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 A210464 Number of bracelets with 2 blue, 2 red, and n black beads. 0
 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 (list; graph; refs; listen; history; text; internal format)
 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 Sequence in context: A008054 A009869 A024989 * A301108 A008061 A304499 Adjacent sequences:  A210461 A210462 A210463 * A210465 A210466 A210467 KEYWORD nonn,easy AUTHOR Ata Aydin Uslu, Jan 22 2013 STATUS approved

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Last modified May 19 17:01 EDT 2019. Contains 323395 sequences. (Running on oeis4.)