## Re: Sierpinski covering sets

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• ... Yes, Jack, you are of course quite right. Thanks for pointing that out so neatly. Stepping back from the code, I realized quite soon after posting that all
Message 1 of 6 , Dec 10, 2009
--- In primenumbers@yahoogroups.com, Jack Brennen <jfb@...> wrote:
>
> Isn't this (for example) a much easier way to show that 10223
> has no such covering set:
>
> 10223*2^137+1 ==
> 1904660910466121 * 935125926286211318869570932217
>
> Implying of course that any covering set for 10223 includes
> one of those two primes...
>

Yes, Jack, you are of course quite right. Thanks for pointing that out so neatly.

Stepping back from the code, I realized quite soon after posting that all it did was find, for each k, any n value that gave a large smallest factor, and that is achievable with lots of off-the-shelf sieving and factorizing programs.

Mike
• ... A Google search of 1904660910466121 finds http://www.math.sc.edu/~filaseta/papers/SierpinskiEtCoPapNew.pdf They reached the same factorization as Jack and
Message 2 of 6 , Dec 10, 2009
Jack Brennen wrote:
> 10223*2^137+1 ==
> 1904660910466121 * 935125926286211318869570932217

A Google search of 1904660910466121 finds
http://www.math.sc.edu/~filaseta/papers/SierpinskiEtCoPapNew.pdf
They reached the same factorization as Jack and listed:

n, smallest prime factor of 10223*2^n+1
77 619033
101 45677096693
137 1904660910466121

The next n with no factor below 10^9 is 509.
What is the smallest factor of 10223*2^509+1?
It is probably above 10^30.

--
Jens Kruse Andersen
• ... In defence of my code, it solved all 6 cases in 15 mins, without my having to depend on referring to web pages of known factorisations, which are the
Message 3 of 6 , Dec 10, 2009
--- In primenumbers@yahoogroups.com, Jack Brennen <jfb@...> wrote:
>
> Isn't this (for example) a much easier way to show that 10223
> has no such covering set:
>
> 10223*2^137+1 ==
> 1904660910466121 * 935125926286211318869570932217
>
> Implying of course that any covering set for 10223 includes
> one of those two primes...
>

In defence of my code, it solved all 6 cases in 15 mins, without my having to depend on referring to web pages of known factorisations, which are the result of significant computational effort, and which you perhaps used to find that n=137 index among the the prima facie hundreds to choose from?

Mike
• No significant computational effort needed for this case. PARI code:
Message 4 of 6 , Dec 10, 2009
No significant computational effort needed for this case. PARI code:

k=10223;for(i=1,200,f=factorint(k*2^i+1,15);if(length(f[,1])==1,print([k,i,factorint(k*2^i+1)])))

Runs with elapsed time about 2 seconds. Output:

[10223, 77, [619033, 1; 2495595681878097381929, 1]]
[10223, 101, [45677096693, 1; 325387086953, 1; 1743849966293, 1]]
[10223, 137, [1904660910466121, 1; 935125926286211318869570932217, 1]]

It's also quick to find one for 22699 which has no small factors:

[22699, 190, [84884846681, 1;
419638892754915061028443044698201980071935256023017, 1]]

Unfortunately, the other 4 candidates don't present such
low-hanging fruit; their k*2^n+1 numbers either have small
factors, or they are too big to factor quickly.

mikeoakes2 wrote:
>
>
> --- In primenumbers@yahoogroups.com, Jack Brennen <jfb@...> wrote:
>> Isn't this (for example) a much easier way to show that 10223
>> has no such covering set:
>>
>> 10223*2^137+1 ==
>> 1904660910466121 * 935125926286211318869570932217
>>
>> Implying of course that any covering set for 10223 includes
>> one of those two primes...
>>
>
> In defence of my code, it solved all 6 cases in 15 mins, without my having to depend on referring to web pages of known factorisations, which are the result of significant computational effort, and which you perhaps used to find that n=137 index among the the prima facie hundreds to choose from?
>
> Mike
>
>
>
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