Re: Small prime divisors of very large numbers
- I wrote:
>There is no need to restrict this to q /prime/, it can be any integer.
> For any prime q, define b(q,n) by the recurrence relation
> b(q,n+1)=q^b(q,n), with initial condition b(q,0)=1.
And the Theorem can be generalized to
Theorem: For k any integer, if a prime p divides (b(q,n)+k) and
(b(q(n+s)+k) for some s > 0, then p divides (b(q,n+s*t)+k) for all t >= 0.
Proof: by induction in a similar way.
In words: Every p appears in the divisor list either zero or one or infinitely many times, and in the latter case at places that are an arithmetic progression of n values.
- Dear all,
I come back to this topic, looking for a working "pow_mod".
but 2^2^2 % 5 = 16 % 5 = 1.
(and idem for 2^2^2^2 - of course.).
David gave some other code, in
Below is a Pari-GP procedure "pmod(a,m)" to compute
a^(a^(a^ ... ^(a[k-1]^a[k]) ... ) modulo m
where the modulus "m" need not be prime.
Although this may work for prime m
(at least, pmod([2,2,2],5) = Mod(1,5) as should),
= Mod(1, 4)
which is most certainly wrong.
So, to put it short, has anyone a working pmod() in his "library" ?
Thanks in advance!
--- In email@example.com, "David Broadhurst"
I consider that his code should not be trusted.