A152036 - OEIS

%I #4 Oct 21 2012 13:40:48

%S 1,1,2,1,2,4,1,2,4,14,1,2,4,16,48,1,2,4,18,56,202,1,2,4,20,64,248,880,

%T 1,2,4,22,72,298,1100,4286,1,2,4,24,80,352,1344,5504,21760,1,2,4,26,

%U 88,410,1612,6914,28336,118898,1,2,4,28,96,472,1904,8528,36096,157472

%N Triangular product sequence based 2^n times the Fibonacci version and 4 replaced with m: t(m,n)=2^n*Product[(1 + m*Cos[k*Pi/n]^2), {k, 1, Floor[(n - 1)/2]}].

%C The row sums are: {1, 3, 7, 21, 71, 283, 1219, 5785, 29071, 156291, 880507,...}. A sequence of sequences with the row numbers m instead of n: and the ratio increases with each row: at (1+Sqrt[5]) for m=4.

%F t(m,n)=2^n*Product[(1 + m*Cos[k*Pi/n]^2), {k, 1, Floor[(n - 1)/2]}].

%e 1;

%e 1, 2;

%e 1, 2, 4;

%e 1, 2, 4, 14;

%e 1, 2, 4, 16, 48;

%e 1, 2, 4, 18, 56, 202;

%e 1, 2, 4, 20, 64, 248, 880;

%e 1, 2, 4, 22, 72, 298, 1100, 4286;

%e 1, 2, 4, 24, 80, 352, 1344, 5504, 21760;

%e 1, 2, 4, 26, 88, 410, 1612, 6914, 28336, 118898;

%e 1, 2, 4, 28, 96, 472, 1904, 8528, 36096, 157472, 675904;

%t f[n_, m_] = 2^n*Product[(1 + m*Cos[k*Pi/n]^2), {k, 1, Floor[(n - 1)/2]}]; Table[Table[FullSimplify[ExpandAll[f[n, m]]], {n, 0, m}], {m, 0, 10}]

%Y Cf. A103435 (row 4), A083694 (row 5)

%K nonn,uned,tabl

%O 0,3

%A _Roger L. Bagula_ and _Gary W. Adamson_, Nov 20 2008