Another observation about pg(43s) primes

[enzocreti]

Pg(k) numbers are numbers so defined:


Pg(k)=(2^k-1)*10^d+2^(k-1)-1 where d is the number of decimal digits of 2^(k-1)-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 r can assume the values 1,10,16,18,37.


Now I consider the primes Pg(215) and Pg(541456)...here k is of the form 41s+10...
is it a coincidence that 215*10-1 and 541456*10-1 are both congruent to 0 mod 307?


541456*10-1=17637*307
215*10-1=7*307


note that 17637=41*43*10+7


In other words 541456=((41*43*10+7)*307+1)/10 and 215=(7*307+1)/10


so 215 and 541456 are numbers of the form ((7+17630m)*307+1)/10 with m a nonnegative integer


so 541456/307=41*43+215/307


I observe that 541456 can be written also as (123459*(307/7)+1)/10


Now i consider (123459^x-1)...this expression is congruent to 0 mod 215 for x multiple of 6.
123459^x-1 is congruent to 0 mod 1763 for x multiple of 60.


541456 can be written as ((111111/7+42^2)*307+1)/10


or 541456=(111111+79*307)*4


or 541456=1763*307+307*7/10+307/((3070+7*307/10^10))


where 307*7=215*10-1




215=5*41+10
541456=(43*307+5)*41+10

[enzocreti]

pg(215), pg(69660), pg(92020) and pg(541456) are probable primes


215, 69660, 92020 and 541456 are multiple of 43.


215 is 10 mod 41
69660 is 1 mod 41
92020 is 1000 mod 41
541456 is again 10 mod 41


I conjecture that the cycle of residues (10,1,1000) mod 41 repeats infinitely many times...so the next pg(43s) probable prime should be such that 43s is 1 mod 41...unfortunally i think that the next pg(43s) probable prime is extremely huge, maybe for exponent >2.000.000

[enzocreti]

pg(215), pg(69660), pg(92020) and pg(541456) are probable primes with k=215,69660, 92020, 541456 multiples of 43.


for k<400.000 there are 38 probale primes pg(k)


and four pg(k) with k multiple of 43...


because the number of pg(k) primes seems to be infinite, also the number of pg(k) primes with k multiple of 43 should be infinite. One would expect 1/43 of the pg(k) primes is a pg(k) prime with k multiple of 43. So it is strange that there are four primes multiples of 43 instead of 1, as one would expect...moreover the k's multiples of 43 are all congruent to 10^m mod 41. So I think that there is an hidden structure.