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%I #10 Feb 25 2023 08:48:58
%S 1,1,3,1,24,3,48,1,9,24,10,3,12,48,24,1,144,9,180,24,48,10,22,3,120,
%T 12,27,48,840,24,320,1,30,144,48,9,36,180,12,24,280,48,308,10,72,22,
%U 46,3,336,120,144,12,936,27,120,48,180,840,29,24,60,320,144,1,24,30,1122,144
%N Pisano period of A002605 modulo n.
%C a(79)=6240. [_John W. Layman_, Aug 10 2010]
%H Michael De Vlieger, <a href="/A175289/b175289.txt">Table of n, a(n) for n = 1..200</a>
%e Reading 0, 1, 2, 6, 16, 44, 120, 328, 896, 2448,.. modulo 12 gives 0, 1, 2, 6, 4, 8, 0, 4, 8, 0, 4, 8 ,.. with period length a(n=12)= 3.
%t a={1};For[n=2,n<=80,n++,{x={{0,1}}; t={1,1}; While[ !MemberQ[x,t], {xl = x[[ -1]]; AppendTo[x,t]; t={Mod[2*(t[[1]]+xl[[1]]),n], Mod[2*(t[[2]] + xl[[2]]),n]};}]; p = Flatten[Position[x,t]][[1]]; AppendTo[a, Length[x] - p+1];}]; Print[a]; (* _John W. Layman_, Aug 10 2010 *)
%Y Cf. A001175, A046738, A175181, A175182, A175183, A175184, A175185, A175286.
%K nonn
%O 1,3
%A _R. J. Mathar_, Mar 24 2010
%E Terms beyond a(28)=48 from _John W. Layman_, Aug 10 2010
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