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A094206
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a(n) = number of consistent orderings of 1..n based only on factorization.
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0
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1, 1, 1, 2, 3, 5, 9, 25, 66, 158, 424, 1048, 2445, 5736, 17069, 88674, 241698, 648786, 1600339, 5379356
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OFFSET
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1,4
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COMMENTS
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Take a set of objects [n] indexed by the positive integers which multiply so that [a] [b] = [ab] (which automatically makes them commute, associate, obey gcd([a],[b])=[gcd(a,b)] etc.) and also partially define a consistent ordering relation < to obey two rules:
Rule 1: p<q ==> [p] < [q], for primes p,q and Rule 2: A<B, C<D ==> AC < BD, for any objects A, B, C, D.
Rule 2 captures certain intuitive requirements for ordering products - for example specializing A=[1] and C=D captures the idea that "multiples are larger", etc. Sequence gives number of ways of consistently ordering [1]..[n].
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LINKS
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EXAMPLE
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Up to n=3 there's only one way: [1], [1][2], [1][2][3], but then for n=4=2^2 the rules do not say whether [3]<[4] or [4]<[3], although they do say that [2]<[4], so we get two orderings [1][2][3][4], [1][2][4][3].
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CROSSREFS
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KEYWORD
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nonn,nice,more
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AUTHOR
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STATUS
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approved
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