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A032242 Number of identity bracelets of n beads of 5 colors. 4
5, 10, 10, 45, 252, 1120, 5270, 23475, 106950, 483504, 2211650, 10148630, 46911060, 217863040, 1017057256, 4767774375, 22438419120, 105960830300, 501928967930, 2384170903140, 11353241255900 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,1
COMMENTS
For n>2 also number of asymmetric bracelets with n beads of five colors. - Herbert Kociemba, Nov 29 2016
LINKS
C. G. Bower, Transforms (2)
F. Ruskey, Necklaces, Lyndon words, De Bruijn sequences, etc. [Cached copy, with permission, pdf format only]
FORMULA
"DHK" (bracelet, identity, unlabeled) transform of 5, 0, 0, 0...
From Herbert Kociemba, Nov 29 2016: (Start)
More generally, gf(k) is the g.f. for the number of asymmetric bracelets with n beads of k colors.
gf(k): Sum_{n>=1} mu(n)*( -log(1-k*x^n)/n - Sum_{i=0..2} binomial(k,i)x^(n*i)/(1-k*x^(2*n)) )/2. (End)
MAPLE
N:= 50: # for a(1)..a(N)
G:= add(1/2*numtheory:-mobius(n)*(-log(1-5*x^n)/n - add(binomial(5, i)*x^(n*i)/(1-5*x^(2*n)), i=0..2)), n=1..N):
S:= series(G, x, N+1):
5, 10, seq(coeff(S, x, j), j=3..N); # Robert Israel, Jun 24 2019
MATHEMATICA
m=5; (* asymmetric bracelets of n beads of m colors *) Table[Sum[MoebiusMu[d](m^(n/d)/n - If[OddQ[n/d], m^((n/d+1)/2), ((m+1)m^(n/(2d))/2)]), {d, Divisors[n]}]/2, {n, 3, 20}] (* Robert A. Russell, Mar 18 2013 *)
mx=40; gf[x_, k_]:=Sum[MoebiusMu[n]*(-Log[1-k*x^n]/n-Sum[Binomial[k, i]x^(n i), {i, 0, 2}]/(1-k x^(2n)))/2, {n, mx}]; ReplacePart[Rest[CoefficientList[Series[gf[x, 5], {x, 0, mx}], x]], {1->5, 2->10}] (* Herbert Kociemba, Nov 29 2016 *)
PROG
(PARI) a(n)={if(n<3, binomial(5, n), sumdiv(n, d, moebius(n/d)*(5^d/n - if(d%2, 5^((d+1)/2), 3*5^(d/2))))/2)} \\ Andrew Howroyd, Sep 12 2019
CROSSREFS
Column k=5 of A309528 for n >= 3.
Sequence in context: A201033 A242894 A256641 * A208541 A324593 A107975
KEYWORD
nonn
AUTHOR
STATUS
approved

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Last modified June 15 03:12 EDT 2024. Contains 373402 sequences. (Running on oeis4.)