%I #26 Jul 19 2015 08:56:48
%S 0,-1,-1,-2,2,1,1,1,-3,-4,-2,-3,1,1,-1,-2,6,6,6,6,3,2,4,3,3,3,1,1,5,4,
%T 4,4,4,3,3,2,3,3,3,2,8,7,9,9,6,6,8,8,3,3,1,0,4,3,1,1,-3,-3,-1,-1,3,3,
%U 3,2,2,1,3,3,0,-1,1,1,7,7,5,4,4,4,4,4,4,3
%N a(n)= Sum_{2 < prime p <= n} c_p - Sum_{n < prime p < 2*n} c_p, where 2^c_p is the greatest power of 2 dividing p-1.
%C It is known that, for n>10, pi(2*n) < 2*pi(n), where pi(n) is the number of primes not exceeding n (A000720). Thus, for n>10, in the interval (1,n] we have more primes than in the interval (n,2*n).
%C In connection with this, it is natural to conjecture that there exists a number N such that a(n)>0 for all n >= N.
%H Peter J. C. Moses, <a href="/A259922/b259922.txt">Table of n, a(n) for n = 1..2000</a>
%t Map[Total[Flatten[Map[IntegerExponent[Select[#,PrimeQ]-1,2]&,{Range[3,#],Range[#+1,2#-1]}]{1,-1}]]&,Range[50]]
%Y Cf. A007814, A060208, A259788, A259897.
%K sign
%O 1,4
%A _Vladimir Shevelev_, Jul 09 2015
%E More terms from _Peter J. C. Moses_, Jul 09 2015