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Functional substitution on {1,2,3}.
0

%I #7 Mar 12 2014 16:36:47

%S 1,2,3,3,1,1,2,2,1,1,1,1,3,2,1,2,1,1,1,2,1,3,1,1,2,2,1,1,1,1,3,2,1,2,

%T 1,1,1,2,1,3,1,1,2,2,1,1,1,1,3,2,1,2,1,1,1,2,1,3,1,1,2,2,1,1,1,1,3,2,

%U 1,2,1,1,1,2,1,3,1,1,2,2,1,1,1,1,3,2,1,2,1,1,1,2,1,3,1,1,2,2,1,1,1,1,3,2,1

%N Functional substitution on {1,2,3}.

%C Based on a prime digits function modulo 10, reduced here to modulo 3.

%C Appears to be 12-periodic for n>3.

%F ar={1, 2, 3} a(n)=b(m, i) = 1 + Mod[ar[[i]]*(1 + Mod[m, 3]), 3]

%t ar = {1, 2, 3} f[n_] := 1 + Mod[ar*(1 + Mod[n, 3]), 3] br = NestList[f, ar, Floor[200/3]] a = Flatten[br]

%K nonn,uned

%O 1,2

%A _Roger L. Bagula_, Apr 12 2005