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The number of divisors of the smallest number whose cube is divisible by n.
5

%I #17 Sep 15 2024 12:32:36

%S 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,

%T 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,

%U 4,8,2,4,2,4,4,4,4,8,2,6,3,4,2,8,4,4,4

%N The number of divisors of the smallest number whose cube is divisible by n.

%C First differs from A365171 at n = 32.

%C The number of divisors of the smallest cube divisible by n, A053149(n), is A365489(n).

%H Amiram Eldar, <a href="/A365488/b365488.txt">Table of n, a(n) for n = 1..10000</a>

%H Vaclav Kotesovec, <a href="/A365488/a365488.jpg">Graph - the asymptotic ratio (100000 terms)</a>

%F a(n) = A000005(A019555(n)).

%F Multiplicative with a(p^e) = ceiling(e/3) + 1.

%F a(n) <= A000005(n) with equality if and only if n is squarefree (A005117).

%F Dirichlet g.f.: zeta(s) * zeta(3*s) * Product_{p prime} (1 + 1/p^s - 1/p^(3*s)).

%F From _Vaclav Kotesovec_, Sep 06 2023: (Start)

%F 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)).

%F Let f(s) = Product_{p prime} (1 - 1/p^(2*s) - 1/p^(3*s) + 1/p^(4*s)).

%F 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 f(1) = Product_{p prime} (1 - 1/p^2 - 1/p^3 + 1/p^4) = 0.5358961538283379998085026313185459506482223745141452711510108346133288...,

%F 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...

%F and gamma is the Euler-Mascheroni constant A001620. (End)

%t f[p_, e_] := Ceiling[e/3] + 1; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]

%t With[{c=Range[200]^3},Table[DivisorSigma[0,Surd[SelectFirst[c,Mod[#,n]==0&],3]],{n,90}]] (* _Harvey P. Dale_, Sep 15 2024 *)

%o (PARI) a(n) = vecprod(apply(x -> (x-1)\3 + 2, factor(n)[, 2]));

%Y Cf. A000005, A005117, A019555, A053149, A365171, A365489, A365498.

%K nonn,easy,mult

%O 1,2

%A _Amiram Eldar_, Sep 05 2023