- 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 > Let k be a positive odd number and define P(k) to be the least value of

Yes, it's almost certainly a waste of time, and at least for k=37, there

> 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?

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... :-(

>