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

Author

Paul D. Hanna, Jan 23 2024

Keywords

• nonn

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.

Crossrefs

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

Programs

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 *)

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),", "))

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