sum of 1/A007504(n)

[Max Alekseyev]

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
>