|
%I #24 Dec 29 2016 16:04:53
%S 1,1,2,5,17,62,258,1050,4365,18012,73945,301073,1214876,4852899,
%T 19187598,75070201,290659230,1113785613,4224773811,15866483556,
%U 59011553910,217410395916,793635925091,2871246090593,10297627606547,36620869115355,129166280330900
%N Number of partitions of n^3 into cubes.
%H Alois P. Heinz and Vaclav Kotesovec, <a href="/A259792/b259792.txt">Table of n, a(n) for n = 0..173</a> (terms 0..120 from Alois P. Heinz)
%H H. L. Fisher, <a href="/A027601/a027601.pdf">Letter to N. J. A. Sloane, Mar 16 1989</a>
%H G. H. Hardy and S. Ramanujan, <a href="http://ramanujan.sirinudi.org/Volumes/published/ram33.html">Asymptotic formulae in combinatory analysis</a>, Proceedings of the London Mathematical Society, 2, XVI, 1917, p. 373.
%F a(n) = [x^(n^3)] Product_{j>=1} 1/(1-x^(j^3)). - _Alois P. Heinz_, Jul 10 2015
%F a(n) = A003108(n^3). - _Vaclav Kotesovec_, Aug 19 2015
%F a(n) ~ exp(4 * (Gamma(1/3)*Zeta(4/3))^(3/4) * n^(3/4) / 3^(3/2)) * (Gamma(1/3)*Zeta(4/3))^(3/4) / (24*Pi^2*n^(15/4)) [after Hardy & Ramanujan, 1917]. - _Vaclav Kotesovec_, Dec 29 2016
%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-> b(n^3, n):
%p seq(a(n), n=0..26); # _Alois P. Heinz_, Jul 10 2015
%t $RecursionLimit = 1000; b[n_, i_] := b[n, i] = If[n==0 || i==1, 1, b[n, i-1] + If[ i^3>n, 0, b[n-i^3, i]]]; a[n_] := b[n^3, n]; Table[a[n], {n, 0, 26}] (* _Jean-François Alcover_, Jul 15 2015, after _Alois P. Heinz_ *)
%Y A row of the array in A259799.
%Y Cf. A279329.
%Y Cf. A001156, A003108, A046042.
%Y Cf. A037444, A259793.
%K nonn
%O 0,3
%A _N. J. A. Sloane_, Jul 06 2015
%E More term from _Alois P. Heinz_, Jul 10 2015
|
