[seqfan] Re: Sum n/prime(n)^2

[Charles Greathouse]

My current estimate of the constant is 1.1490642... based on a summation of
the primes up to 10^10
1.10463086777...
and the integral from pi(10^10) + 1/2 to infinity of x/ali(x)^2 where
ali(x) is the inverse logarithmic integral
0.04443338335...

I'm fairly confident in the first seven decimals but I'll try to extend the
sum before making any claims of correctness. Still, 1.1096 seems to be
wrong from the first digit after the decimal.


Charles Greathouse
Analyst/Programmer
Case Western Reserve University

On Thu, Dec 4, 2014 at 3:07 PM, Charles Greathouse <
charles.greathouse at case.edu> wrote:

> Richard Mathar is the expert here, I think. I tried to get some decimals
> but I can't convince myself that I have any correct. For example, I think
> (integrating n over the square of the inverse logarithmic integral) that
> the tail error in CAMI's 1.1012478923014213 is at least 0.04, and it may be
> quite a bit more. Maybe using the Cipolla series for prime(n) would help,
> but then you have a whole bunch of problems no easier than A115563 and the
> remaining terms still won't be that small.
>
> Charles Greathouse
> Analyst/Programmer
> Case Western Reserve University
>
> On Thu, Dec 4, 2014 at 12:25 PM, Neil Sloane <njasloane at gmail.com> wrote:
>
>> Someone added a remark that
>> A097906 = A115563, which is nonsense. The
>> latter is Sum 1/(n*log(n)^2) and we have it to many places, whereas the
>> former is  Sum n/prime(n)^2 and we have 3 places.
>>
>> Can anyone get Sum n/prime(n)^2 to more (any!?) places?
>>
>> _______________________________________________
>>
>> Seqfan Mailing list - http://list.seqfan.eu/
>>
>
>