A380395 The number of unitary divisors of n that are cubes.
1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
Offset: 1
Examples
a(8) = 2 since 8 has 2 unitary divisors that are cubes, 1 = 1^3 and 8 = 2^3. a(216) = 4 since 216 has 4 unitary divisors that are cubes, 1 = 1^3, 8 = 2^3, 27 = 3^3 and 216 = 6^3.
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000
Crossrefs
Programs
-
Mathematica
f[p_, e_] := 2^If[Divisible[e, 3], 1, 0]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
-
PARI
a(n) = {my(f = factor(n)); prod(i = 1, #f~, 2^if(f[i, 2]%3, 0, 1));}
Formula
a(n) = Sum_{d|n, gcd(d, n/d) = 1} [d is cube], where [] is the Iverson bracket.
Multiplicative with a(p^e) = 2 is e is divisible by 3, and 1 otherwise.
a(n) = abs(A307427(n)).
a(n) >= 1, with equality if and only if n is not in A366761.
Dirichlet g.f.: zeta(s)*zeta(3*s)/zeta(4*s).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = zeta(3)/zeta(4) = 1.11062653532614811717... .
In general, the asymptotic mean of the number of unitary divisors of n that are m-th powers is zeta(m)/zeta(m+1), for m >= 2.
Comments