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Re: Small prime divisors of very large numbers

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  • Mike Oakes
    ... 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, "richard_in_reading" <richard_in_reading@...> wrote:
      >
      > --- 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
    • Maximilian F. Hasler
      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
        https://groups.yahoo.com/neo/groups/primenumbers/conversations/topics/20526
        :
        "
        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" ?

        Thanks in advance!
        Maximilian


        --- In primenumbers@yahoogroups.com, "David Broadhurst"
        <d.broadhurst@...> wrote:

        > fail

        (...)

        I consider that his code should not be trusted.

        David
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