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A097500
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Consider the succession of single digits of A008585 (multiples of 3): 3 6 9 1 2 1 5 1 8 2 1 2 4 2 7 3 0 .... This sequence gives the lexicographically earliest derangement of A001651 (non-multiples of 3) that produces the same succession of digits.
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2
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3691, 2, 1, 5, 182, 124, 27303336394, 245, 4, 8, 515, 457, 606366697, 275, 7, 88, 184, 879093969910, 2105, 10, 811, 11, 14, 1171, 20, 1231, 26, 1291, 32, 13, 5138, 1411, 44, 1471, 50, 1531, 56, 1591, 62, 16, 5168, 17, 1174, 1771, 80, 1831, 86, 1891, 92, 19, 5198, 20120, 4207, 2102, 1321, 62192, 22
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OFFSET
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1,1
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COMMENTS
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Derangement here means a(n) != A008585(n) for all n.
Original name: "Write each non-multiple of 3 integer >0 on a single label. Put the labels in numerical order to form an infinite sequence L. Now consider the succession of single digits of A008585 (multiples of 3): 3,6,9,1,2,1,5,1,8,2,1,2,4,2,7,3,0,3,3,3,6,3,9,4,2,4,5,4,8,5,1,5,4,5,7,6,0... The sequence S gives a rearrangement of the labels that reproduces the same succession of digits, subject to the constraint that the smallest label must be used that does not lead to a contradiction."
This could be roughly rephrased like this: "Rewrite in the most economical way the 'multiples-of-3 pattern' using only non-multiples of 3. Do not use any non-multiple of 3 twice."
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LINKS
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EXAMPLE
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We must begin with "3,6,9,12,..." and we cannot represent "3" with 3, 36, or 369, because they are all multiples of 3. So the first possibility for a(1) is 3691.
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MATHEMATICA
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f[lst_List, k_] := Block[{L = lst, g, a = {}, m = 0}, g[] := {Set[m, First@ FromDigits@ Append[IntegerDigits@ m, First@ #]], Set[L, Last@ #]} &@ TakeDrop[L, 1]; Do[g[]; While[Or[Mod[m, 3] == 0, First@ L == 0, MemberQ[a, m]], g[]]; AppendTo[a, m]; m = 0, {k}]; a]; f[Flatten@ Map[IntegerDigits, Array[3 # &, {120}]], 57] (* Michael De Vlieger, Nov 30 2015, Version 10.2 *)
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CROSSREFS
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KEYWORD
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base,easy,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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