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%I #31 Nov 16 2023 15:42:38
%S 1,3,8,6,2,24,6,12,8,6,24,24,12,6,8,24,36,24,40,6,24,24,22,24,10,12,8,
%T 6,116,24,64,48,24,36,6,24,76,120,24,12,28,24,88,24,8,66,96,24,42,30,
%U 72,12,52,24,24,12,40,348,58,24,124,192,24,96,12,24,66,36,88,6,70,24,148
%N Pisano period of the 5-Fibonacci numbers, A052918 preceded by 0.
%C Period of the sequence defined by reading A052918 modulo n.
%H Vincenzo Librandi, <a href="/A175184/b175184.txt">Table of n, a(n) for n = 1..1000</a>
%H Sergio Falcon and Ángel Plaza, <a href="http://dx.doi.org/10.1016/j.chaos.2008.02.014">k-Fibonacci sequences modulo m</a>, Chaos, Solit. Fractals 41 (2009), 497-504.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/PisanoPeriod.html">Pisano period</a>.
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Pisano_period">Pisano period</a>.
%p F := proc(k,n) option remember; if n <= 1 then n; else k*procname(k,n-1)+procname(k,n-2) ; end if; end proc:
%p Pper := proc(k,m) local cha, zer,n,fmodm ; cha := [] ; zer := [] ; for n from 0 do fmodm := F(k,n) mod m ; cha := [op(cha),fmodm] ; if fmodm = 0 then zer := [op(zer),n] ; end if; if nops(zer) = 5 then break; end if; end do ; if [op(1..zer[2],cha) ] = [ op(zer[2]+1..zer[3],cha) ] and [op(1..zer[2],cha)] = [ op(zer[3]+1..zer[4],cha) ] and [op(1..zer[2],cha)] = [ op(zer[4]+1..zer[5],cha) ] then return zer[2] ; elif [op(1..zer[3],cha) ] = [ op(zer[3]+1..zer[5],cha) ] then return zer[3] ; else return zer[5] ; end if; end proc:
%p k := 5 ; seq( Pper(k,m),m=1..80) ;
%t Table[s = t = Mod[{0, 1}, n]; cnt = 1; While[tmp = Mod[5*t[[2]] + t[[1]], n]; t[[1]] = t[[2]]; t[[2]] = tmp; s!= t, cnt++]; cnt, {n, 100}] (* _Vincenzo Librandi_, Dec 20 2012, after _T. D. Noe_ *)
%Y Cf. A001175, A052918, A175181, A175182, A175183, A175185.
%K nonn,easy
%O 1,2
%A _R. J. Mathar_, Mar 01 2010
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