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.

A321554 a(n) = Sum_{d|n} (-1)^(n/d+1)*d^9.

Original entry on oeis.org

1, 511, 19684, 261631, 1953126, 10058524, 40353608, 133955071, 387440173, 998047386, 2357947692, 5149944604, 10604499374, 20620693688, 38445332184, 68584996351, 118587876498, 197981928403, 322687697780, 510998308506, 794320419872, 1204911270612, 1801152661464, 2636771617564, 3814699218751
Offset: 1

Views

Author

N. J. A. Sloane, Nov 23 2018

Keywords

Crossrefs

Cf. A321543 - A321565, A321807 - A321836 for similar sequences.
Cf. A013668.

Programs

  • Mathematica
    f[p_, e_] := (p^(9*e + 9) - 1)/(p^9 - 1); f[2, e_] := (255*2^(9*e + 1) + 1)/511; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 50] (* Amiram Eldar, Nov 11 2022 *)
  • PARI
    apply( A321554(n)=sumdiv(n, d, (-1)^(n\d-1)*d^9), [1..30]) \\ M. F. Hasler, Nov 26 2018

Formula

G.f.: Sum_{k>=1} k^9*x^k/(1 + x^k). - Seiichi Manyama, Nov 24 2018
From Amiram Eldar, Nov 11 2022: (Start)
Multiplicative with a(2^e) = (255*2^(9*e+1)+1)/511, and a(p^e) = (p^(9*e+9) - 1)/(p^9 - 1) if p > 2.
Sum_{k=1..n} a(k) ~ c * n^10, where c = 511*zeta(10)/5120 = 0.0999039... . (End)