OFFSET
0,4
COMMENTS
It appears that Limit[Sqrt[a[n+2]/a[n]],n->Infinity]=1+(Sqrt[5]+1)/2.
MATHEMATICA
f[n_] = 2*Product[(1 + 4*Cos[k*Pi/n]^2)*(1 + 4*Sin[k*Pi/n]^2), {k, 1, Floor[(n - 1)/2]}] - Product[(1 + 4*Sin[k*Pi/n]^2), {k, 1, Floor[(n - 1)/2]}]; Table[N[f[n]], {n, 0, 30}]; Floor[%]
CROSSREFS
KEYWORD
nonn
AUTHOR
Roger L. Bagula, Nov 28 2008
STATUS
approved