A307609 Number of partitions of n^3 into consecutive positive cubes.

1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1

Offset: 1

Author

Ilya Gutkovskiy, Apr 18 2019

Keywords

• nonn

Examples

			20^3 = 11^3 + 12^3 + 13^3 + 14^3, so a(20) = 2.
2856^3 = 213^3 +...+ 555^3 = 273^3 +...+ 560^3, so a(2856) = 3. See also _Donovan Johnson_'s comment in A097811. - _Antti Karttunen_, Aug 22 2019
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..16861
• Index entries for sequences related to sums of cubes

Crossrefs

Cf. A000578, A030272, A097811, A131643, A217843, A259792, A297199.

Programs

PARI

A297199(n) = { my(s=0, k=1, c); while((c=k^3) <= n, my(u=n-c, i=k); while(u>0, i++; c = i^3; u=u-c); s += (!u); k++); (s); };
A307609(n) = A297199(n^3); \\ Antti Karttunen, Aug 22 2019

Formula

a(n) = [x^(n^3)] Sum_{i>=1} Sum_{j>=i} Product_{k=i..j} x^(k^3).

a(n) = A297199(A000578(n)).

a(n) >= 2 for n in A097811.