?:Continued Fractions Producing Primes

[Leroy Quet]

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