|
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
Contribution from Farideh Firoozbakht (mymontain(AT)yahoo.com), 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)
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 first 2*10^9 primes), we
conclude that the sum up to prime(2000,000,000)is less than s2+1500*(s2-s1).
But since s2-s1 < 10^(-15) so 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)
| 