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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... |
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... |
a typo
sorry 24390=29^3+1
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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) |
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. |
other observation
[QUOTE=enzocreti;506198]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.[/QUOTE] 215*271, 69660*271, 92020*271, 541456*271 are all congruent to plus or minus 1 mod 13. |
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? |
215,51456,69660,92020,541456
215, 69660, 92020, 541456 are 0 mod 43
51456 is 71 (a prime) mod 43 |
searching another pg(43k) prime
i am currently searching for another pg(43k) prime
guessing that 43k has the form 41s+r |
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
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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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