2^k - 2m+1 = prime: (Was: Sum Of Primes Is Power Of Primes: Question)

[Leroy Quet]

--- Leroy Quet <q1qq2qqq3qqqq at yahoo.com> wrote:

I wrote in part:
>... 
> For these sequences to be infinite, a necessary
> (unless forbidden m's are never obtained) but
> not
> sufficient condition is that there is always at
> least one prime p equal to:
> 
> p = 2^k - m
> 
> for some positive integer(s) k, for any fixed
> odd
> integer m.
> 
> But is there always a k that works for all odd
> m?
> 
> This question might have already been answered
> by
> application of some variation on Dirichlet's
> theorem, perhaps.
 

Could someone please calculate the sequence:

a(n) = the smallest positive integer k such that
2^k - 2n+1 is prime

?

And maybe this should be added to the sequence
definition:

a(n) = 0 if no k exists.

(Of course, proving no k exists means doing
something other than checking all positive
integers k...)

And a suggestion: The last time I asked for terms
to be calculated I received submissions from
several people. (Thanks to all who sent me
those.)
Maybe it is a good idea that if you calculate
terms of the sequence, you send them to seq.fan
instead of just to me. That way, other seq.fan
members won't also submit the same terms if they
don't have to.

Thank you,
Leroy Quet




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