A061391 a(n) = t(n,3) = Sum_{d|n} tau(d^3), where tau(n) = number of divisors of n, cf. A000005.

1, 5, 5, 12, 5, 25, 5, 22, 12, 25, 5, 60, 5, 25, 25, 35, 5, 60, 5, 60, 25, 25, 5, 110, 12, 25, 22, 60, 5, 125, 5, 51, 25, 25, 25, 144, 5, 25, 25, 110, 5, 125, 5, 60, 60, 25, 5, 175, 12, 60, 25, 60, 5, 110, 25, 110, 25, 25, 5, 300, 5, 25, 60, 70, 25, 125, 5, 60, 25, 125, 5, 264

Offset: 1

Author

Vladeta Jovovic, Apr 29 2001

Keywords

• nonn
• mult

Comments

Inverse Mobius transform of A048785. - R. J. Mathar, Feb 09 2011

Links

• Antti Karttunen, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. t(n, 0) = A000005(n), t(n, 1) = A007425(n), t(n, 2) = A035116(n).

Cf. A048691.

Programs

Mathematica

f[p_, e_] := (3*e^2 + 5*e + 2)/2; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 16 2020 *)

PARI

A061391 = n -> sumdiv(n, d, numdiv(d^3));
for(n=1, 10000, write("b061391.txt", n, " ", A061391(n)));
\\ Antti Karttunen, Jan 17 2017

PARI

for(n=1, 100, print1(direuler(p=2, n, (1 + 2*X)/(1 - X)^3)[n], ", ")) \\ Vaclav Kotesovec, May 15 2021

PARI

for(n=1, 100, print1(direuler(p=2, n, (1 - 3*X^2 + 2*X^3)/(1 - X)^5)[n], ", ")) \\ Vaclav Kotesovec, Aug 20 2021

Formula

t(n, k) = Sum_{d|n} tau(d^k) is multiplicative: if the canonical factorization of n = Product p^e(p) over primes then t(n, k) = Product t(p^e(p), k), t(p^e(p), k) = (1/2) *(k*e(p)+2)*(e(p)+1).

For k=2 we get an interesting identity: Sum_{d|n} tau(d^2)=(tau(n))^2, cf. A048691, A035116.

a(n) = Sum_{d|n} tau(n*d). - Benoit Cloitre, Nov 30 2002

G.f.: Sum_{n>=1} tau(n^3)*x^n/(1-x^n). - Joerg Arndt, Jan 01 2011

Dirichlet g.f.: zeta(s)^3 * Product_{primes p} (1 + 2/p^s). - Vaclav Kotesovec, May 15 2021

Dirichlet g.f.: zeta(s)^5 * Product_{primes p} (1 - 3/p^(2*s) + 2/p^(3*s)). - Vaclav Kotesovec, Aug 20 2021