20190611, 10:41  #1 
Mar 2018
17·31 Posts 
PG primes and possible residues of 10^n mod 41
Pg(k) numbers are numbers so defined:
Pg(k)=(2^k1)*10^d+2^(k1)1 where d is the number of decimal digits of 2^(k1)1. I found that when k is multiple of 43, as in the cases Pg(215), Pg(69660), Pg(92020) and Pg(541456) and Pg(k) is prime/probable prime, then k is of the form 41s+r, where r are the possible residues of 10^n mod 41, that is 1,10,16,18,37. Do you believe that this is only coincidence or there is a connection with the base ten? Last fiddled with by enzocreti on 20190611 at 12:11 
20190611, 12:42  #2 
Mar 2018
17·31 Posts 
REFINEMENT OF THE CONJECTURE
I think that the conjecture can maybe refined...r maybe cannot assume the values 18 and 37...so at least for the four probable primes known...r assumes only the values 1,10,16...these are Always residues 10^n mod 41 but not all of them...18 and 37 are not present...I don't know if r=18 and r=37 will pop up...

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