%I #14 Apr 17 2013 14:16:48
%S 3,271,5195969,1801241230056600467
%N a(n) = largest prime p such that Sum_{primes q = 2, ..., p} 1/q does not exceed n.
%C Since Sum_{prime q} 1/q diverges, the sequence is infinite.
%C In fact, by the Prime Number Theorem Prime(k) ~ k log k as k -> infinity, and by integration Sum_{k <= n} 1/(k log k) ~ log log n, so a(n) ~ Prime(Floor(e^e^n)).
%C a(4) = A000040(A046024(4)-1) = Prime[43922730588128389], but Mathematica 7.0.0 cannot compute this prime on a Mac computer running OS X.
%C Instead, using a(4) = largest prime < A016088(4) = 1801241230056600523, Mathematica's PrimeQ function finds that a(4) = 1801241230056600467.
%C See A016088 for other relevant comments, references, links, and programs.
%F a(n) = A000040(A046024(n)-1) = largest prime < A016088(n).
%F a(n) ~ Prime(Floor(e^e^n)) = A000040(A096232(n)) as n -> infinity.
%e a(1) = 3 because 1/2 + 1/3 < 1 < 1/2 + 1/3 + 1/5.
%Y Cf. A016088, A046024, A024451, A096232.
%K nonn,hard,more
%O 1,1
%A _Jonathan Sondow_, Apr 16 2013