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Re: Small prime divisors of very large numbers

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  • Mike Oakes
    ... There is no need to restrict this to q /prime/, it can be any integer. And the Theorem can be generalized to Theorem: For k any integer, if a prime p
    Message 1 of 70 , Jul 1, 2009
      I wrote:
      >
      > 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.
      >

      There is no need to restrict this to q /prime/, it can be any integer.

      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.

      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
        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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