[seqfan] Re: Code

[Hans Havermann]

As others have already deconstructed the code, let me just add that the first thing I did (in Mathematica) was to fully simplify the expression of which we were asking the first ten terms. Mathematica suggested that it was equivalent to cos(pi*n)^2 * (F(n)+F(n+1)) - F(n+1). But for positive integer n, cos(pi*n)^2 = 1, so the complicated expression becomes simply F(n).