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A113780 Number of solutions to 24*n+1 = x^2+24*y^2, x a positive integer, y an integer.

%I A113780 #35 Jul 09 2025 04:25:36
%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,
%T A113780 0,3,4,4,4,2,3,4,2,2,4,0,0,2,2,4,2,9,2,0,2,2,4,1,4,0,4,4,2,0,4,4,4,2,
%U A113780 0,2,1,8,0,2,2,2,6,1,2,4,0,4,4,2,2,0,8,2,2,2,2,0,1,8,0,2,4,0,0,2,5,6,4,2,4
%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.
%C A113780 Ramanujan theta functions: f(q) (see A121373), phi(q) (A000122), psi(q) (A010054), chi(q) (A000700).
%H A113780 G. C. Greubel, <a href="/A113780/b113780.txt">Table of n, a(n) for n = 0..1000</a>
%H A113780 Michael Somos, <a href="/A010815/a010815.txt">Introduction to Ramanujan theta functions</a>
%H A113780 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/RamanujanThetaFunctions.html">Ramanujan Theta Functions</a>
%F A113780 Expansion of phi(x) * phi(-x^3) / chi(-x) in powers of x where phi(), chi() are Ramanujan theta functions. - _Michael Somos_, Jun 08 2012
%F A113780 Expansion of f(x, x) * f(x, x^2) in powers of x where f(, ) is Ramanujan's general theta function. - _Michael Somos_, Jun 08 2013
%F A113780 Expansion of eta(q^2)^6 * eta(q^3)^2 / (eta(q)^3 * eta(q^4)^2 * eta(q^6)) in powers of q. - _Michael Somos_, Jun 08 2012
%F A113780 Euler transform of period 12 sequence [ 3, -3, 1, -1, 3, -4, 3, -1, 1, -3, 3, -2, ...]. - _Michael Somos_, Jun 08 2012
%F A113780 a(n) = A128580(12*n) = A129402(12*n) = A134177(12*n) = A190615(12*n). - _Michael Somos_, Jun 08 2012
%e A113780 If n=51, the solutions (x,y) are: (7,+-7), (19,+-6), (25,+-5), (29,+-4), (35,0) so a(51)=9.
%e A113780 G.f. = 1 + 3*x + 3*x^2 + 2*x^3 + 2*x^4 + 3*x^5 + 4*x^6 + x^7 + 2*x^8 + 4*x^9 + ...
%e A113780 G.f. = q + 3*q^25 + 3*q^49 + 2*q^73 + 2*q^97 + 3*q^121 + 4*q^145 + q^169 + 2*q^193 + ...
%t A113780 a[ n_] := If[ n < 0, 0, With[{m = 24 n + 1}, Sum[ KroneckerSymbol[ -12, d] KroneckerSymbol[ 2, m/d], {d, Divisors @ m}]]]; (* _Michael Somos_, Jun 08 2013 *)
%t A113780 a[ n_] := SeriesCoefficient[ EllipticTheta[ 4, 0, x^3] EllipticTheta[ 3, 0, x] / QPochhammer[ x, x^2], {x, 0, n}]; (* _Michael Somos_, Jun 08 2013 *)
%o A113780 (PARI) {a(n) = if( n<0, 0, n = 24*n + 1; sumdiv( n, d, kronecker( -12, d) * kronecker( 2, n/d)))}; /* _Michael Somos_, Mar 11 2007 */
%o A113780 (PARI) {a(n) = my(A); if( n<0, 0, A = x * O(x^n); polcoeff( eta(x^2 + A)^6 * eta(x^3 + A)^2 / (eta(x + A)^3 * eta(x^4 + A)^2 * eta(x^6 + A)), n))}; /* _Michael Somos_, Jun 08 2012 */
%Y A113780 Cf. A001318 generalized pentagonal numbers, indices of odd values of a(n) and A000009.
%Y A113780 Cf. A114913 = values k such that A000009(k) == 2 (mod 4) and such that a(k) == 2 (mod 4).
%Y A113780 Cf. A128580, A129402, A134177, A190615.
%K A113780 nonn
%O A113780 0,2
%A A113780 _Christian G. Bower_, Jan 20 2006, based on a message from _Dean Hickerson_