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

[Farideh Firoozbakht]

> Richard Mathar
> snapshots of results:
> prime(900000)  0.23725177594...
> prime(1290000) 0.23725177616...
> ...
> might test whether the 0.23725177 may turn to 0.23725178 later on.


No, 0.23725177 may not turn to 0.23725178.
Because,

prime(100,000,000)     0.237251776574074...
prime(110,000,000)     0.237251776574342...
prime(120,000,000)     0.237251776574562...
prime(130,000,000)     0.237251776574746...

So the constant c is less than
    0.237251776574747 + lim(sum(1/k^2,{k, prime(130,000,001), n}, n ->  
infinity)
  < 0.237251776574747 + 3.72376*10^(-10) < 0.237251776947124

Hence 0.237251776574746 < c < 0.237251776947124 and we conclude that the
first nine terms of the sequence are:  2, 3, 7, 2, 5, 1, 7, 7, 6

Rigards,
Farideh




Quoting Richard Mathar <mathar at strw.leidenuniv.nl>:

>
> The current entry of  0.237251058
> http://research.att.com/~njas/sequences/A160910
> seems to be incorrect in at least three decimal digits.
> I guess it is defined as the sum over all 1/A001359(n)^2+1/A006512(n)^2.
> Gathering twin primes up to prime(1070000) the constant
> is at least 0.23725177606 .
> Someone with some spare computer cycles---not the cycles that spin the
> computer and the earth around the sun in a year but the other,     
> quicker ones---
> might test whether the 0.23725177 may turn to 0.23725178 later on.
>
> Richard Mathar
> snapshots of results:
> prime(900000)  0.23725177594...
> prime(1290000) 0.23725177616...
>
> Digits := 40 ;
> x := 0.0 ;
> for n from 1 do
>         p := ithprime(n) ;
>         if isprime(p+2) then
>                 x := evalf(x+1/p^2+1/(p+2)^2) ;
>         fi;
>         if n mod 10000 = 0 then
>         print(n,x) ;
>         fi;
> od:
>
>
>
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