OFFSET
1,2
COMMENTS
Apparently the inverse Mobius transform of A056624 (and therefore multiplicative). - R. J. Mathar, Feb 07 2011
LINKS
Antti Karttunen, Table of n, a(n) for n = 1..10000
FORMULA
a(n) = A000005(n) iff n is squarefree.
From Werner Schulte, Feb 19 2018: (Start)
Multiplicative with a(p^e) = floor((3*e+2)/2) = A001651(e+1), p prime and e >= 0.
Dirichlet g.f.: Sum_{n>0} a(n)/n^s = (zeta(s))^2 * zeta(2*s) / zeta(3*s). (End)
Sum_{k=1..n} a(k) ~ Pi^2 * n/(6*zeta(3)) * (log(n) - 1 + 2*gamma + 12*zeta'(2)/Pi^2 - 3*zeta'(3)/zeta(3)) + zeta(1/2)^2 * sqrt(n) / zeta(3/2), where gamma is the Euler-Mascheroni constant A001620. - Vaclav Kotesovec, Feb 08 2019
EXAMPLE
For n=12, the divisors of 12^3 = 1728 are 1 = 1^2, 2, 3, 4 = 2^2, 6, 8, 9 = 3^2, 12, 16 = 4^2, 18, 24, 27, 32, 36 = 6^2, 48, 54, 64 = 8^2, 72, 96, 108, 144 = 12^2, 192, 216, 288, 432, 576 = 24^2, 864 and 1728: eight of them are squares, therefore a(12) = 8.
MAPLE
MATHEMATICA
a[n_] := Count[Divisors[n^3], d_ /; IntegerQ[Sqrt[d]]]; Array[a, 100] (* Jean-François Alcover, Feb 13 2018 *)
f[p_, e_] := Floor[(3*e+2)/2]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 15 2020 *)
PROG
(PARI) A046951(n) = factorback(apply(e->e\2+1, factor(n)[, 2])) \\ This function from Charles R Greathouse IV, Sep 17 2015
(PARI) a(n) = sumdiv(n^3, d, issquare(d)); \\ Michel Marcus, Apr 08 2018
(PARI) a(n) = vecprod(apply(x->(3*x+2)\2, factor(n)[, 2])); \\ Amiram Eldar, Aug 18 2024
CROSSREFS
KEYWORD
nonn,mult,easy
AUTHOR
Reinhard Zumkeller, Apr 06 2004
STATUS
approved