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A179921 a(n) = prime(n) if n<=3; for n>3, a(n) is the smallest prime >a(n-1), such that the denominator of fraction (a(n-1)-a(n-2))/(a(n)-a(n-1)) did not appear earlier.

%I A179921 #10 Mar 30 2012 18:52:59
%S A179921 2,3,5,7,13,23,31,53,67,79,113,131,151,193,233,271,307,353,379,409,
%T A179921 457,557,613,691,761,809,883,907,1013,1069,1123,1181,1213,1279,1361,
%U A179921 1423,1483,1571,1657,1709,1811,1933,1997,2087,2179,2273,2341,2459
%N A179921 a(n) = prime(n) if n<=3; for n>3, a(n) is the smallest prime >a(n-1), such that the denominator of fraction (a(n-1)-a(n-2))/(a(n)-a(n-1)) did not appear earlier.
%C A179921 Using Dirichlet's theorem on arithmetic progressions, it is easy to prove that the sequence is infinite.  The sequence of the corresponding denominators begins with 2,1,3,5,4,11,7,6,17, ...
%e A179921 The first four terms 2,3,5,13 give three denominators: 2,1,3. Then a(5) is not in {17, 19}, since (13-5)/(17-13) = 2/1, (13-5)/(19-13) = 4/3 and denominators 1 and 3 already appeared earlier. Since (13-5)/(23-13) = 4/5 and 5 is not yet in the denominator sequence, a(5) = 23.
%Y A179921 Cf. A168253, A178942, A179210, A179234, A179256, A179328.
%K A179921 nonn
%O A179921 1,1
%A A179921 _Vladimir Shevelev_, Jan 12 2011
%E A179921 Edited by _Alois P. Heinz_, Jan 12 2011