There are infinitely many probable primes of the form $(n-1)^2+1$.

This generalizes to more polynomial since Fermat numbers satisfy $F_n=(F_{n-1}-1)^2+1$.

Second answer.

Define the linear recurrence $a(0)=1,a(n)=2a(n-1)+1$.

Then $a(n)=2^n-1=M_n$. So we get linear recurrence which is always probable prime.

Also setting $a(n)=x$, we get the tuples $(x,2x+1,4x+3,...)$ which are all probable primes.

There are infinitely many probable primes of the form $(n-1)^2+1$.

This generalizes to more polynomial since Fermat numbers satisfy $F_n=(F_{n-1}-1)^2+1$.

There are infinitely many probable primes of the form $(n-1)^2+1$.

This generalizes to more polynomial since Fermat numbers satisfy $F_n=(F_{n-1}-1)^2+1$.

There are infinitely many probable primes of the form $(n-1)^2+1$.

This generalizes to more polynomial since Fermat numbers satisfy $F_n=(F_{n-1}-1)^2+1$.

Second answer.

Define the linear recurrence $a(0)=1,a(n)=2a(n-1)+1$.

Then $a(n)=2^n-1=M_n$. So we get linear recurrence which is always probable prime.

Also setting $a(n)=x$, we get the tuples $(x,2x+1,4x+3,...)$ which are all probable primes.