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

Robert G. Wilson, v rgwv at rgwv.com
Wed Jun 3 19:14:06 CEST 2009

```Seq. Fans,

Here are the partial sums to 64 decimals up to 10^k, k is the first
number in parenthesis and the sum is the second.

{ 1, 0.2115192743764172335600907029478458049886621315192743764172335601}
{ 2, 0.2362385924524670829013613789754682069943385757467958344217207676}
{ 3, 0.2372055599710870423370176800457876035334200633564086869752709814}
{ 4, 0.2372492660182123262653761802065117005961398895904169885978597094}
{ 5, 0.2372516057897405689878427120429396865698609342590217726341897254}
{ 6, 0.2372517642213452514702629520604252299216348196823717517418485243}
{ 7, 0.2372517756682939981077218099169625939398841251892729771263012051}
{ 8, 0.2372517765061919663119729774028233698105539917995612519644924169}
{ 9, 0.2372517765710358159350110748777318158842188919259177572926430824}
{10, 0.2372517765762034510704083224746640344084016190159084662884267692}

Bob.

Hagen von Eitzen wrote:
>>>/ Richard Mathar
>>
>>
>>
>>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
>>
>>
>
>
> The error term can be cut down to about 1/3 its size by summing only
> over k= +-1 mod 6
> and to about 1/5 its size by summing only over the relevant residues
> -1,1, 11,13, 17,19 mod 30.
> This should make c < 0.23725177665, I think; better, but unfortunately
>
> Hagen
>
>
>>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

```