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A125147
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a(0)=1; for n >= 1, a(n) is the smallest positive integer not occurring earlier in the sequence such that Sum_{k=0..n} a(k) is a multiple of n.
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1
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1, 2, 3, 6, 4, 9, 5, 12, 14, 7, 17, 8, 20, 22, 10, 25, 11, 28, 30, 13, 33, 35, 15, 38, 16, 41, 43, 18, 46, 19, 49, 51, 21, 54, 56, 23, 59, 24, 62, 64, 26, 67, 27, 70, 72, 29, 75, 77, 31, 80, 32, 83, 85, 34, 88, 90, 36, 93, 37, 96, 98, 39, 101, 40, 104, 106, 42, 109, 111, 44
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OFFSET
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0,2
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COMMENTS
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Sequence A019444 is a similar sequence, but it has an offset of 1 and a(1) =1 instead.
This sequence is a permutation of the positive integers. This can be proved using the theorem-prover Walnut. - Jeffrey Shallit, Aug 12 2023
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LINKS
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EXAMPLE
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a(5) = 9 because 9 is the smallest positive integer m which does not occur earlier in the sequence and which is such that 5 divides m + Sum_{k=0..4} a(k).
So Sum_{k=0..5} a(k) = 25, which is divisible by 5.
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MATHEMATICA
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f[l_List] := Block[{k = 1, n = Length[l], s = Plus @@ l}, While[MemberQ[l, k] || Mod[s + k, n] > 0, k++ ]; Append[l, k]]; Nest[f, {1}, 70] (* Ray Chandler, Jan 23 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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