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Re: Small prime divisors of very large numbers

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  • richard_in_reading
    ... For a prime p to occur twice with a gap in between then p must divide x^m^n+c and x+c but not x^n+c. It must therefore also divide the difference x^m^n-x
    Message 1 of 70 , Jul 2, 2009
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      --- In primenumbers@yahoogroups.com, "David Broadhurst" <d.broadhurst@...> wrote:
      >
      > --- In primenumbers@yahoogroups.com,
      > "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"?

      For a prime p to occur twice with a gap in between then p must
      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.

      Richard Heylen
    • 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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