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

1,1

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

Table of n, a(n) for n=1..46.

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 A339214 A008061

Adjacent sequences: A210461 A210462 A210463 * A210465 A210466 A210467

Keyword

nonn,easy

Author

Ata Aydin Uslu, Jan 22 2013

Status

approved