(Almost)Consecutive-Integer Egyptian-Fraction Sum =1

[Roland Bacher]

On Mon, May 24, 2004 at 04:58:43PM +0200, Ralf Stephan wrote:
> > For example,
> > 
> > 1/3 + 1/4 + 1/5 + 1/6 + 1/20 = 1.
> > 
> > So, my question is,
> > 
> > which integers n are such that:
> > 
> > 1/k + 1/(k+1) + 1/(k+2) +....+1/(k+m) + 1/n = 1 ?
> > 
> > What is the sequence of such n's? Is it in the EIS?
> 
> Sure there is more than your example plus 1/2+1/2 and 1/2+1/3+1/6?
> For k up to 1000 my Kiste could find no more, and anyway, I feel
> there is an easy proof but cannot substantiate it further.
> 
> 
> ralf
> 
The Euler-Maclaurin summation formula could be helpful:

Indeed, 1/k+1/(k+1)+.....+1/(k+m-1)+1/(k+m) is roughly given by
log(m+k)-log(k) and the number

  1-log(m+k)+log(k)-1/(2k)-1/(2m+2k)-1/n 

can be expressed using the Euler Mac-Laurin formula as an
infinite (and rapidly converging) sum involving Bernoulli numbers 
and factorials.

Roland Bacher