Generating a specific list of intsgers

[Musatov]

On Sunday, August 26, 2018 at 3:07:41 PM UTC-5, Oscar Benjamin wrote:
> On Sun, 26 Aug 2018 at 20:32, Musatov <tomusatov at gmail.com> wrote:
> >
> > On Sunday, August 26, 2018 at 2:14:29 PM UTC-5, Oscar Benjamin wrote:
> > > > > > > > >> On Fri, 24 Aug 2018 14:40:00 -0700, tomusatov wrote:
> > > > > > > > >>
> > > > > > > > >>> I am looking for a program able to output a set of integers meeting the
> > > > > > > > >>> following requirement:
> > > > > > > > >>>
> > > > > > > > >>> a(n) is the minimum k > 0 such that n*2^k - 3 is prime, or 0 if no such
> > > > > > > > >>> k exists
> > > > > > > > >>>
> > > > > > > > >>> Could anyone get me started? (I am an amateur)
> > >
> > > Fair enough. So finding a(n) when a(n)!=0 is straight-forward (simply
> > > loop through testing k=1,2...) but the issue is determining for any
> > > given n whether a(n)=0 i.e. that there does not exist k such that
> > > n*2^k-3 is prime.
> > >
> > > Perhaps if you explain how you know that
> > >    a(72726958979572419805016319140106929109473069209) = 0
> > > then that would suggest a way to code it.
> > >
> > Oscar, I simply asked someone and they provided me the number. I know they often use Maple, but I was interested in Python.
> > He also said some of the n are prime by Dirichlet's theorem. One is 8236368172492875810638652252525796530412199592269.
> 
> If it is possible at all then it is certainly possible to do this in
> Python but only for someone who knows the necessary maths. The purpose
> of computers in these kinds of problems is that they are much faster
> at number-crunching. You still need to know how (at least in
> principle) you would do this by hand in order to program it in Python
> or most likely anything else.
> 
> I don't think anyone here knows the answer to the mathematical
> question "how do I prove that a(n)=0 for some n?". If you knew the
> answer to that question then I'm sure many people could help you write
> code for it.
> 
> Without that I think you need to go back to your mathematician friends
> or do some more reading.
> 
> Are you sure that the problem you have posed here is solvable (i.e.
> that whether or not a(n)=0 is decidable for any n)?
> 
> --
> Oscar

My understanding is this: there are an infinite number of n's that are not multiples of three, and yet will always be divisible by at least one of 22 primes for all values of k.

i.e. certain n values make the equation produce only composite numbers for all values of k.