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%I #32 Jun 18 2017 02:26:43
%S 4,11,18,33,48,74,100,140,180,237,294,371,448,548,648,774,900,1055,
%T 1210,1397,1584,1806,2028,2288,2548,2849,3150,3495,3840,4232,4624,
%U 5066,5508,6003,6498,7049,7600,8210,8820,9492,10164,10901,11638,12443,13248,14124
%N Number of bracelets with 2 blue, 2 red, and n black beads.
%C As n=1,2,3,... and F(2,2,4n) is the number of bracelets with 2 blue, 2 red and 4n black beads,
%C 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,
%C 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.
%H Ata A. Uslu and Hamdi G. Ozmenekse, <a href="http://commons.wikimedia.org/wiki/File:Bracelet_problem.pdf">F(1,6,n)</a>
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (2,1,-4,1,2,-1).
%F F(2,2,4n) = 8n^3 + 14n^2 + 9n + 2.
%F F(2,2,2n+1) = n^3 + 5n^2 + 8n + 4.
%F F(2,2,4n+2) = 8n^3 + 26n^2 + 29n + 11.
%F 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]
%e 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.
%e 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.
%e 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.
%t 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 *)
%K nonn,easy
%O 1,1
%A _Ata Aydin Uslu_, Jan 22 2013
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