A062367 Multiplicative with a(p^e) = (e+1)*(e+2)*(2*e+3)/6.

1, 5, 5, 14, 5, 25, 5, 30, 14, 25, 5, 70, 5, 25, 25, 55, 5, 70, 5, 70, 25, 25, 5, 150, 14, 25, 30, 70, 5, 125, 5, 91, 25, 25, 25, 196, 5, 25, 25, 150, 5, 125, 5, 70, 70, 25, 5, 275, 14, 70, 25, 70, 5, 150, 25, 150, 25, 25, 5, 350, 5, 25, 70, 140, 25, 125, 5, 70, 25, 125, 5

Offset

1,2

Links

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Formula

a(n) = Sum_{i|n, j|n} tau(gcd(i, j)) = Sum_{d|n} tau(d)^2.

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

Dirichlet convolution of A035116 and A000012 (i.e., inverse Mobius transform of A035116). Dirichlet g.f.: zeta^5(s)/zeta(2s). - R. J. Mathar, Feb 03 2011

G.f.: Sum_{n>=1} A000005(n)^2*x^n/(1-x^n). - Mircea Merca, Feb 26 2014

L.g.f.: -log(Product_{k>=1} (1 - x^k)^(tau(k)^2/k)) = Sum_{n>=1} a(n)*x^n/n. - Ilya Gutkovskiy, May 23 2018

Dirichlet convolution of A007426 and A008966. Dirichlet convolution of A007425 and A034444. - R. J. Mathar, Jun 05 2020

Let b(n), n > 0, be Dirichlet inverse of a(n). Then b(n) is multiplicative with b(p^e) = (-1)^e*(Sum_{i=0..e} binomial(4,i)) for prime p and e >= 0, where binomial(n,k)=0 if n < k; abs(b(n)) is multiplicative and has the Dirichlet g.f.: (zeta(s))^5/(zeta(2*s))^4. - Werner Schulte, Feb 07 2021

a(n) = Sum_{d divides n} tau(d^2)*tau(n/d). - Peter Bala, Jan 26 2024

Mathematica

{1}~Join~Array[Times @@ Map[((# + 1) (# + 2) (2 # + 3))/6 &, FactorInteger[#][[All, -1]] ] &, 70, 2] (* or *)
Array[DivisorSum[#, DivisorSigma[0, #]^2 &] &, 71] (* Michael De Vlieger, Mar 05 2021 *)

Prog

(PARI) a(n) = sumdiv(n, d, numdiv(d)^2) \\ Michel Marcus, Jun 17 2013

Crossrefs

Cf. A000005, A000012, A000330, A029939, A035116, A048691, A060648, A062368.

Sequence in context: A076903 A192987 A252768 * A286257 A168418 A266440

Adjacent sequences: A062364 A062365 A062366 * A062368 A062369 A062370

Keyword

nonn,mult

Author

Vladeta Jovovic, Jul 07 2001

Status

approved