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A105244
Functional substitution on {1,2,3}.
0
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
OFFSET
1,2
COMMENTS
Based on a prime digits function modulo 10, reduced here to modulo 3.
Appears to be 12-periodic for n>3.
FORMULA
ar={1, 2, 3} a(n)=b(m, i) = 1 + Mod[ar[[i]]*(1 + Mod[m, 3]), 3]
MATHEMATICA
ar = {1, 2, 3} f[n_] := 1 + Mod[ar*(1 + Mod[n, 3]), 3] br = NestList[f, ar, Floor[200/3]] a = Flatten[br]
CROSSREFS
Sequence in context: A005135 A290003 A139460 * A257451 A209007 A145854
KEYWORD
nonn,uned
AUTHOR
Roger L. Bagula, Apr 12 2005
STATUS
approved