mersenneforum.org Exponents leading to pg primes
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 2019-01-16, 12:16 #1 enzocreti   Mar 2018 10228 Posts Exponents leading to pg primes The pg(k) numbers are formed by the concatenation of two consecutive Mersenne numbers... pg(k)=(2^k-1)*10^d+2^(k-1)-1 where d is the number of decimal digits of 2^(k-1)-1 example pg(1)=10 pg(2)=31 pg(3)=73... Some values k for which pg(k) is prime are 215, 92020, 69660, 541456, 51456... i noticed that (215/41), (92020/41), (69660/41), (541456/41), (51456/41) have all the same periodic decimal expansion 24390=29^3+1. Is there any reason? Example 215/41=5.2439024390... Last fiddled with by enzocreti on 2019-01-16 at 13:15
 2019-01-16, 13:09 #2 enzocreti   Mar 2018 2·5·53 Posts exponents leading to a probable prime In particular all the four exponents leading to a probable prime which are multiples of 43: 215, 69660, 92020, 541456 have this property. Infact 215/41=5.2439024390... 69660/41=1699,02439024390... 92020/41=2244,3902439024390... 541456/41=13206,2439024390...
 2019-01-16, 13:15 #3 enzocreti   Mar 2018 21216 Posts a typo sorry 24390=29^3+1
 2019-01-16, 19:48 #4 enzocreti   Mar 2018 10228 Posts 10^n mod 41 somebody on mathexchange pointed me out that this is equivalent to say that the exponents are congruent to 10^n mod 41 with n>=0 Infact 215, 69660, 92020, 541456 are all congruent to 10^n (mod 41)
 2019-01-17, 08:10 #5 enzocreti   Mar 2018 2·5·53 Posts SOME ADDITIONAL CONSIDERATION Somebody on mathexchange told me that if x=41*a+r (with r=1,10,16,18,37), then x/41 will have a repeating term of 24390... so when pg(k) is prime and k is a multiple of 43, then k=41*a+r.
2019-01-17, 13:03   #6
enzocreti

Mar 2018

2·5·53 Posts
other observation

Quote:
 Originally Posted by enzocreti Somebody on mathexchange told me that if x=41*a+r (with r=1,10,16,18,37), then x/41 will have a repeating term of 24390... so when pg(k) is prime and k is a multiple of 43, then k=41*a+r.

215*271, 69660*271, 92020*271, 541456*271 are all congruent to plus or minus 1 mod 13.

 2019-01-21, 09:41 #7 enzocreti   Mar 2018 2×5×53 Posts exponents leading to a pg prime exponents leading to a pg prime of the form 41s+r with r=1,10,16,18,37 are 215, 51456, 69660, 92020, 541456. 215, 69660, 92020 and 541456 are congruent to 0 mod 43 51456 instead has the factorization (2^8*3*67) where 2^8 is congruent to 41 mod 43. Is there some explanation? Last fiddled with by enzocreti on 2019-01-21 at 11:56
 2019-01-21, 11:51 #8 enzocreti   Mar 2018 2·5·53 Posts 215,51456,69660,92020,541456 215, 69660, 92020, 541456 are 0 mod 43 51456 is 71 (a prime) mod 43 Last fiddled with by enzocreti on 2019-01-21 at 11:56
 2019-01-21, 14:15 #9 enzocreti   Mar 2018 2·5·53 Posts searching another pg(43k) prime i am currently searching for another pg(43k) prime guessing that 43k has the form 41s+r Last fiddled with by enzocreti on 2019-01-21 at 14:17
 2019-01-22, 13:29 #10 enzocreti   Mar 2018 2×5×53 Posts pg(215), pg(51456), pg(69660), pg(92020), pg(541456) it seems that there are infinitely many pg(1763n+d)'s primes, where d=215, 329, 344, 387, 903, 1677 with d congruent to (1,10,16,18,37) mod 41 Last fiddled with by enzocreti on 2019-01-22 at 13:48
 2019-01-31, 07:50 #11 enzocreti   Mar 2018 21216 Posts pg(69660) and pg(92020) I wonder if the primality of pg(69660) and pg(92020) can be proven, I mean I know that they are probable primes but do you think that with Primo I can proof them surely prime?

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