A294333 Number of partitions of n into cubes dividing n.

1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 2, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 3, 1, 8, 1, 1, 1, 1, 1, 1, 1, 10, 1, 1, 1, 1, 1, 1, 1, 10, 1, 1, 1, 1, 1, 1, 1, 11, 4, 1, 1, 1, 1, 1, 1, 12, 1, 1, 1, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 14, 1

Offset: 0

Author

Ilya Gutkovskiy, Oct 28 2017

Keywords

• nonn

Examples

			a(8) = 2 because 8 has 4 divisors {1, 2, 4, 8} among which 2 are cubes {1, 8} therefore we have [8] and [1, 1, 1, 1, 1, 1, 1, 1].
		

Links

• Antti Karttunen, Table of n, a(n) for n = 0..6640
• Index entries for sequences related to partitions
• Index entries for sequences related to sums of cubes

Crossrefs

Cf. A018818, A003108, A004709, A030078, A061704, A284345.

Programs

Mathematica

Table[SeriesCoefficient[Product[1/(1 - Boole[Mod[n, k] == 0 && IntegerQ[k^(1/3)]] x^k), {k, 1, n}], {x, 0, n}], {n, 0, 105}]

PARI

cubes_dividing(n) = select(d -> ispower(d,3),divisors(n));
partitions_into(n,parts,from=1) = if(!n,1,my(k = #parts, s=0); for(i=from,k,if(parts[i]<=n, s += partitions_into(n-parts[i],parts,i))); (s));
A294333(n) = if(n<2,1,partitions_into(n,vecsort(cubes_dividing(n), , 4))); \\ Antti Karttunen, Jul 21 2018

Formula

a(n) = 1 if n is a cubefree.

a(n) = 2 if n is a cube of prime.