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On Saturday, August 25, 2018 at 1:52:17 PM UTC-5, Oscar Benjamin wrote: > On Sat, 25 Aug 2018 at 18:12, <tomusatov at gmail.com> wrote: > > > > On Saturday, August 25, 2018 at 9:46:21 AM UTC-5, Richard Damon wrote: > > > On 8/25/18 10:27 AM, Dennis Lee Bieber wrote: > > > > On Sat, 25 Aug 2018 03:56:28 +0000 (UTC), Steven D'Aprano > > > > <steve+comp.lang.python at pearwood.info> declaimed the following: > > > > > > > >> 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. > > > > I'd want more punctuation in that just to ensure I'm interpreting it > > > > properly -- I'm presuming it is meant to be parsed as: > > > > (n * (2 ^ k)) - 3 > > > > > > > > Suspect this needs to be attacked in the reverse direction -- generate > > > > a list of primes, add 3, determine if it is a multiple of powers of two. > > > > Though in that case, k = 1 would fit all since if it is a multiple 2^2 (4) > > > > it would also be a multiple of 2^1 (2), for all greater powers of 2.. > > > > > > > > prime 5 > > > > + 3 => 8 > > > > log_2 8 => 3 <<< integral k > > > > 8 => 1 * (2 ^ 3) > > > > 2 * (2 ^ 2) > > > > 4 * (2 ^ 1) > > > > > > > > n=4, k=1 > > > > > > > > OTOH, if it is supposed to be (n*2) ^ k, or even worse (n*2) ^ (k-3) > > > > the solution becomes more difficult. > > > > > > > > > > > I think the issue is given n, find k. > > > > > > a(1): 1*2-3=-1 no, 1*4-3=1 no, 1*8-3 - 5 YES, a(1) = 3 > > > > > > a(2) 2*2-3 = 1, no 2*4-3=5 YES a(2) = 2 > > > > > > a(3) 3*2-3 - 3 YES, a(3) = 1 > > > > > > and so on. > > > > > > One path to solution is to just count up the k values and test each > > > result for being prime, except that will never return 0 to say no such k > > > exists. That will require some higher level of math to detect (or an > > > arbitrary cap on k, and we say we can say 0 if the only k is too big, > > > but with big nums, that would be VERY large and take a very long time. > > > > > > -- > > > Richard Damon > > Here is a sample output: > > 3, 2, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 2, 0, 1, 1, 0, 2, 1, 0, 1, 1, 0, 1, 2, 0, 1, 2, 0, 1, 1, 0, 3, 1, 0, 1, 1, 0, 2, 1, 0, 1, 2, 0, 1, 1, 0, 2, 1, 0, 1, 1, 0, 1, 1, 0, 1, 2, 0, 2, 1, 0, 3, 1, 0, 1, 2, 0, 1, 1, 0, 5, 2, 0, 1, 1, 0, 2, 1, 0, 3, 1, 0, 1 > > 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? such n is 72726958979572419805016319140106929109473069209 (which is not divisible by 3) For that you need the converse of the above: > whenever n is not a multiple of 3 then a(n) != 0 > I haven't put much thought into it but that might be easy to prove. > > -- > Oscar

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