- 2011/8/6 djbroadhurst <d.broadhurst@...>

> **

N=28998978021284765 is good with strength~3.92795.

>

>

>

>

> --- In primenumbers@yahoogroups.com,

> "Jens Kruse Andersen" <jens.k.a@...> wrote:

>

> > There are some computational results for 3 and 4 integers in

> > http://www.mersenneforum.org/showthread.php?t=5647

>

> 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.

>

> Example: The sequence of 5 consecutive integers

> beginning with N = 612130418113087340 has strength

> s = log(N)/log(48437) =~ 3.7964, as shown here:

>

> for(k=0,4,print(factor(612130418113087340+k)[,1]~));

>

> [2, 5, 17, 263, 347, 929, 1237, 17167]

> [3, 13, 31, 71, 157, 757, 1367, 4877]

> [2, 7, 37, 1481, 2069, 14591, 26431]

> [19, 23, 467, 3257, 19013, 48437]

> [2, 3, 43, 479, 1109, 13007, 42923]

>

> Puzzle: Find a sequence of 5 consecutive integers with

> strength s > 3.80.

>

> Comment: A solution may be found in less than 200 seconds.

>

> David Broadhurst

>

>

>

? N=28998978021284765;for(i=0,4,print(factor(N+i)))

[5, 1; 13, 1; 2663, 1; 4261, 1; 4919, 1; 7993, 1]

[2, 1; 3, 1; 11, 1; 59, 1; 313, 1; 337, 1; 4547, 1; 15527, 1]

[47, 1; 241, 2; 863, 1; 1129, 1; 10903, 1]

[2, 5; 7, 1; 1549, 1; 1613, 1; 6871, 1; 7541, 1]

[3, 6; 19, 2; 137, 2; 2423, 2]

The code in PARI-GP (note that it would be much faster to use sieve), it ran

about 150 minutes: (prints the record strength also for these type of

numbers)

T(n)=f=factor(n);return(log(n)/log(f[matsize(f)[1],1]))

r=0;n=3;while(1,n++;if(T(n)>r/2&&T(n-1)>r/2&&T(n+1)>r/2&&\

T(n-2)>r/2&&T(n+2)>r/2&&T(n*n-2)>r&&T(n*n-3)>r,\

r=100.0;for(i=0,4,r=min(r,T(n*n-i)));print(r" "n*n-4)))

In this way N=n*n-4, so N;N+3 and N+4 has got an algebraic factorization,

increasing our chances to find a large strength. (I've used a slightly

different strength definition.)

[Non-text portions of this message have been removed] - --- In primenumbers@yahoogroups.com, Andrey Kulsha <Andrey_601@...> wrote:
>

Too hard!

> >> 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

>

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.