OFFSET
1,1
COMMENTS
As n=1,2,3,... and F(2,2,4n) is the number of bracelets with 2 blue, 2 red and 4n black beads,
n=0,1,2... and F(2,2,2n+1) is the number of bracelets with 2 blue, 2 red and 2n+1 black beads,
n=0,1,2... and F(2,2,4n+2) is the number of bracelets with 2 blue, 2 red and 4n+2 black beads.
LINKS
Ata A. Uslu and Hamdi G. Ozmenekse, F(1,6,n)
Index entries for linear recurrences with constant coefficients, signature (2,1,-4,1,2,-1).
FORMULA
F(2,2,4n) = 8n^3 + 14n^2 + 9n + 2.
F(2,2,2n+1) = n^3 + 5n^2 + 8n + 4.
F(2,2,4n+2) = 8n^3 + 26n^2 + 29n + 11.
a(n) = (25+7*(-1)^n+3*(11+(-1)^n)*n+14*n^2+2*n^3)/16. a(n) = 2*a(n-1)+a(n-2)-4*a(n-3)+a(n-4)+2*a(n-5)-a(n-6). G.f.: -x*(2*x^5-4*x^4-2*x^3+8*x^2-3*x-4) / ((x-1)^4*(x+1)^2). [Colin Barker, Feb 06 2013]
EXAMPLE
For 12 black beads, number of possible bracelets: a(12), 4n=12, n=3 so a(12) = F(2,2,12) = 8*3^3 + 14*3^2 + 9*3+2 = 371.
For 19 black beads: a(19), 2n+1=19, n=9 so a(19) = F(2,2,19) = 9^3 + 5*9^2 + 8*9 + 4 = 1210.
For 10 black beads: a(10), 4n+2=10, n=2 so a(10) = F(2,2,10) = 8*2^3 + 26*2^2 + 29*2 + 11 = 237.
MATHEMATICA
Table[If[Mod[m, 4] == 0, n = m/4; 8*n^3 + 14*n^2 + 9*n + 2, If[Mod[m, 2] == 1, n = (m - 1)/2; n^3 + 5*n^2 + 8*n + 4, If[Mod[m, 4] == 2, n = (m - 2)/4; 8*n^3 + 26*n^2 + 29*n + 11]]], {m, 46}] (* T. D. Noe, Jan 24 2013 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Ata Aydin Uslu, Jan 22 2013
STATUS
approved