# [seqfan] Re: Another BBP formula for ln(2) ?

Robert Israel israel at math.ubc.ca
Sun Jul 5 19:20:43 CEST 2009

```Or equivalently:

sum_{k=0}^infty x^k/binomial(k+L-1,k) = hypergeom([1,1],[L],x)
= (L-1) sum_{j=0}^{L-3} (x-1)^j/((L-j-2) x^(j+1))
- (L-1) (x-1)^(L-2) ln(1-x)/x^(L-1)

for integers L >= 2.

Cheers,
Robert Israel

On Sun, 5 Jul 2009, Richard Mathar wrote:

>
> On behalf of the message of
> http://list.seqfan.eu/pipermail/seqfan/2009-July/001831.html
>
> ap> Return-Path: <seqfan-bounces at list.seqfan.eu>
> ap> Date: Sat, 4 Jul 2009 19:35:44 -0400
> ap> From: Alexander Povolotsky <apovolot at gmail.com>
> ap> To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> ap> Cc: lagarias at umich.edu, Marc Chamberland <chamberl at ramanujan.math.grinnell.edu>,
> ap> 	David H Bailey <dhbailey at lbl.gov>
> ap> Subject: [seqfan]  Another BBP formula for ln(2) ?
> ap> ...
> ap> FYI - I came up with the following exact BBP formula for ln(2)
> ap>
> ap> log(2)=
> ap> (230166911/9240 -
> ap> - Sum((1/2)^k*
> ap> (11/k+10/(k+1)+9/(k+2)+8/(k+3)+7/(k+4)+6/(k+5)-6/(k+7)-7/(k+8)-8/(k+9)-9/(k+10)-10/(k+11)),
> ap>  k = 1 .. infinity))/35917
> ap>
> ap> Could it be reduced to already known BBP variants for ln(2) ?
> ap> ...
>
> we can say that there is an infinitude of formulas of this type by inserting
> x=1/2 into equation what currently is equation (1.26) of
> http://www.strw.leidenuniv.nl/~mathar/public/mathar20071105.pdf
> and then using a partial fraction decomposition.
>
> Richard Mathar
>
>
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>

```