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Feb 15 at 8:43 comment added Peter Taylor @Voile, now that you mention it I have been a bit cavalier in the switch from $\mathbb{Z}/(p^2-p-1)$ treating $p$ as a variable to specific values of $p$. We might require $F(n+1)p + F(n) < p^2 - p - 1$, so which gives a bound of approximately $\frac12 F(n+2) + F(\frac n2)$.
Feb 15 at 8:10 comment added Voile I tested some values of $p$; it definitely works for any $p$ above a certain bound, but $p > F(n)$ is not enough. $p > F(n+2)$ would work, but I haven't proven the actual bound yet.
Feb 15 at 3:45 vote accept Voile
Feb 15 at 10:34
Feb 14 at 8:38 history answered Peter Taylor CC BY-SA 4.0