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Old 2019-01-16, 12:16   #1
enzocreti
 
Mar 2018

10228 Posts
Default 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
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Old 2019-01-16, 13:09   #2
enzocreti
 
Mar 2018

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Default 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...
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Old 2019-01-16, 13:15   #3
enzocreti
 
Mar 2018

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Default a typo

sorry 24390=29^3+1
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Old 2019-01-16, 19:48   #4
enzocreti
 
Mar 2018

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Default 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)
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Old 2019-01-17, 08:10   #5
enzocreti
 
Mar 2018

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Default 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.
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Old 2019-01-17, 13:03   #6
enzocreti
 
Mar 2018

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Default other observation

Quote:
Originally Posted by enzocreti View Post
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.
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Old 2019-01-21, 09:41   #7
enzocreti
 
Mar 2018

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Default 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
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Old 2019-01-21, 11:51   #8
enzocreti
 
Mar 2018

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Default 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
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Old 2019-01-21, 14:15   #9
enzocreti
 
Mar 2018

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Default 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
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Old 2019-01-22, 13:29   #10
enzocreti
 
Mar 2018

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Default 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
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Old 2019-01-31, 07:50   #11
enzocreti
 
Mar 2018

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Default 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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