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• ## Re: consecutive p-smooth integers

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• ... May we see your data for length 5 with N between 20 and 30 digits? Here is mine :-) [N, s = log(N)/log(p)] [5787885182600784208790, 3.826403843] in 70
Message 1 of 42 , Aug 7, 2011
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Andrey Kulsha <Andrey_601@...> wrote:

> > 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's little sense to use log(N)/log(p)

May we see your data for length 5 with N between 20 and 30 digits?

Here is mine :-)

[N, s = log(N)/log(p)]
[5787885182600784208790, 3.826403843] in 70 seconds
[195110934522453734763104, 4.009147298] in 169 seconds
[3108993777544846030873003, 4.106736160] in 229 seconds
[877496832307054822313934323, 4.173526105] in 3033 seconds
[21299560799614314335258375426, 4.194840344] in 8660 seconds
[51196989520720340392524462575, 4.217628079] in 5324 seconds

Best regards, as ever

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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