19819Re: [PrimeNumbers] Find non-integer A such that floor(A^n) is never prime?
- Jan 16, 2009Ah, 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.
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
>> 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.
- << Previous post in topic