- --- In primenumbers@yahoogroups.com,

Andrey Kulsha <Andrey_601@...> wrote:

> Some related data is collected there:

Your straight-line fits of log(N) against sqrt(p),

> http://www.primefan.ru/stuff/math/maxs.xls

for k=6 (i.e. 7 consecutive integers) and k=7

(i.e. 8) were at very small p. If you were to

extrapolate them to larger p, you would be led to

the absurdity that 8 consecutive p-smooth integers

are easier to find than 7, not so :-?

David - --- In primenumbers@yahoogroups.com, Andrey Kulsha <Andrey_601@...> wrote:
>

Too hard!

> >> http://www.primefan.ru/stuff/math/maxs.xls

> >> http://www.primefan.ru/stuff/math/maxs_plots.gif

> >

> > Thanks, Andrey. The gradients are fanning out better now:

>

> The files were updated again.

>

> Puzzle: find a chain of 13 consecutive p-smooth integers, starting at N,

> with log(N)/log(p) greater than

>

> log(8559986129664)/log(58393) = 2.71328

>

> Best regards,

>

> Andrey

>

It was difficult getting just over 2 for the first time with

log(287946949)/log(15823).

My only consolation is that the above is also good for 15 consecutive 15823-smooth integers.