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Oct 7, 2020 at 17:07 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Sep 11, 2020 at 13:14 comment added Phil Ramsden p = NextPrime[10^100]; Mod[(Algebra`MatrixPowerMod[{{0, 1}, {1, 1}}, p -2 - JacobiSymbol[p, 5], p].{1, 1})[[2]], p] returns 0, very quickly; exactly what I was trying to do. That's Fibonacci; adapting to other recurrence relations will be a breeze. Thanks again.
Sep 11, 2020 at 12:29 comment added Phil Ramsden Oh, wow--I think that's almost certainly going to do the trick. Thanks so much!
Sep 10, 2020 at 11:17 comment added J. M.'s missing motivation There isn't a built-in function, but you might be interested in the undocumented function Algebra`MatrixPowerMod[], which might help in your implementation.
Sep 7, 2020 at 15:20 answer added Daniel Huber timeline score: 1
Sep 7, 2020 at 14:27 comment added Daniel Lichtblau A concrete example, with input and desired output, would be useful here.
Sep 7, 2020 at 13:01 review First posts
Sep 7, 2020 at 15:37
Sep 7, 2020 at 12:59 history asked Phil Ramsden CC BY-SA 4.0