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A094840
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a(1) = 1; for n > 1, a(n) = curling number of (b(1),...,b(n-1)), where b() = Linus sequence A006345.
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2
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1, 1, 1, 1, 2, 1, 2, 1, 1, 2, 2, 1, 2, 2, 2, 1, 2, 1, 1, 2, 3, 1, 2, 1, 1, 2, 2, 1, 2, 1, 2, 3, 1, 1, 2, 2, 2, 1, 1, 2, 2, 1, 2, 2, 2, 1, 2, 1, 1, 2, 3, 1, 2, 1, 1, 2, 2, 1, 2, 1, 2, 2, 3, 1, 2, 1, 1, 2, 2, 1, 2, 1, 2, 3, 1, 1, 2, 2, 2, 1, 1, 2, 2, 1, 2, 2, 2, 1, 2, 1, 1, 2, 3, 1, 2, 1, 1, 2, 2, 1
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OFFSET
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1,5
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COMMENTS
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The curling number of a finite string S = (s(1),...,s(n)) is the largest integer k such that S can be written as xy^k for strings x and y (where y has positive length).
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LINKS
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F. J. van de Bult, D. C. Gijswijt, J. P. Linderman, N. J. A. Sloane and Allan Wilks, A Slow-Growing Sequence Defined by an Unusual Recurrence [pdf, ps].
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MAPLE
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fd := fopen("b006345.txt", READ) : a006345 := [] : bf := fscanf(fd, "%d %d") : while nops(bf) <> 0 do a006345 := [op(a006345), op(2, bf) ] ; bf := fscanf(fd, "%d %d") ; od: curlN := proc(L) local a, k, klen, Llen, y ; a := 1 ; Llen := nops(L) ; for klen from 1 to floor(Llen/2) do y := op(Llen-klen+1..Llen, L) ; for k from 2 to floor(Llen/klen) do if op(Llen-k*klen+1..Llen-(k-1)*klen, L) = y then if k > a then a := k ; fi ; else break ; fi ; od: od: RETURN(a) ; end: A094840 := proc(n) global a006345 ; if n = 1 then 1; else curlN( [op(1..n-1, a006345)] ) ; fi ; end: for n from 1 to 100 do printf("%d, ", A094840(n)) ; od: # R. J. Mathar, Dec 07 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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STATUS
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approved
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