Exponents leading to pg primes

enzocreti · 2019-01-16, 12:16

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...

enzocreti · 2019-01-16, 13:09

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...

enzocreti · 2019-01-16, 13:15

sorry 24390=29^3+1

enzocreti · 2019-01-16, 19:48

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)

enzocreti · 2019-01-17, 08:10

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.

enzocreti · 2019-01-17, 13:03

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.

enzocreti · 2019-01-21, 09:41

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?

enzocreti · 2019-01-21, 11:51

215, 69660, 92020, 541456 are 0 mod 43
51456 is 71 (a prime) mod 43

enzocreti · 2019-01-21, 14:15

i am currently searching for another pg(43k) prime

guessing that 43k has the form 41s+r

enzocreti · 2019-01-22, 13:29

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

enzocreti · 2019-01-31, 07:50

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?

2019-01-16, 12:16	#1
enzocreti Mar 2018 2×5×53 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 2·5·53 Posts	a typo sorry 24390=29^3+1

2019-01-16, 19:48	#4
enzocreti Mar 2018 2·5·53 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 212₁₆ Posts	SOME ADDITIONAL CONSIDERATION Somebody on mathexchange told me that if x=41a+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=41a+r.

2019-01-21, 09:41	#7
enzocreti Mar 2018 1000010010₂ 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 (2^8367) 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 212₁₆ 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 1022₈ 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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