A384785 The number of unordered factorizations of the n-th cubefull number into 1 and prime powers p^e where p is prime and e >= 3 (A246549).

1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 3, 1, 1, 4, 1, 1, 2, 1, 1, 5, 1, 1, 2, 1, 1, 6, 2, 1, 1, 1, 1, 1, 1, 2, 1, 1, 9, 1, 1, 2, 1, 2, 3, 1, 3, 1, 2, 10, 1, 1, 1, 2, 1, 1, 2, 1, 4, 1, 2, 2, 2, 13, 1, 1, 2, 1, 1, 4, 1, 3, 1, 2, 2, 1, 1, 1, 5, 1, 1, 1, 1, 2, 3, 17, 2

Offset: 1

Author

Amiram Eldar, Jun 10 2025

Keywords

• nonn
• easy

Comments

The positive values of the multiplicative function f(n) with f(p^e) = A008483(e). Or, equivalently, a(n) is the value of this function at A036966(n).

Examples

			a(6) = 2 since the 6th cubefull number, A036966(6) = 64, has 2 factorizations: 2^3 * 2^3 and 2^6.
a(12) = 3 since the 12th cubefull number, A036966(12) = 256, has 3 factorizations: 2^3 * 2^5, 2^4 * 2^4, and 2^8.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A008483, A036966, A246549, A384783 (powerful analog), A384786.

Programs

Mathematica

f[p_, e_] := PartitionsP[e] - PartitionsP[e-1] - PartitionsP[e-2] + PartitionsP[e-3]; s[1] = 1; s[n_] := Times @@ f @@@ FactorInteger[n]; seq[lim_] := Module[{cub = Union[Flatten[Table[i^3*j^4*k^5, {k, 1, Surd[lim, 5]}, {j, 1, Surd[lim/k^5, 4]}, {i, 1, Surd[lim/(j^4*k^5), 3]}]]]}, Select[s /@ cub, # > 0 &]]; seq[10^5]

PARI

s(n) = vecprod(apply(x -> numbpart(x)-numbpart(x-1)-numbpart(x-2)+numbpart(x-3), factor(n)[, 2]));
cubs(lim) = {my(c = List()); for(k = 1, sqrtnint(lim, 5), for(j = 1, sqrtnint(lim \ k^5, 4), for(i = 1, sqrtnint(lim \ (j^4*k^5), 3), listput(c, i^3*j^4*k^5)))); Set(c); }
list(lim) = {my(c = cubs(lim), v = List(), s1); for(k = 1, #c, s1 = s(c[k]); if(s1 > 0, listput(v, s1))); Vec(v);}