## Re: consecutive p-smooth integers

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Message 1 of 42 , Aug 9, 2011

> > Some related data is collected there:
> > http://www.primefan.ru/stuff/math/maxs.xls
>
> Your straight-line fits of log(N) against sqrt(p),
> 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 :-?

seem to increase in an unusual manner:

I guess that may just be because your primes are so small?

David (guilty of careless error in previous message: sorry!)
• ... Too hard! 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
Message 42 of 42 , Nov 8, 2011
--- In primenumbers@yahoogroups.com, Andrey Kulsha <Andrey_601@...> wrote:
>
> >> 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
>

Too hard!

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.
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