A287991 Expansion of Jacobi theta constant (theta_2/2)^48.

1, 48, 1128, 17344, 196836, 1764192, 13051008, 82244736, 452197434, 2210431056, 9753024192, 39328459968, 146436844568, 507826976160, 1652238451200, 5074887938688, 14794635174459, 41126600601168, 109456398969568, 279899944411776, 689873759134308

Offset: 0

Author

Seiichi Manyama, Jun 04 2017

Keywords

• nonn

Comments

Number of ways of writing n as the sum of 48 triangular numbers.

Examples

			4*1 + 2*1 + 1*1 = 1 + 6. So a(1) = (4*2*1)^3*((16-1)*(16-4)*(4-1))^2 / 3110400 = 48.
		

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..10000

Crossrefs

Column k=48 of A286180.

Cf. A007331 (k=4*1*2), A014809 (k=4*2*3), this sequence (k=4*3*4).

Programs

Mathematica

a002129[n_]:=-Sum[(-1)^d*d, {d, Divisors[n]}]; a[n_]:=a[n]=If[n==0, 1, 48 Sum[a002129[k] a[n - k], {k, n}]/n]; Table[a[n], {n, 0, 100}] (* Indranil Ghosh, Aug 02 2017 *)

Python

from sympy import divisors
from sympy.core.cache import cacheit
def a002129(n): return -sum((-1)**d*d for d in divisors(n))
@cacheit
def a(n): return 1 if n==0 else 48*sum(a002129(k)*a(n - k) for k in range(1, n + 1))//n
print([a(n) for n in range(101)]) # Indranil Ghosh, Aug 02 2017

Formula

a(0) = 1, a(n) = (48/n)*Sum_{k=1..n} A002129(k)*a(n-k) for n > 0.

a(n) = 1/3110400 * Sum_{a, b, c, x, y, z > 0, a*x + b*y + c*z = n + 6, x == y == z == 1 mod 2 and a > b > c} (a*b*c)^3*((a^2 - b^2)*(a^2 - c^2)*(b^2 - c^2))^2.

G.f.: exp(48*Sum_{k>=1} (x^k/k)/(1 + x^k)). - Ilya Gutkovskiy, Aug 02 2017