OFFSET
0,7
COMMENTS
a(n-1) = number of aperiodic necklaces (Lyndon words) with 5 black beads and n-5 white beads.
REFERENCES
J. M. Borwein, D. H. Bailey and R. Girgensohn, Experimentation in Mathematics, A K Peters, Ltd., Natick, MA, 2004. x+357 pp. See p. 147.
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..1000
D. J. Broadhurst, On the enumeration of irreducible k-fold Euler sums and their roles in knot theory and field theory, arXiv:hep-th/9604128, 1996.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1,1,-4,6,-4,1).
FORMULA
G.f.: x^5*(1+x^3)/((1-x)^3*(1-x^2)*(1-x^5)) = x^5*(1-x+x^2)/((1-x)^5*(1+x+x^2+x^3+x^4)).
a(n) = floor(binomial(n+1,5)/(n+1)). - Gary Detlefs, Nov 23 2011
MAPLE
seq(floor(binomial(n, 4)/5), n=0.. 70); # Zerinvary Lajos, Jan 12 2009
MATHEMATICA
CoefficientList[Series[x^5(1+x^3)/((1-x)^3(1-x^2)(1-x^5)), {x, 0, 70}], x] (* Vincenzo Librandi, Jun 19 2012 *)
CoefficientList[Series[x^4/5 (1/(1-x)^5-1/(1- x^5)), {x, 0, 70}], x] (* Herbert Kociemba, Oct 16 2016 *)
PROG
(Magma) [Floor(Binomial(n+1, 5)/(n+1)): n in [0..70]]; // Vincenzo Librandi Jun 19 2012
(PARI) a(n)=binomial(n, 4)\5 \\ Charles R Greathouse IV, Oct 07 2015
(SageMath) [binomial(n, 4)//5 for n in range(71)] # G. C. Greubel, Oct 20 2024
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
STATUS
approved