A062368 Multiplicative with a(p^e) = (e+1)*(e+2)*(4*e+3)/6.

1, 7, 7, 22, 7, 49, 7, 50, 22, 49, 7, 154, 7, 49, 49, 95, 7, 154, 7, 154, 49, 49, 7, 350, 22, 49, 50, 154, 7, 343, 7, 161, 49, 49, 49, 484, 7, 49, 49, 350, 7, 343, 7, 154, 154, 49, 7, 665, 22, 154, 49, 154, 7, 350, 49, 350, 49, 49, 7, 1078, 7, 49, 154, 252, 49, 343, 7, 154

Offset: 1

Author

Vladeta Jovovic, Jul 07 2001

Keywords

• nonn
• easy
• mult

Comments

Conjecture: this is the third inverse Mobius transform of the sequence 4^A001221(n). - R. J. Mathar, Aug 09 2012

Links

• Antti Karttunen, Table of n, a(n) for n = 1..16384
• Index entries for sequences computed from exponents in factorization of n.

Crossrefs

Cf. A000005, A001221, A002412, A007425, A060648, A062380.

Programs

Mathematica

f[p_, e_] := (e+1)*(e+2)*(4*e+3)/6; a[n_] := Times @@ f @@@ FactorInteger[n]; a[1] = 1; Array[a, 100] (* Amiram Eldar, Sep 05 2023 *)

Formula

a(n) = Sum_{i|n, j|n} tau(i)*tau(j)/tau(gcd(i, j)), where tau(n) = number of divisors of n, cf. A000005.

Also a(n) = Sum_{i|n, j|n} tau(lcm(i, j)).

a(n) = Sum_{d|n} tau_3(d^2) = Sum_{d|n} A007425(d^2). - Enrique Pérez Herrero, Jan 17 2013