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.

A001868 Number of n-bead necklaces with 4 colors.

Original entry on oeis.org

1, 4, 10, 24, 70, 208, 700, 2344, 8230, 29144, 104968, 381304, 1398500, 5162224, 19175140, 71582944, 268439590, 1010580544, 3817763740, 14467258264, 54975633976, 209430787824, 799645010860, 3059510616424, 11728124734500, 45035996273872, 173215372864600, 667199944815064
Offset: 0

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Comments

From Richard L. Ollerton, May 07 2021: (Start)
Here, as in A000031, turning over is not allowed.
(1/n) * Dirichlet convolution of phi(n) and 4^n, n>0. (End)

References

  • J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 162.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Problem 7.112(a).

Crossrefs

Column 4 of A075195.
Cf. A054611.

Programs

  • Maple
    A001868 := proc(n) local d,s; if n = 0 then RETURN(1); else s := 0; for d in divisors(n) do s := s+phi(d)*4^(n/d); od; RETURN(s/n); fi; end;
  • Mathematica
    a[n_] := (1/n)*Total[ EulerPhi[#]*4^(n/#) &  /@ Divisors[n]]; a[0] = 1; Table[a[n], {n, 0, 24}] (* Jean-François Alcover, Oct 21 2011 *)
    mx=40;CoefficientList[Series[1-Sum[EulerPhi[i] Log[1-4*x^i]/i,{i,1,mx}],{x,0,mx}],x] (* Herbert Kociemba, Nov 01 2016 *)
    k=4; Prepend[Table[DivisorSum[n, EulerPhi[#] k^(n/#) &]/n, {n, 1, 30}], 1] (* Robert A. Russell, Sep 21 2018 *)
  • PARI
    a(n) = if (n, sumdiv(n, d, eulerphi(d)*4^(n/d))/n, 1); \\ Michel Marcus, Nov 01 2016

Formula

a(n) = (1/n)*Sum_{d|n} phi(d)*4^(n/d) = A054611(n)/n, n>0.
G.f.: 1 - Sum_{n>=1} phi(n)*log(1 - 4*x^n)/n. - Herbert Kociemba, Nov 01 2016
a(0) = 1; a(n) = (1/n) * Sum_{k=1..n} 4^gcd(n,k). - Ilya Gutkovskiy, Apr 17 2021
a(0) = 1; a(n) = (1/n)*Sum_{k=1..n} 4^(n/gcd(n,k))*phi(gcd(n,k))/phi(n/gcd(n,k)). - Richard L. Ollerton, May 07 2021