A294335 Number of compositions (ordered partitions) of n into cubes dividing n.
1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 11, 1, 1, 1, 1, 1, 1, 1, 64, 1, 1, 2, 1, 1, 1, 1, 345, 1, 1, 1, 1, 1, 1, 1, 1824, 1, 1, 1, 1, 1, 1, 1, 9661, 1, 1, 1, 1, 1, 30, 1, 51284, 1, 1, 1, 1, 1, 1, 1, 272334, 1, 1, 1, 1, 1, 1, 1, 1445995, 1, 1, 1, 1, 1, 1, 1, 7677250, 463, 1, 1, 1, 1
Offset: 0
Keywords
Examples
a(16) = 11 because 16 has 5 divisors {1, 2, 4, 8, 16} among which 2 are cubes {1, 8} therefore we have [8, 8], [8, 1, 1, 1, 1, 1, 1, 1, 1], [1, 8, 1, 1, 1, 1, 1, 1, 1], [1, 1, 8, 1, 1, 1, 1, 1, 1], [1, 1, 1, 8, 1, 1, 1, 1, 1], [1, 1, 1, 1, 8, 1, 1, 1, 1], [1, 1, 1, 1, 1, 8, 1, 1, 1], [1, 1, 1, 1, 1, 1, 8, 1, 1], [1, 1, 1, 1, 1, 1, 1, 8, 1], [1, 1, 1, 1, 1, 1, 1, 1, 8] and [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1].
Links
Programs
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Mathematica
Table[SeriesCoefficient[1/(1 - Sum[Boole[Mod[n, k] == 0 && IntegerQ[k^(1/3)]] x^k, {k, 1, n}]), {x, 0, n}], {n, 0, 85}]
Formula
a(m)=1 when m is cubefree (A004709) and a(m)<>1 when m is not cubefree (A046099). - Michel Marcus, Oct 29 2017