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A067992
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a(0)=1 and, for n > 0, a(n) is the smallest positive integer such that the ratios min(a(k)/a(k-1), a(k-1)/a(k)) for 0 < k <= n are all distinct.
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11
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1, 1, 2, 3, 1, 4, 3, 5, 1, 6, 5, 2, 7, 1, 8, 3, 7, 4, 5, 7, 6, 11, 1, 9, 2, 11, 3, 10, 1, 12, 5, 8, 7, 9, 4, 11, 5, 9, 8, 11, 7, 10, 9, 11, 10, 13, 1, 14, 3, 13, 2, 15, 1, 16, 3, 17, 1, 18, 5, 13, 4, 15, 7, 12, 11, 13, 6, 17, 2, 19, 1, 20, 3, 19, 4, 17, 5, 14, 9, 13, 7, 16, 5, 19, 6, 23, 1, 21, 2
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OFFSET
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0,3
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COMMENTS
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Every positive rational number appears exactly once as the ratio of adjacent terms (in either order). Conjecture: adjacent terms are always relatively prime. - Franklin T. Adams-Watters, Sep 13 2006
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LINKS
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Neil Calkin and Herbert S. Wilf, Recounting the rationals, The American Mathematical Monthly, Vol. 107, No. 4 (2000), 360-363.
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FORMULA
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a(6)=3, since 1/4 and 2/4 = 1/2 have already occurred as ratios of adjacent terms.
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EXAMPLE
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The sequence of all rational numbers between 0 and 1 obtained by taking ratios of sorted consecutive terms begins: 1/2, 2/3, 1/3, 1/4, 3/4, 3/5, 1/5, 1/6, 5/6, 2/5, 2/7, 1/7, 1/8, 3/8, 3/7, 4/7, 4/5, 5/7, 6/7. - Gus Wiseman, Aug 30 2018
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MATHEMATICA
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Nest[Function[seq, Append[seq, NestWhile[#+1&, 1, MemberQ[Divide@@@Sort/@Partition[seq, 2, 1], Min[Last[seq]/#, #/Last[seq]]]&]]], {1}, 100] (* Gus Wiseman, Aug 30 2018 *)
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PROG
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(PARI) seen = Set([]); other(p) = for (v=1, oo, my (r = min(v, p)/max(v, p)); if (!set search(seen, r), seen = set union(seen, Set([r])); return (v)))
for (n=0, 88, v = if (n==0, 1, other(v)); print1 (v ", ")) \\ Rémy Sigrist, Aug 07 2017
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CROSSREFS
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See A066720 for a somewhat similar sequence.
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KEYWORD
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AUTHOR
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STATUS
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approved
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