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%I #18 Aug 01 2019 09:30:38
%S 2,3,2,3,5,3,5,2,5,2,3,7,3,7,2,5,11,5,11,2,3,13,17,13,17,3,2,19,2,19,
%T 2,5,2,5,2,5,2,5,23,3,23,2,23,29,3,29,2,7,2,7,2,3,2,3,31,3,31,2,31,2,
%U 3,2,11,2
%N Primes arising in A=A256189 as gcd(A(n-2), A(n)), n > 3.
%C Conjecture: the sequence contains all primes such that the first appearances of them are in the natural order.
%C Provided all primes appear, they must appear in the natural order. Suppose otherwise: For A=A256189 and some n and primes p < q, q divides a = A(n) but not A(k) for k < n and p does not divide A(k) for k <= n. But then A(n) should be pa/q. - _Danny Rorabaugh_, Mar 31 2015
%H Peter J. C. Moses, <a href="/A256366/b256366.txt">Table of n, a(n) for n = 2..1001</a>
%Y Cf. A256189.
%K nonn
%O 2,1
%A _Vladimir Shevelev_, Mar 26 2015
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