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%I #20 Mar 30 2012 18:52:54
%S 5,11,17,19,29,41,47,67,73,97,101,359,367,379,383,389,397,419,421,449,
%T 467,547,613,631,647,683,691,733,769,797,811,929,941,1021,1087,1153,
%U 1181,1193,1249,1709,1721,1747,1847,1889,2017,2153,2357
%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 equal to A006843(n) (or 0, if such a prime does not exist).
%C Conjecture: a(n) > 0 for all n.
%e For n = 1..3, A006843(n) = 1, and p,q,r have to obey the condition
%e r-q | q-p. Thus a(1) = 5, a(2) = 11, a(3) = 17.
%Y Cf. A006843, A168253, A179210, A179234, A179256, A001223
%K nonn
%O 1,1
%A _Vladimir Shevelev_, Jan 06 2011
%E More terms from _Alois P. Heinz_, Jan 06 2011