A384783 The number of unordered factorizations of the n-th powerful number into 1 and prime powers p^e where p is prime and e >= 2 (A025475).
1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 4, 1, 2, 1, 1, 1, 1, 4, 2, 1, 1, 1, 1, 1, 2, 7, 2, 1, 2, 1, 1, 1, 2, 2, 1, 1, 1, 8, 1, 4, 2, 2, 1, 1, 4, 2, 2, 1, 2, 1, 1, 1, 2, 1, 12, 1, 1, 4, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 2, 4, 1, 4, 1, 1, 1, 2, 2, 2, 2, 14, 1, 4, 1, 1, 7, 1, 2
Offset: 1
Examples
a(5) = 2 since the 5th powerful number, A001694(5) = 16, has 2 factorizations: 2^2 * 2^2 and 2^4. a(11) = 4 since the 11th powerful number, A001694(11) = 64, has 4 factorizations: 2^2 * 2^2 * 2^2, 2^2 * 2^4, 2^3 * 2^3, and 2^6.
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000
Programs
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Mathematica
f[p_, e_] := PartitionsP[e] - PartitionsP[e-1]; s[1] = 1; s[n_] := Times @@ f @@@ FactorInteger[n]; seq[lim_] := Module[{pow = Union[Flatten[Table[i^2*j^3, {j, 1, Surd[lim, 3]}, {i, 1, Sqrt[lim/j^3]}]]]}, Select[s /@ pow, # > 0 &]]; seq[10^4] -
PARI
s(n) = vecprod(apply(x -> numbpart(x)-numbpart(x-1), factor(n)[, 2])); pows(lim) = {my(p = List()); for(j = 1, sqrtnint(lim, 3), for(i = 1, sqrtint(lim \ j^3), listput(p, i^2 * j^3))); Set(p); } list(lim) = {my(p = pows(lim), v = List(), s1); for(k = 1, #p, s1 = s(p[k]); if(s1 > 0, listput(v, s1))); Vec(v);}
Comments