A094550 Numbers n such that there are integers a < b with a+(a+1)+...+(n-1) = (n+1)+(n+2)+...+b.

4, 6, 9, 11, 14, 15, 16, 17, 19, 21, 22, 23, 24, 25, 26, 29, 30, 31, 32, 34, 35, 36, 38, 39, 40, 41, 43, 44, 46, 48, 49, 50, 51, 52, 53, 54, 56, 57, 59, 61, 64, 66, 68, 69, 70, 71, 72, 74, 76, 77, 79, 81, 82, 83, 84, 86, 87, 89, 91, 93, 94, 95, 96, 97, 98, 99, 100, 101, 104

Offset: 1

Author

T. D. Noe, May 10 2004

Keywords

• nonn
• easy

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)

Examples

			6 is in this sequence because 1+2+3+4+5 = 7+8.
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..1000
• R. J. Liljestrom and Richard Zucker, Numerical Fulcrums (PowerPoint Format)

Crossrefs

Cf. A094551, A094552, A094553.

Cf. A001912.

Programs

Magma

[n: n in [1..100] |not IsPrime(4*n^2 + 1)]; // Vincenzo Librandi, Sep 27 2012

Mathematica

lst={}; Do[i1=n-1; i2=n+1; s1=i1; s2=i2; While[i1>1 && s1!=s2, If[s1T. D. Noe, Nov 12 2010 *)