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A127461
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a(0)=1. a(n) = number of earlier terms a(k), 0<=k<=n-1, such that (k+a(k)) divides n.
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2
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1, 1, 2, 1, 4, 1, 3, 1, 6, 2, 2, 2, 6, 2, 3, 2, 6, 3, 5, 1, 6, 1, 5, 2, 8, 2, 4, 4, 5, 1, 6, 2, 7, 4, 4, 1, 10, 2, 3, 4, 8, 2, 4, 3, 7, 3, 6, 1, 11, 1, 4, 4, 7, 1, 9, 3, 7, 1, 4, 3, 11, 1, 6, 4, 7, 2, 8, 3, 7, 2, 4, 4, 12, 1, 6, 5, 5, 2, 7, 2, 10, 6, 3, 1, 9, 5, 4, 2, 9, 2, 11, 3, 8, 3, 3, 1, 14, 3, 3, 5, 10, 3
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OFFSET
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0,3
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LINKS
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EXAMPLE
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(a(0)+0) divides 6; (a(1)+1) divides 6; and (a(5)+5) divides 6. These 3 cases are the only cases where (a(k)+k) divides 6, for 0<=k<=5. So a(6)=3.
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MATHEMATICA
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f[l_List] := Block[{n = Length[l]}, Append[l, Count[Table[Mod[n, k - 1 + l[[k]]], {k, n}], 0]]]; Nest[f, {1}, 101] (* Ray Chandler, Jan 22 2007 *)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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