# A question about prime numbers

## Recommended Posts

Hey everybody, new here...but don't worry, I don't have a "magic" equation for finding prime numbers, I do it the old fashioned way...through brute force.

I've been dabbling with the series:

2*b^n+1

Where:

b=11+12*c (c is any integer)

When b is prime and 2*b^n+1 is also prime then 2*b^n+1 divides phi(b^n,2), according to proth.exe at least. I'm sure there's a simple explanation for this, it's been true in every case I've tested so far, but that's not the question...

What I'm puzzling over are the values of b for which there is no "small" n. (n <1000) Has anyone pondered the same question? Is there ever "no n" for a given value?

Of course, there are simple cases of k*b^n+1 where all numbers will covered by a small factor or small set of factors. But I've sieved my "no n" cases into the billions and still have candidates left, and tested some of them to 100,000 digits still without luck.

My question is...is this fools errand? When there is no small set of factors, is it possible to prove there is or is not a value of n to make 2*b^n+1 prime? Not necessarily to find the prime itself, rather a proof as to whether or not it exists in the first place. Is it a simple a matter of needing a faster computer?

Clearly, as n gets larger you are fighting probability of intersecting the set of prime numbers and the possibility must exist that some n values never produce a prime...but is there a proof?

##### Share on other sites

What I'm puzzling over are the values of b for which there is no "small" n.

As in there is no "small" n for which 2*b^n + 1 is prime?

##### Share on other sites

Small is entirely relative to the speed of your computer to process this.

So far, observations have shown if there isn't a prime for n<1000, then there is also none for n<10,000, and my puzzle is: is there ever a case of no n, no matter how large.

Currently trying to take one of the "no n" bases out to 100k but that takes time.

##### Share on other sites

So you're asking if there exists a number c such that $2(11+12c)^n + 1$ is not prime for any n? I'm confused, since as far as I can see, $2(11+12c)^n + 1$ will always be divisible by three, since it is of the form $2k+1$. So it is never prime.

##### Share on other sites

$2k+1$ will not be divisible by three if k is a multiple of three. ie if k =6 2k+1 =13, k =9 2k+1 =19

##### Share on other sites

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.

## Create an account

Register a new account