Re: Small prime divisors of very large numbers
- --- In email@example.com, "David Broadhurst" <d.broadhurst@...> wrote:
>For a prime p to occur twice with a gap in between then p must
> --- In firstname.lastname@example.org,
> "Mike Oakes" <mikeoakes2@> wrote:
> > 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.
> Very nice!
> Can you give an example where the arithmetic
> progression is different from the so-far-observed
> case of "every powering after the first appearance"?
divide x^m^n+c and x+c but not x^n+c.
It must therefore also divide the difference x^m^n-x but not x^n-x.
This is fine, however when we consider the next one x^m^n^n+c which it also must not divide for it to alternate or to occur in an arithmetic progression, then we're asking for it not to divide the difference x^m^n^n-x which is impossible as this is always divisible by x^m^n-x which we have agreed is divisible by p.
Complex behaviour can be constructed such as with 77783
3+77756 = 11 * 7069
3^3+77756 = 77783
3^3^3+77756 = 31 * 245987018153
3^3^3^3+77756 has factors 31 and 194819
3^3^3^3^3+77756 has factors 31, 13697, 31013
3^3^3^3^3^3+77756 has factors 31, 13697, 31013, 77783
3^3^3^3^3^3^3+77756 has factors 31, 13697, 31013, 77783 and these continue.
So the rules seem to be that a prime can occur once in isolation but that if it occurs again, it's in every subsequent time.
- 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.