A210464 Number of bracelets with 2 blue, 2 red, and n black beads.
4, 11, 18, 33, 48, 74, 100, 140, 180, 237, 294, 371, 448, 548, 648, 774, 900, 1055, 1210, 1397, 1584, 1806, 2028, 2288, 2548, 2849, 3150, 3495, 3840, 4232, 4624, 5066, 5508, 6003, 6498, 7049, 7600, 8210, 8820, 9492, 10164, 10901, 11638, 12443, 13248, 14124
Offset: 1
Examples
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.
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).
Programs
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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 *)
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]
Comments