%I #29 Jan 26 2024 07:03:35
%S 1,1,2,7,39,289,2514,24625,262768,2998410,36104634,454307830,
%T 5930683832,79871973688,1104858227039,15642564963293,226021630964469,
%U 3325082850102429,49705611691011214,753760884548732440,11578963116012317128,179963257865062829685,2826949471456067702388,44840817231090333901577
%N Number of partitions of the square of the n-th triangular number into cubes.
%C 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.
%H Vaclav Kotesovec, <a href="/A369087/b369087.txt">Table of n, a(n) for n = 0..100</a>
%H G. H. Hardy and S. Ramanujan, <a href="http://ramanujan.sirinudi.org/Volumes/published/ram36.pdf">Asymptotic formulae in combinatory analysis</a>, Proceedings of the London Mathematical Society, 2, XVI, 1917, p. 373.
%F 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.
%F a(n) = A003108(A000537(n)) for n >= 0.
%F 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.
%F 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
%p b:= proc(n, i) option remember; `if`(n=0 or i=1, 1,
%p b(n, i-1) +`if`(i^3>n, 0, b(n-i^3, i)))
%p end:
%p a:= n-> (t-> b(t, iroot(t, 3)))((n*(n+1)/2)^2):
%p seq(a(n), n=0..23); # _Alois P. Heinz_, Jan 24 2024
%t 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 *)
%o (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)}
%o for(n=0,30, print1(a(n),", "))
%Y Cf. A000217, A000578, A003108, A000537, A218495.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Jan 23 2024