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a(0)=3; for n>0, a(n) is the smallest odd prime q not already in the sequence such that the n-th prime p(n) divides a(n-1)+q. If no such prime q exists, the sequence terminates.
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%I #13 Dec 12 2015 12:32:09

%S 3,5,7,13,29,37,41,61,53,131,43,19,277,379,137,239,79,157,331,71

%N a(0)=3; for n>0, a(n) is the smallest odd prime q not already in the sequence such that the n-th prime p(n) divides a(n-1)+q. If no such prime q exists, the sequence terminates.

%C a(20) does not exist, so the sequence terminates. A134204 is a similar sequence for which the termination question is unresolved.

%e After a(3)=13, to find a(4) we look for an odd prime q such that the fourth prime, 7, divides 13+q, and q=29 works, since 7 divides 13+29 = 42.

%e After a(19)=71 we look for an odd prime q such that p(20)=71 divides 71+q. The only candidate is q=71. Since 71 is already in the sequence, the sequence terminates.

%Y Cf. A134204, A224222.

%K nonn,fini,full

%O 0,1

%A _Daniel Drucker_ and _N. J. A. Sloane_, Apr 05 2013