sum of 1/A007504(n)

[Jack Brennen]

This updated value of A:

   A ~ 1.628166907946785888

is the value of A such that the integral of

    -1/(log(x)*eint1(-2*log(x))) dx

from A to 2^34 is equal to the sum in question over the
primes up to 2^34.  In other words, it's a "fudge factor"
number chosen to fit the data... but let me explain another
way...

For each integer 24 <= N <= 33, the finite integral of
the term above from 2^N to 2^(N+1) lies within +/-0.01%
of the partial sum of the series over the same range.
(In fact, often much closer than that, but we'll take
+/-0.01% as a reasonable error estimate.)

The infinite integral from 2^34 to +infinity of the term
above is:

   0.00000000011399...

So I'm pretty confident in saying that the infinite sum
of our series over the primes > 2^34 is equal to

   0.00000000011399 +/- 1 in the least significant digit

And the finite sum up to 2^34 is:

   1.0234763238061374...

Giving the infinite sum as:

   1.02347632392013...  with some uncertainty in the last digit.


So there's the same answer based on the actual sum over the
primes < 2^34, plus an experimentally accurate integral
approximation for the rest of the infinite sum.

Max Alekseyev wrote:
> Jack,
> 
> Where this A comes from? Are all its listed digits correct?
> It is important as the result depends on the (numerical) value of A.
> 
> Max
> 
> On 5/14/07, Jack Brennen <jb at brennen.net> wrote:
>> The sum in question appears to be asymptotic to:
>>
>>   integral from A to N of
>>
>>    -1/(log(x)*eint1(-2*log(x)))  dx
>>
>>   where A ~ 1.6281669079467875
>>
>> And the integral appears to converge (as N -> infinity) to:
>>
>>    1.0234763239201...
>>
>>
>> (All of which assumes that I trust the PARI/GP implementation of numeric
>> integration...)
>>
>>
>>   Jack
>>
> 
>