 or 4) mod 5].
A092464 is a subset of this sequence [4*n^2+1 is divisible by 13 if n is (4 or 9) mod 13].
The above are for divisibility by 5, 13; notation (1,4,5), (4,9,13). Divisibility by p for a and p-a; notation (a,p-a,p). These are the next tuples: (2,15,17), (6,23,29), (3,34,37), (16,25,41), ... . The corresponding sequences are a subset of this sequence [ 2,15,19,32,36,49,... for (2,15,17) ]. These sequences have no entries in the OEIS yet. For any prime of the form 4*k+1 there is exactly one of these tuples/sequences [solution to 4*a^2+1=0 (mod p)].
For n>1, A000290 (squares) is a subset of this sequence (4,9,16,25,...) [ 4*(m^2)^2+1 is divisible by m^2+(m+1)^2, tuple (m^2, (m+1)^2, m^2+(m+1)^2) ].
(End)
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..1000
R. J. Liljestrom and Richard Zucker, Numerical Fulcrums (PowerPoint Format)
EXAMPLE
6 is in this sequence because 1+2+3+4+5 = 7+8.
MATHEMATICA
lst={}; Do[i1=n-1; i2=n+1; s1=i1; s2=i2; While[i1>1 && s1!=s2, If[s1<s2, i1--; s1=s1+i1, i2++; s2=s2+i2]]; If[s1==s2, AppendTo[lst, n]], {n, 2, 140}];
lst (* end of program *)
Select[Range[1000], !PrimeQ[4#^2+1]&] (* T. D. Noe, Nov 12 2010 *)
PROG
(Magma) [n: n in [1..100] |not IsPrime(4*n^2 + 1)]; // Vincenzo Librandi, Sep 27 2012
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
T. D. Noe, May 10 2004
STATUS
approved