cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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

Original entry on oeis.org

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

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Author

Vladeta Jovovic, Jul 07 2001

Keywords

Crossrefs

Programs

  • Maple
    A062367 := proc(n)
        add(numtheory[tau](d)^2,d=numtheory[divisors](n)) ;
    end proc:
    seq(A062367(n),n=1..40) ; # R. J. Mathar, May 15 2025
  • 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 *)
  • PARI
    a(n) = sumdiv(n, d, numdiv(d)^2) \\ Michel Marcus, Jun 17 2013

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), Dirichlet convolution of A048691 and A000005. - Peter Bala, Jan 26 2024