## Re: Small prime divisors of very large numbers

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• ... No, it s not actually. If you set x=b(q,n), then b(q,n+1)=q^x, NOT x^y for some y, so your observation does not apply. But I do agree I didn t need an
Message 1 of 70 , Jul 2, 2009
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>
> --- In primenumbers@yahoogroups.com, "Mike Oakes" <mikeoakes2@> wrote:
> >
> > Here's a Lemma that should be both true and useful.
> >
> > For any non-zero integer q, define b(q,n) by the recurrence relation
> > b(q,n+1)=q^b(q,n), with initial condition b(q,0)=1.
> > [So b(q,1)=q, b(q,2)=q^q, and so on.]
>
> Isn't this equivalent to the not very interesting observation that x divides x^y when x and y are nonzero integers?

No, it's not actually.
If you set x=b(q,n), then b(q,n+1)=q^x, NOT x^y for some y, so your observation does not apply.
But I do agree I didn't need an inductive proof: David's observation is the one to use.

Mike
• Dear all, I come back to this topic, looking for a working pow_mod . This : {mypmod(b,n,p)=local(m=[p],f=0); while(n=n-1,m=concat(eulerphi(m[1]),m));
Message 70 of 70 , Mar 23, 2014
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Dear all,
I come back to this topic, looking for a working "pow_mod".
This :
{mypmod(b,n,p)=local(m=[p],f=0);
while(n=n-1,m=concat(eulerphi(m[1]),m));
for(p=n=1,#m,n=lift(Mod(b,m[p])^n);
if(f=f+(n*log(b)>=log(m[p])),n=n+m[p]));n%m[#m];}

yields:
mypmod(2,3,5)
= 2
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[1]^(a[2]^(a[3]^ ... ^(a[k-1]^a[k]) ... ) modulo m
where the modulus "m" need not be prime.
(...)
{pmod(a,m)=local(k,q,t);k=#a;q=[m];t=a[k];
for(j=2,k-1,q=concat(eulerphi(q[1]),q));
for(j=1,k-1,t=Mod(a[k-j],q[j])^lift(t));t}
"
Although this may work for prime m
(at least, pmod([2,2,2],5) = Mod(1,5) as should),
it gives:

pmod([2,2,2],4)
= Mod(1, 4)
which is most certainly wrong.

So, to put it short, has anyone a working pmod() in his "library" ?

Maximilian