Ah, great -- that's just the kind of trick I was looking for.

Thanks for the example; I'm sure it's just one example of a

whole family of solutions.

Jack

Robert Gerbicz wrote:

> 2009/1/16 jbrennen <jfb@...>

>

>> Just a silly little thought I had...

>>

>> It is almost certainly possible to find a positive non-integer A > 1

>> such that for all integer n, floor(A^n) is not prime.

>>

>> Question -- does any such number A exist with a simple closed-form

>> expression?

>>

>>

>

>> A=5+sqrt(19) is good. To prove it use that

>> g(n)=(5+sqrt(19))^n+(5-sqrt(19))^n is an integer sequence for that

>> g(n)=10*g(n-1)-6*g(n-2), using this it's easy by induction that g(n)==1 mod

>> 3 so floor(A^n)=g(n)-1 is divisible by 3 for all n>0 and isn't 3, so not

>> prime. For n<=0 floor(A^n)=0,1 so not prime.

>>