## Re: [PrimeNumbers] Re: consecutive p-smooth integers

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• Hello David, ... as I wrote in http://tech.groups.yahoo.com/group/primenumbers/message/22884, it seems that for fixed p maximal N grows nearly as
Message 1 of 42 , Aug 7 1:30 AM
Hello David,

> Here is a puzzle about sequences of 5 consecutive integers.
>
> Definition: A sequence of consecutive positive
> integers, beginning with N, has strength
> s = log(N)/log(p), where p is the largest prime
> dividing any integer in the sequence.

it seems that for fixed p maximal N grows nearly as exp(b*sqrt(p)), where b
is a constant depending on the length of the chain, so it's little sense to
use log(N)/log(p) as a measure of the "strength". The better would be
log(N)/sqrt(p), and the largest currently known for 5 integers is

log(1517)/sqrt(41) = 1.14389...

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
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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