## Twins

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• If there is a web page somewhere dealing with exactly this then I would appreciate it if someone would point me in its direction... Let k be a positive odd
Message 1 of 2 , Sep 2, 2003
If there is a web page somewhere dealing with exactly this then I would
appreciate it if someone would point me in its direction...

Let k be a positive odd number and define P(k) to be the least value of
n such that 3*k*2^n-1 and 3*k*2^n+1 are both prime.

e.g. P(49)=44 since 3*49*2^44+-1 are both prime.

I've been looking at small values of k (<100) and got stuck with (so
far) 37, 41 and 51. I know that for some values of k there are no
solutions e.g. k=79 has a covering set of (5,7,13,17,241), but is there
a good way of at least looking to see whether k=37,41,51 etc. might
actually have covering sets and thus make looking for P(k) in these
cases a waste of time?

Thanks,

Richard
• ... Yes, it s almost certainly a waste of time, and at least for k=37, there is likely no finite covering set.
Message 2 of 2 , Sep 2, 2003
> Let k be a positive odd number and define P(k) to be the least value of
> n such that 3*k*2^n-1 and 3*k*2^n+1 are both prime.
>
> e.g. P(49)=44 since 3*49*2^44+-1 are both prime.
>
> I've been looking at small values of k (<100) and got stuck with (so
> far) 37, 41 and 51. I know that for some values of k there are no
> solutions e.g. k=79 has a covering set of (5,7,13,17,241), but is there
> a good way of at least looking to see whether k=37,41,51 etc. might
> actually have covering sets and thus make looking for P(k) in these
> cases a waste of time?

Yes, it's almost certainly a waste of time, and at least for k=37, there
is likely no finite covering set.

As I posted to this list on 18 November '02:

> The interesting value of k is k=111, which is divisible by 3.
>
> The probability that there exist twin primes of the form
> 111*2^n+/-1 is vanishingly small, but apparently there is
> no finite covering set... If such twin primes do exist,
> it is easy to show that n == 11 (mod 36), but beyond that,
> it gets tough.
>
> So k=111 is almost certainly the answer to the twin-prime form of
> the Sierpinski problem, but may prove exceedingly difficult to
> prove... :-(
>
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