[seqfan] Re: more digits of sum over squared inverse twin primes

[Hagen von EItzen]

>  submitted the following comment.
>
>   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.
>   
Again, your error term sum_{k > N} 1/k^2 < 1/(N-1) can be slightliy 
improved without using additional primes:
Modulo 2*3*5*...*23, only a proportion of 405/5681 of residue classes 
can contain twin primes.
By summing only over these classes, the error term can be redueced to 
405/5681 * 1/(N-2*3*...*23)
If I'm not wrong, this makes c < 0.2725177657771

Hagen
>   ---Farideh
>