Re: Small prime divisors of very large numbers
- --- In firstname.lastname@example.org, "David Broadhurst" <d.broadhurst@...> wrote:
>Indeed there is a nonsequitor, David :-(
> --- In email@example.com,
> "Mike Oakes" <mikeoakes2@> wrote:
> > Then b(q,n+1) = b(q,n) mod p,
> > so q^b(q,n+1) = q^b(q,n) mod p,
> Err, Mike, how does the second line follow from the first?
> If x = y mod (p-1)
> then q^x = q^y mod p,
> in my book.
> Has one of us made a big boob?
While I try and mend the proof (I still believe the Theorem to be true - HELP, anyone?), here are the results of a nontrivial run that has just finished (17.2 hrs @3.79GHz).
I used Jens's pari script, with the prime limit upped from 10^8 to 4*10^9, which is just about as high as you can go with 32-bit gp.exe, on your original puzzle form of 137^^n+73.
Here are the sets of prime divisors up to that limit:-
n=1: 2, 3, 5, 7 [done manually, since 137+73=210]
n=2: 2, 3, 5, 821, 71757331, 152555243
n=3: 2, 3, 5, 29, 821
n=4: 2, 3, 5, 29, 821
n=5: 2, 3, 5, 29, 821, 23339, 67525153, 224354393
6<=n<=15: 2, 3, 5, 29, 821, 23339, 67525153
They still support my Theorem.
Can anyone find a counterexample to it, for any (q^^n+k sequence)?
- 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 firstname.lastname@example.org, "David Broadhurst"
I consider that his code should not be trusted.