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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