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A179479
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a(n) is the smallest prime q > a(n-1) such that, for the previous prime p and the following prime r, the fraction (q-p)/(r-q) has denominator 2, for odd n and 1 for even n (or 0, if such a prime does not exist).
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0
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3, 5, 7, 11, 13, 17, 19, 29, 43, 53, 67, 71, 79, 97
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OFFSET
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1,1
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COMMENTS
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Conjecture: a(n) > 0 for all n.
Since, as it is accepted in the OEIS, we consider the uncancelled fractions, then, by the condition, for even n, we have (r-q)|(q-p).
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LINKS
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EXAMPLE
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If n=1, then denominator should be 2. Thus a(1)=3, since (3-2)/(5-3)=1/2. If n=2, then denominator should be 1. Thus a(2)=5, since (5-3)/(7-5)=1/1, etc.
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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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