2k+1 not always divisible by 3
Perhaps it wasn't entirely clear: for this sequence b=11+12c and that I'm looking at cases of 2*b^n+1 where b itself is a prime as well a member of the set 11+12c, also not divisible by three.
Try 2*467^n+1 , (467=11+38*12) the first value of n that produces a prime is 126775, other values for b can produce quite a number of primes quickly.
It's more a questions of philosophy, probability and patience. Looking for thoughts or ideas on proving there is no n for which 2*b^n+1 is prime, I have observed differences in the behavior of some bases but not enough data gathered yet to get a grasp on why.
I'm currently working on a 5 digit base, it's been sieved for 3<n<3000000. So far I've eliminated 98.9% of all candidates as composite...that still leaves a whopping 32,825 candidates, a life's work. (Would take my computer roughly 100 years to complete the task) I have two computers running PRP tests on the small end of the file, while the sieve continues to work out ahead of them on the big end. Only reached n<38975 so far as confirmed composites and looking ahead, one starts to ponder, is it possible there is no n? Can this expression have an infinite number of different prime factors as it grows forever, but never be prime itself?
I know it may seem like fantasy, but there is a possibility...just haven't been able to get my mind around the programing to work on it yet...
You know 1 divided by any integer gives you a repeating decimal, the length of the period for 1/p will divide p-1 if p is prime. If you could show for a given b the period of 1/(2*b^n+1) will always fall into a set class that will never divide 2*b^n without a remainder even as n grows, then you might have the makings of a proof. I have no work towards this yet, just suggesting there might be ways to look at the question beyond brute force number crunching for all time.