A369087 Number of partitions of the square of the n-th triangular number into cubes.

1, 1, 2, 7, 39, 289, 2514, 24625, 262768, 2998410, 36104634, 454307830, 5930683832, 79871973688, 1104858227039, 15642564963293, 226021630964469, 3325082850102429, 49705611691011214, 753760884548732440, 11578963116012317128, 179963257865062829685, 2826949471456067702388, 44840817231090333901577

Offset

0,3

Comments

The sum of the first n cubes equals the square of the n-th triangular number: (1^3 + 2^3 + 3^3 + ... + n^3) = (1 + 2 + 3 + ... + n)^2 = (n*(n+1)/2)^2.

Links

Vaclav Kotesovec, Table of n, a(n) for n = 0..100

G. H. Hardy and S. Ramanujan, Asymptotic formulae in combinatory analysis, Proceedings of the London Mathematical Society, 2, XVI, 1917, p. 373.

Formula

a(n) equals the coefficient of x^A000537(n) in Product_{k>=1} 1/(1 - x^(k^3)) for n >= 0, where A000537(n) = (n*(n+1)/2)^2.

a(n) = A003108(A000537(n)) for n >= 0.

Limit a(n)^(1/n) = exp( (Gamma(1/3) * Zeta(4/3))^(3/4) * (2/3)^(3/2) ) = 19.6784468967446... - see formula by Vaclav Kotesovec in A003108.

a(n) ~ (Gamma(1/3)*zeta(4/3))^(3/4) * exp(sqrt(2) * 3^(-3/2) * (Gamma(1/3)*zeta(4/3))^(3/4) * (2*n+1)) / (3 * sqrt(2) * Pi^2 * n^5). - Vaclav Kotesovec, Jan 25 2024

Maple

b:= proc(n, i) option remember; `if`(n=0 or i=1, 1,
      b(n, i-1) +`if`(i^3>n, 0, b(n-i^3, i)))
    end:
a:= n-> (t-> b(t, iroot(t, 3)))((n*(n+1)/2)^2):
seq(a(n), n=0..23);  # Alois P. Heinz, Jan 24 2024

Mathematica

Table[SeriesCoefficient[Product[1/(1 - x^(k^3)), {k, 1, (n*(n + 1)/2)^(2/3) + 1}], {x, 0, (n*(n + 1)/2)^2}], {n, 0, 20}] (* Vaclav Kotesovec, Jan 24 2024 *)

Prog

(PARI) {a(n) = my(M=(n*(n+1)/2)^2); polcoeff(1/prod(k=1, ceil(M^(1/3)), 1-x^(k^3)+x*O(x^M)), M)}
for(n=0, 30, print1(a(n), ", "))

Crossrefs

Cf. A000217, A000578, A003108, A000537, A218495.

Sequence in context: A364336 A266310 A032118 * A125660 A112944 A060073

Adjacent sequences: A369084 A369085 A369086 * A369088 A369089 A369090

Keyword

nonn

Author

Paul D. Hanna, Jan 23 2024

Status

approved