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A097488
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Write the positive multiples of 3 on labels in numerical order, forming an infinite sequence L. Now consider the succession of single digits of L: 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 .... This sequence is a derangement of L that produces the same succession of digits, subject to the constraint that the smallest unused label must be used that does not lead to a contradiction.
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2
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36, 9, 12, 15, 18, 21, 24, 27, 30, 3, 336, 39, 42, 45, 48, 51, 54, 57, 60, 6, 36, 669, 72, 75, 79, 81, 84, 87, 90, 93, 96, 99, 102, 105, 108, 111, 114, 117, 120, 123, 126, 129, 132, 135, 138, 141, 144, 147, 150, 153, 156, 159, 162, 165
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OFFSET
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1,1
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COMMENTS
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This could be roughly rephrased like this: Rewrite in the most economical way the "multiples-of-3 pattern" using only multiples of 3, but rearranged. No term in the sequence can appear more than once.
Derangement here means the n-th element of L is not the n-th element of this sequence, so a(n) != 3n.
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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 have a(1) = 3, so the first possibility is the label "36". The next term must be the smallest available label not leading to a contradiction, thus "9". The next one will be "12", etc. After the label "30" the smallest available label is "3". After this "3" we cannot have a(11) = 33 -- we thus take the smallest available label which is "336". No label is allowed to start with a leading zero. - Eric Angelini, Aug 12 2008
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CROSSREFS
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Cf. A097481 for this sequence with multiples of 2.
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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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