cp's OEIS Frontend

A217884 Let c(m)=prime(m), m=1,2,3,4. For m>=5, suppose that c(m)/e is in the interval [c(k),c(k+1)). Then let c(m+1)=e*c(k+1) if e*c(k+1) < prime(m+1), and otherwise let c(m+1) = prime(m+1). Then a(n) is the n-th prime in {c(m)}.

%I A217884 #27 Nov 02 2013 17:50:03
%S A217884 2,3,5,7,13,19,31,43,47,67,71,73,79,83,103,107,109,113,137,139,157,
%T A217884 163,173,179,181,197,211,229,239,241,251,257,269,271,283,313,317,337,
%U A217884 347,353,359,367,397,401,409,419
%N A217884 Let c(m)=prime(m), m=1,2,3,4. For m>=5, suppose that c(m)/e is in the interval [c(k),c(k+1)). Then let c(m+1)=e*c(k+1) if e*c(k+1) < prime(m+1), and otherwise let c(m+1) = prime(m+1). Then a(n) is the n-th prime in {c(m)}.
%C A217884 The preliminary sequence begins 2,3,5,7,3*e,13,5*e,19,7*e,3*e^2,31,...
%C A217884 with terms of the form p*e^k, where p is prime, k>=0.
%F A217884 If A(n)is the number of terms not exceeding n, then heuristically A(n)~pi(n). Practically, an approximation is given by formula A(n) ~ n/log(n*log(n)).
%Y A217884 Cf. A217689, A217691.
%K A217884 nonn
%O A217884 1,1
%A A217884 _Vladimir Shevelev_, Oct 14 2012
%E A217884 Terms a(1)-a(20) confirmed and terms a(21)-a(46) added by _John W. Layman_, Oct 24 2012