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A365488
The number of divisors of the smallest number whose cube is divisible by n.
5
1, 2, 2, 2, 2, 4, 2, 2, 2, 4, 2, 4, 2, 4, 4, 3, 2, 4, 2, 4, 4, 4, 2, 4, 2, 4, 2, 4, 2, 8, 2, 3, 4, 4, 4, 4, 2, 4, 4, 4, 2, 8, 2, 4, 4, 4, 2, 6, 2, 4, 4, 4, 2, 4, 4, 4, 4, 4, 2, 8, 2, 4, 4, 3, 4, 8, 2, 4, 4, 8, 2, 4, 2, 4, 4, 4, 4, 8, 2, 6, 3, 4, 2, 8, 4, 4, 4
OFFSET
1,2
COMMENTS
First differs from A365171 at n = 32.
The number of divisors of the smallest cube divisible by n, A053149(n), is A365489(n).
FORMULA
a(n) = A000005(A019555(n)).
Multiplicative with a(p^e) = ceiling(e/3) + 1.
a(n) <= A000005(n) with equality if and only if n is squarefree (A005117).
Dirichlet g.f.: zeta(s) * zeta(3*s) * Product_{p prime} (1 + 1/p^s - 1/p^(3*s)).
From Vaclav Kotesovec, Sep 06 2023: (Start)
Dirichlet g.f.: zeta(s)^2 * zeta(3*s) * Product_{p prime} (1 - 1/p^(2*s) - 1/p^(3*s) + 1/p^(4*s)).
Let f(s) = Product_{p prime} (1 - 1/p^(2*s) - 1/p^(3*s) + 1/p^(4*s)).
Sum_{k=1..n} a(k) ~ zeta(3) * f(1) * n * (log(n) + 2*gamma - 1 + 3*zeta'(3)/zeta(3) + f'(1)/f(1)), where
f(1) = Product_{p prime} (1 - 1/p^2 - 1/p^3 + 1/p^4) = 0.5358961538283379998085026313185459506482223745141452711510108346133288...,
f'(1) = f(1) * Sum_{p prime} (-4 + 3*p + 2*p^2) * log(p) / (1 - p - p^2 + p^4) = f(1) * 1.4525924794451595590371439593828547341482465114411929136723476679...
and gamma is the Euler-Mascheroni constant A001620. (End)
MATHEMATICA
f[p_, e_] := Ceiling[e/3] + 1; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
With[{c=Range[200]^3}, Table[DivisorSigma[0, Surd[SelectFirst[c, Mod[#, n]==0&], 3]], {n, 90}]] (* Harvey P. Dale, Sep 15 2024 *)
PROG
(PARI) a(n) = vecprod(apply(x -> (x-1)\3 + 2, factor(n)[, 2]));
KEYWORD
nonn,easy,mult
AUTHOR
Amiram Eldar, Sep 05 2023
STATUS
approved