Re: consecutive p-smooth integers
- --- In email@example.com, "djbroadhurst" <d.broadhurst@...> wrote:
> > Some related data is collected there:No: I misread the gradients. Please scrap the above message.
> > 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 :-?
I am, however, surprised that your gradients
seem to increase in an unusual manner:
k=5: gradient = 0.6042
k=6: gradient = 0.3917
k=7: gradient = 0.3398
I guess that may just be because your primes are so small?
David (guilty of careless error in previous message: sorry!)
- --- In firstname.lastname@example.org, 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,
It was difficult getting just over 2 for the first time with
My only consolation is that the above is also good for 15 consecutive 15823-smooth integers.