Compare Viggo Brun's constant (1/3 + 1/5) + (1/5 + 1/7) + (1/11 + 1/13) + (1/17 + 1/19) + (1/29 + 1/31) + ... (see A065421, A005597).
It appears that c = Sum 1/A001359(n)^2 + 1/A006512(n)^2.  R. J. Mathar, May 30 2009
0.237251776574746 < c < 0.237251776947124.  Farideh Firoozbakht, May 31 2009
c < 0.2725177657771.  Hagen von Eitzen, Jun 03 2009
From Farideh Firoozbakht, Jun 01 2009: (Start)
We can show that a(9)=6, a(10)=5, and a(11) is in the set {7, 8, 9}.
Proof: s1 = 0.237251776576249072... is the sum up to prime(499,000,000) and s2 = 0.237251776576250009... is the sum up to prime(500,000,000).
By using the fact that number of twin primes between the first 10^6*n primes and the first 10^6*(n+1) primes is decreasing (up to the first 2*10^9 primes), we conclude that the sum up to prime(2,000,000,000) is less than s2 + 1500*(s2s1).
But since s2s1 < 10^(15), the sum up to prime(2*10^9) is less than s2 + 1.5*10^(12) = 0.237251776576250009... + 1.5*10^(12) = 0.237251776577550009... .
Hence the constant c is less than
0.237251776577550009... + lim(sum(1/k^2,{k, prime(2,000,000,001), n}, n > infinity)
< 0.237251776577550009... + 2.12514*10^(11)
< 0.237251776598801409.
So we have 0.237251776576250009 < c < 0.237251776598801409, hence a(9)=6, a(10)=5, and a(11) is in the set {7, 8, 9}.
I guess that a(11)=7. (End)
