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%I #9 Feb 22 2019 02:04:14
%S 3,5,7,11,13,17,19,29,43,53,67,71,79,97
%N 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).
%C Conjecture: a(n) > 0 for all n.
%C 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).
%e 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.
%Y Cf. A168253, A179210, A179234, A179240, A179256, A179328.
%K nonn
%O 1,1
%A _Vladimir Shevelev_, Jan 08 2011
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