The 4^n Polynomial conjecture

[Marc LeBrun]

>= Edwin Clark
>>= Cino Hilliard

>>Polynomial numbers of the form 4^n + 4^(n-1) + ... + n mod 2 + 1

> The formula is not clear. I think you mean:
>   (4^n + 4^(n-1) + ... + 4 + 1) + (n mod 2)
> Is this correct?

I agree that this is a much better way to write it--that mod in the
penultimate term is confusing (as is the adjective "polynomial").

And the repunit part of course sums to the expression (4^(n+1)-1)/3
(ie (b^(n+1)-1)/(b-1) for other values of 4)

> Unless I misunderstand something here's a proof of the conjecture:
> If n is odd then a(n) = 4^n + ... + 4^1 + 2 so is even and not prime.
> If n is even then a(n) = 4^n + ... + 4^1 + 1, that is, is a repunit
> to base 4 and according to Chris Caldwell's repunits page at
> http://primes.utm.edu/glossary/page.php?sort=GeneralizedRepunit
> the only one is 4^1 + 1. But since 1 is not even that's not a problem.

This isn't exactly a proof, because that page just claims to cover
numbers of less than 100 digits, while commenting "(You might want to
explain why the lists for b=4, 9, 16 and 25 are so short.)" which
suggests that there may be a theorem for square b.

> It look like another situation of probability - 
> the larger the numbers get the scarcer the primes.

You're shooting logarithmically scarcer fish in an exponentially bigger
barrel.  It's not surprising that we hit very few primes.  On the other
hand, it's also not a proof that you won't accidentally nail one
somewhere on the way to infinity.  But the compositeness doesn't seem
remarkable.