A113780 - OEIS

A113780

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

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, 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, 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

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OFFSET

0,2

COMMENTS

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).

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.

Ramanujan theta functions: f(q) (see A121373), phi(q) (A000122), psi(q) (A010054), chi(q) (A000700).

LINKS

G. C. Greubel, Table of n, a(n) for n = 0..1000

Michael Somos, Introduction to Ramanujan theta functions

Eric Weisstein's World of Mathematics, Ramanujan Theta Functions

FORMULA

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

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

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

Euler transform of period 12 sequence [ 3, -3, 1, -1, 3, -4, 3, -1, 1, -3, 3, -2, ...]. - Michael Somos, Jun 08 2012

a(n) = A128580(12*n) = A129402(12*n) = A134177(12*n) = A190615(12*n). - Michael Somos, Jun 08 2012

EXAMPLE

If n=51, the solutions (x,y) are: (7,+-7), (19,+-6), (25,+-5), (29,+-4), (35,0) so a(51)=9.

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 + ...

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 + ...

MATHEMATICA

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 *)

a[ n_] := SeriesCoefficient[ EllipticTheta[ 4, 0, x^3] EllipticTheta[ 3, 0, x] / QPochhammer[ x, x^2], {x, 0, n}]; (* Michael Somos, Jun 08 2013 *)

PROG

(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 */

(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 */

CROSSREFS

Cf. A001318 generalized pentagonal numbers, indices of odd values of a(n) and A000009.

Cf. A114913 = values k such that A000009(k) == 2 (mod 4) and such that a(k) == 2 (mod 4).

Cf. A128580, A129402, A134177, A190615.

Sequence in context: A303934 A304031 A162235 * A007515 A352455 A293729

Adjacent sequences: A113777 A113778 A113779 * A113781 A113782 A113783

KEYWORD

nonn

AUTHOR

Christian G. Bower, Jan 20 2006, based on a message from Dean Hickerson.

STATUS

approved