Pg(k) primes with k prime

enzocreti · 2019-05-17, 09:30

Pg(k) numbers=(2^k-1)*10^d+2^(k-1)-1 where d is the number of decimal digits of 2^(k-1)-1.

For k>2, these are the values of k such that both Pg(k) and k are prime:

3,7,19,67,79,359,2131,3371,331259

These primes are all of the form 4*s+3, with s>=0.

Can it be proven that if Pg(k) is prime and k is prime, then k must have the form 4*s+3?

enzocreti · 2019-05-17, 11:06

Ok it can be easily proven...never mind!

2019-05-17, 09:30	#1
enzocreti Mar 2018 530₁₀ Posts	Pg(k) primes with k prime Pg(k) numbers=(2^k-1)10^d+2^(k-1)-1 where d is the number of decimal digits of 2^(k-1)-1. For k>2, these are the values of k such that both Pg(k) and k are prime: 3,7,19,67,79,359,2131,3371,331259 These primes are all of the form 4s+3, with s>=0. Can it be proven that if Pg(k) is prime and k is prime, then k must have the form 4*s+3?

2019-05-17, 11:06	#2
enzocreti Mar 2018 2·5·53 Posts	Proof Ok it can be easily proven...never mind!

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