## Re: consecutive p-smooth integers

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• ... My current record is s = 4.1067361598, achieved with x = 20375617 N = (x^2-3^4)*(x^2-2^10)/55440 = 3108993777544846030873003 David
Message 1 of 42 , Aug 6, 2011
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> 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.

> Puzzle: Find a sequence of 5 consecutive integers with
> strength s > 3.80.

> I imagine that Jens might soon beat strength s = 4.009147298.

My current record is s = 4.1067361598, achieved with
x = 20375617
N = (x^2-3^4)*(x^2-2^10)/55440 = 3108993777544846030873003

David
• ... 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
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--- 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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