OFFSET
0,2
COMMENTS
a(n-9)=number of aperiodic necklaces (Lyndon words) with 8 black beads and n-8 white beads.
LINKS
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.
FORMULA
G.f.: (1+x^2)/((1-x)*(1-x^2))^4
a(n) = [C(n+8,7)-(n%2)*C((n+7)/2,3)]/8, where C = binomial, n%2 = parity of n (=1 if odd, 0 else). - M. F. Hasler, May 02 2009
a(0)=1, a(1)=4, a(2)=15, a(3)=40, a(4)=99, a(5)=212, a(6)=429, a(7)=800, a(8)=1430, a(9)=2424, a(10)=3978, a(11)=6288, a(n) = 4*a(n-1)-2*a(n-2)-12*a(n-3)+17*a(n-4)+8*a(n-5)-28*a(n-6)+8*a(n-7)+17*a(n-8)-12*a(n-9)- 2*a(n-10)+4*a(n-11)-a(n-12). - Harvey P. Dale, Jun 20 2011
G.f.: ((-1+x)^-8-(-1+x^2)^-4)/(8*x). - Herbert Kociemba, Oct 16 2016
MATHEMATICA
Table[(Binomial[n+8, 7]-If[OddQ[n], 1, 0]Binomial[(n+7)/2, 3])/8, {n, 0, 40}] (* or *) CoefficientList[Series[(1+x^2)/((1-x)^8 (1+x)^4), {x, 0, 40}], x] (* Harvey P. Dale, Jun 20 2011 *)
PROG
(PARI) A031164(n)=(binomial(n+8, 7)-if(n%2, binomial(n\2+4, 3)))>>3 \\ M. F. Hasler, May 02 2009
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
STATUS
approved