A294072 Number of noncube divisors of n^3.

0, 2, 2, 4, 2, 12, 2, 6, 4, 12, 2, 22, 2, 12, 12, 8, 2, 22, 2, 22, 12, 12, 2, 32, 4, 12, 6, 22, 2, 56, 2, 10, 12, 12, 12, 40, 2, 12, 12, 32, 2, 56, 2, 22, 22, 12, 2, 42, 4, 22, 12, 22, 2, 32, 12, 32, 12, 12, 2, 100, 2, 12, 22, 12, 12, 56, 2, 22, 12, 56, 2, 58, 2, 12, 22, 22, 12, 56, 2, 42, 8, 12, 2, 100, 12

Offset: 1

Author

Ilya Gutkovskiy, Feb 07 2018

Keywords

• nonn

Comments

All terms are even. a(n)=2 if and only if n is prime. - Robert Israel, Jan 16 2020

Examples

			a(4) = 4 because 4^3 = 64 has 7 divisors {1, 2, 4, 8, 16, 32, 64} among which 4 divisors {2, 4, 16, 32} are noncubes.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000
• Index entries for sequences computed from exponents in factorization of n

Crossrefs

Cf. A000005, A000578, A001221, A007412, A048785, A055205, A061704, A291208.

Programs

Maple

f:= proc(n) local F;
  F:= map(t -> t[2],ifactors(n)[2]);
  mul(1+3*t,t=F) - mul(1+t,t=F)
end proc:
map(f, [$1..100]; # Robert Israel, Jan 16 2020

Mathematica

nmax = 85; Rest[CoefficientList[Series[Sum[(3^PrimeNu[k] - 1) x^k/(1 - x^k), {k, 1, nmax}], {x, 0, nmax}], x]]
a[n_] := Length[Select[Divisors[n], ! IntegerQ[#^(1/3)] &]]; Table[a[n^3], {n, 1, 85}]
Table[DivisorSigma[0, n^3] - DivisorSigma[0, n], {n, 1, 85}]

Formula

G.f.: Sum_{k>=1} (3^omega(k) - 1)*x^k/(1 - x^k), where omega(k) is the number of distinct primes dividing k (A001221).

a(n) = [x^(n^3)] Sum_{k>=1} x^A007412(k)/(1 - x^A007412(k)).

a(n) = A291208(A000578(n)).

a(n) = A048785(n) - A000005(n).