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Numbers n such that 1/p(2n+1)*sum(k=1,n,p(2k+1)-p(2k)) >= 1/p(2*n)*sum(k=1,n,p(2k)-p(2k-1)) where p(k) denotes the k-th prime.
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%I #7 Mar 30 2012 18:39:15

%S 1,2,3,4,5,7,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,

%T 28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,

%U 51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74

%N Numbers n such that 1/p(2n+1)*sum(k=1,n,p(2k+1)-p(2k)) >= 1/p(2*n)*sum(k=1,n,p(2k)-p(2k-1)) where p(k) denotes the k-th prime.

%C Conjectured to be finite with last term = 314. Other conjecture log(n)^2*(1/p(2n+1)*sum(k=1,n,p(2k+1)-p(2k)) - 1/p(2*n)*sum(k=1,n,p(2k)-p(2k-1))) -> constant. Weaker : previous formula is bounded.

%K nonn

%O 1,2

%A _Benoit Cloitre_, Feb 25 2003