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Aug 19, 2020 at 14:22 comment added Max Alekseyev In fact, the sequence $S_k$ satisfy a second-order recurrence: $$S_k = 2S_{k-1} - 3S_{k-2}.$$
Aug 19, 2020 at 2:10 comment added Pedja @MaxAlekseyev Thanks!
Aug 18, 2020 at 12:16 comment added Max Alekseyev Notice that $\gcd(a,b) = \gcd(a\bmod b,b)$. That is, you can restrict computation of $S_k$ only to modulo $M_p$ (or $F_n$) and still compute the gcd.
Aug 18, 2020 at 9:18 history asked Pedja CC BY-SA 4.0