A152036 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]}].
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, 1, 2, 4, 22, 72, 298, 1100, 4286, 1, 2, 4, 24, 80, 352, 1344, 5504, 21760, 1, 2, 4, 26, 88, 410, 1612, 6914, 28336, 118898, 1, 2, 4, 28, 96, 472, 1904, 8528, 36096, 157472
Offset: 0
Examples
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; 1, 2, 4, 22, 72, 298, 1100, 4286; 1, 2, 4, 24, 80, 352, 1344, 5504, 21760; 1, 2, 4, 26, 88, 410, 1612, 6914, 28336, 118898; 1, 2, 4, 28, 96, 472, 1904, 8528, 36096, 157472, 675904;
Programs
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Mathematica
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}]
Formula
t(m,n)=2^n*Product[(1 + m*Cos[k*Pi/n]^2), {k, 1, Floor[(n - 1)/2]}].
Comments