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A105244
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Functional substitution on {1,2,3}.
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0
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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, 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, 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
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OFFSET
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1,2
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COMMENTS
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Based on a prime digits function modulo 10, reduced here to modulo 3.
Appears to be 12-periodic for n>3.
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LINKS
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FORMULA
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ar={1, 2, 3} a(n)=b(m, i) = 1 + Mod[ar[[i]]*(1 + Mod[m, 3]), 3]
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MATHEMATICA
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ar = {1, 2, 3} f[n_] := 1 + Mod[ar*(1 + Mod[n, 3]), 3] br = NestList[f, ar, Floor[200/3]] a = Flatten[br]
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CROSSREFS
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KEYWORD
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nonn,uned
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AUTHOR
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STATUS
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approved
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