%I A113780
%S A113780 1,3,3,2,2,3,4,1,2,4,2,4,1,2,2,1,8,2,2,2,0,4,1,4,2,2,5,4,2,0,4,4,2,0,0,
%T A113780 3,4,4,4,2,3,4,2,2,4,0,0,2,2,4,2,9,2,0,2,2,4,1,4,0,4,4,2,0,4,4,4,2,0,2,
%U A113780 1,8,0,2,2,2,6,1,2,4,0,4,4,2,2,0,8,2,2,2,2,0,1,8,0,2,4,0,0,2,5,6,4,2,4
%N A113780 Number of solutions to 24*n+1 = x^2+24*y^2, x a positive integer, y an 
               integer.
%C A113780 If 24*n+1 is not a square or if sqrt(24*n+1) == 1 or 11 (mod 12), then 
               A000009(n) == a(n) (mod 4), otherwise A000009(n) == a(n) + 2 (mod 
               4).
%C A113780 Implied by the arithmetic of Q[sqrt(-6)]: Let 24*n+1 = p_1^e_1 * ... 
               * p_r^e_r * q_1^f_1 * ... * q_s^f_s, where the p_i's are distinct 
               primes == 1, 5, 7, or 11 (mod 24) and the q_i's are distinct primes 
               == 13, 17, 19, or 23 (mod 24). If some f_i is odd, then a(n) = 0. 
               Otherwise, a(n) = (e_1 + 1) * ... * (e_r + 1). a(n) == 2 (mod 4) 
               iff all of the f_i's are even and all but one of the e_i's are even 
               and the one e_i which is odd is == 1 (mod 4). Since A000009(n) and 
               a(n) are both odd if 24*n+1 is a square, we can replace a by A000009 
               in this.
%e A113780 If n=51, the solutions (x,y) are: (7,+-7), (19,+-6), (25,+-5), (29,+-4), 
               (35,0) so a(51)=9.
%o A113780 (PARI) {a(n)= if(n<0, 0, n=24*n+1; sumdiv(n, d, kronecker(-12, d)* kronecker(8, 
               n/d)))} /* Michael Somos Mar 11 2007 */
%Y A113780 Cf. A001318 generalized pentagonal numbers, indices of odd values of 
               a(n) and A000009.
%Y A113780 A128580(12n)= a(n).
%Y A113780 Cf. A114913 = values k such that A000009(k) == 2 (mod 4) and such that 
               a(k) == 2 (mod 4).
%Y A113780 Sequence in context: A021305 A075788 A162235 this_sequence A007515 A014967 
               A120992
%Y A113780 Adjacent sequences: A113777 A113778 A113779 this_sequence A113781 A113782 
               A113783
%K A113780 nonn
%O A113780 0,2
%A A113780 Christian G. Bower (bowerc(AT)usa.net), Jan 20 2006, based on a message 
               from Dean Hickerson (dean.hickerson(AT)yahoo.com).