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.

A288420 a(n) = Sum_{d|n} d^4*A000593(n/d).

Original entry on oeis.org

1, 17, 85, 273, 631, 1445, 2409, 4369, 6898, 10727, 14653, 23205, 28575, 40953, 53635, 69905, 83539, 117266, 130341, 172263, 204765, 249101, 279865, 371365, 394406, 485775, 558778, 657657, 707311, 911795, 923553, 1118481, 1245505, 1420163, 1520079, 1883154
Offset: 1

Views

Author

Seiichi Manyama, Jun 09 2017

Keywords

Comments

Multiplicative because this sequence is the Dirichlet convolution of A000583 and A000593 which are both multiplicative. - Andrew Howroyd, Jul 20 2018

Crossrefs

Sum_{d|n} d^k*A000593(n/d): A288417 (k=0), A109386 (k=1), A288418 (k=2), A288419 (k=3), this sequence (k=4).

Programs

  • Mathematica
    f[p_, e_] :=  (p^(4*e+7) - (p^3+p^2+p+1)*p^(e+1) + p^2 + p + 1)/(p^7 - (p^3+p^2+p+1)*p + p^2 + p + 1); f[2, e_] := (16^(e+1)-1)/15; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 50] (* Amiram Eldar, Nov 13 2022 *)

Formula

From Amiram Eldar, Nov 13 2022: (Start)
a(n) = A027848(n) for odd n.
Multiplicative with a(2^e) = (16^(e+1)-1)/15 and a(p^e) = (p^(4*e+7) - (p^3+p^2+p+1)*p^(e+1) + p^2 + p + 1)/(p^7 - (p^3+p^2+p+1)*p + p^2 + p + 1) for p > 2.
Sum_{k=1..n} a(k) ~ c * n^5, where c = Pi^4*zeta(5)/480 = (3/16)*zeta(4)*zeta(5) = 0.210429... . (End)

Extensions

Keyword:mult added by Andrew Howroyd, Jul 23 2018