Statistics on Pg(43k) primes
 

Pg(k) numbers are numbers of the form (2^k-1)*10^d+2^(k-1)-1 where k is a integer and d is the number of decimal digits of 2^(k-1)-1.


Some of these numbers are primes as 31, 73, 12763, 255127


Up to k=541456 I found four primes/probable primes with k multiple of 43


These are:


pg(215), pg(69660), pg(92020), pg(541456).


215, 69660, 92020, 541456 are of the form 41s+r, where r can assume only 5 values: 1,10,16,18,37.
Now my question is:
is correct to calculate the probability that this happens only by mere chance is:
1/(5/41)^4, which is about 0.02 per cent? This because the possible residues mod 41 are 1,10,16,18,37. These residues are the only possible such that (41s+r=43k)/41 has a repeating decimal term 02439.

number of digits of the pg(43)'s prp
 

pg(215),pg(69660),pg(92020),pg(541456) are prp


215,69660,92020,541456 are multiple of 43


pg(215) is (2^215-1)*10^65+2^214-1
pg(69660) is (2^...)*10^20970...
pg(92020) is 2^...*10^27701...
pg(541456) is 2^...*10^162995...


65,20970,27701,162995 are the number of digits of 2^(k-1)-1


could it be a mere chance that the number of digits or is a prime (27701) or is a multiple of 5?