?:Continued Fractions Producing Primes
Reiner Martin
reinermartin at nyc.rr.com
Tue Dec 24 04:18:37 CET 2002
Leroy,
I think the sequence starts 0,1,1,3,3,20,126,694,...
Superseeker does not come up with anything on this.
- Reiner
On Monday, December 23, 2002 8:20 PM, Leroy Quet <qqquet at mindspring.com>
wrote:
>
> How many permutations of {1,2,3,...,m}->{a(1),a(2),a(3),...,a(m)} give
> continued fractions, [a(1); a(2), a(3),...,a(m)] where both the
> numerator and denominator of the resulting fraction are both primes?
>
> For example: for m=3, I get that there is only one fraction:
> {3,1,2} ->
>
> 3 + 1/(1 +1/2) = 11/3,
>
> where 3 and 11 are both primes.
>
> Two examples for m=4:
>
> [2;3,1,4] = 43/19
> and
> [4;3,1,2] = 47/11
>
> The sequence of number of solutions begins:
> 0, 1, 1, ...
>
> (The 4th term being >= 2.)
>
> This might be a trivially-answered question, or it might (most-likely,
> I believe) be only solvable with brute-force computer search.
>
> Disclaimer: Above calculations done by hand. So, error (sic) may have
> been made...
>
> Thanks,
> Leroy Quet
>
More information about the SeqFan
mailing list