## Find non-integer A such that floor(A^n) is never prime?

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• 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.
Message 1 of 6 , Jan 15, 2009
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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?

Or, along those lines... is floor(sqrt(18)^n) ever prime? :)
• ... Heuristically, it ought to be prime for some large odd power n. There is no obvious factor in that case and the integral of 1/(2*n*log(18)) diverges for
Message 2 of 6 , Jan 16, 2009
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--- In primenumbers@yahoogroups.com, "jbrennen" <jfb@...> wrote:

> is floor(sqrt(18)^n) ever prime? :)

Heuristically, it ought to be prime for some large odd power n.
There is no obvious factor in that case and the integral of
1/(2*n*log(18)) diverges for large n.

Since there is no PRP for n < 10^4, we might need to go up
to something like n = 10^4*18^2 to get a half-way decent
chance of a prime, which would explain the smile sign.

As for the larger question: I have no good idea for a
closed form. Did Mills address the question of the
existence, in principle, of such a number A?

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• ... How about n = 10675 for a BPSW probable prime? It seems that Poisson can be cheated by posting ... and then Poisson s gremlins almost immediately oblige
Message 3 of 6 , Jan 16, 2009
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--- In primenumbers@yahoogroups.com, "jbrennen" <jfb@> wrote:

> is floor(sqrt(18)^n) ever prime? :)

How about n = 10675 for a BPSW probable prime?

It seems that Poisson can be cheated by posting
a remark like this:

> Since there is no PRP for n < 10^4, we might need to go up
> to something like n = 10^4*18^2 to get a half-way decent
> chance of a prime

and then Poisson's gremlins almost immediately oblige
with a hit, before I can turn off the process:

? print(ispseudoprime(sqrtint(18^10675)));
1

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• 2009/1/16 jbrennen ... [Non-text portions of this message have been removed]
Message 4 of 6 , Jan 16, 2009
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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.
>
>
>

[Non-text portions of this message have been removed]
• Thanks Robert for the nice closed form example. ... No, his note is very short, contains no numerics, and is very focused on his result. However, it would be
Message 5 of 6 , Jan 16, 2009
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Thanks Robert for the nice closed form example.

> As for the larger question: I have no good idea for a
> closed form. Did Mills address the question of the
> existence, in principle, of such a number A?

No, his note is very short, contains no numerics,
and is very focused on his result. However, it
would be trivial to modify his methods of proof
for composites, and of course the exponent 3 can
be reduced (for composites). CC
• 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
Message 6 of 6 , Jan 16, 2009
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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.
>>
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