On Sat, 25 Aug 2018 at 20:27, Musatov <tomusa...@gmail.com> wrote: > > On Saturday, August 25, 2018 at 2:18:09 PM UTC-5, Musatov wrote: > > On Saturday, August 25, 2018 at 1:52:17 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) > > > > > >> > > > > > >> That's more a maths question than a programming question. Find out > > > > > >> how to > > > > > >> tackle it mathematically, and then we can code it. > > > > > > Looks like it's zero for any multiple of 3 (apart from 3 itself). This > > > makes sense since if n is a equal to b*3 for some integer b then > > > n*2^k - 3 = b*3*2^k - 3 = (b*2^k - 1)*3 > > > which can only be prime if > > > b*2^k - 1 = 1 > > > which can only be true if b=1 (since k>0) implying that n=3. So for > > > any *other* multiple of 3 you must necessarily have a(n) = 0. > > > > > > The above means that you can handle all multiples of 3 but how do you > > > know that you won't hit an infinite loop when n is not a multiple of > > > 3? > > Rather, I should say one such n is > 72726958979572419805016319140106929109473069209 (which is not divisible by 3)
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 -- https://mail.python.org/mailman/listinfo/python-list
