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A224762
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Define a sequence of rationals by S(1)=1; for n>=1, write S(1),...,S(n) as XY^k, Y nonempty, where the fractional exponent k is maximized, and set S(n+1)=k; sequence gives numerators of S(1), S(2), ...
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4
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1, 1, 2, 1, 3, 1, 3, 2, 6, 1, 5, 1, 3, 4, 1, 4, 3, 5, 8, 1, 6, 13, 1, 4, 5, 8, 9, 1, 6, 5, 6, 3, 16, 1, 7, 1, 3, 6, 8, 14, 1, 6, 5, 16, 1, 5, 4, 24, 1, 5, 3, 15, 1, 5, 3, 7, 1, 5, 3, 7, 2, 54, 1, 7, 31, 1, 4, 21, 1, 4, 5, 1, 4, 5, 2, 15, 25, 1, 7, 17, 1, 4, 11, 1, 4, 5, 5, 30, 1, 6, 25, 15, 17, 1, 6, 7, 1, 4, 15, 1, 4, 5, 19
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OFFSET
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1,3
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COMMENTS
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k is the "fractional curling number" of S(1),...,S(n). The infinite sequence S(1), S(2), ... is a fractional analog of Gijswijt's sequence A090822.
For the first 1000 terms, 1 <= S(n) <= 2. Is this always true?
The fractional curling number k of S = (S(1), S(2), ..., S(n)) is defined as follows. Write S = X Y Y ... Y Y' where X may be empty, Y is nonempty, there are say i copies of Y, and Y' is a prefix of Y. There may be many ways to do this. Choose the version in which the ratio k = (i|Y|+|Y'|)/|Y| is maximized; this k is the fractional curling number of S.
For example, if S = (S(1), ..., S(6)) = (1, 1, 2, 1, 3/2, 1), the best choice is to take X = 1,1,2, Y = 1,3/2, Y' = 1, giving k = (2+1)/2 = 3/2 = S(7).
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LINKS
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EXAMPLE
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The sequence S(1), S(2), ... begins 1, 1, 2, 1, 3/2, 1, 3/2, 2, 6/5, 1, 5/4, 1, 3/2, 4/3, 1, 4/3, 3/2, 5/4, 8/7, 1, 6/5, 13/12, 1, 4/3, 5/4, 8/7, 9/7, 1, 6/5, 5/4, 6/5, 3/2, 16/15, 1, 7/6, 1, 3/2, 6/5, 8/7, 14/13, 1, 6/5, 5/4, 16/13, 1, 5/4, 4/3, 24/23, 1, 5/4, 3/2, 15/14, 1, 5/4, 3/2, 7/4, ...
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MAPLE
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See link.
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CROSSREFS
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KEYWORD
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nonn,frac
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AUTHOR
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Conference dinner party, Workshop on Challenges in Combinatorics on Words, Fields Institute, Toronto, Apr 22 2013, entered by N. J. A. Sloane
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STATUS
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approved
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