On Sunday, August 26, 2018 at 3:21:08 PM UTC-5, Musatov wrote: > On Sunday, August 26, 2018 at 3:07:41 PM UTC-5, Oscar Benjamin wrote: > > On Sun, 26 Aug 2018 at 20:32, Musatov <tomusa...@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.
Just to be clear it is not the n I was referring to being composite but the result when certain n are fed into the n*2^k - 3 -- https://mail.python.org/mailman/listinfo/python-list
