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Re: zetaprime puzzle

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  • djbroadhurst
    ... Here zeta is Riemann s zeta function, which vanishes at the even negative integers and is rational at the odd negative integers, with zeta(-1)=-1/12. Primo
    Message 1 of 9 , Sep 2, 2002
      I wrote:

      > zeta(-k)/zeta(-1) is prime for
      > k=25,33,37,73,117,673,
      > what, if anything, comes next?

      "ttpi314159" wrote:

      > I am puzzled,... probabilities?

      Here zeta is Riemann's zeta function,
      which vanishes at the even negative integers
      and is rational at the odd negative integers,
      with zeta(-1)=-1/12.

      Primo has proven that zeta(-673)/zeta(-1) is prime.
      In terms of the Bernoulli number Bern(674) it is

      > 5604 2*3*Bern(674)/337 1077 c4 2002 Irregular, ECPP (notes)

      with notes at

      http://groups.yahoo.com/group/primenumbers/message/4197

      I was wondering: what is the next k=2*p-1 for which

      zeta(-k)/zeta(-1)=6*Bern(2*p)/p

      is probably prime?

      Yesterday, I set a Pari-GP job running on a fast unix box
      with 500MB of core:

      nm=3000;b=bernvec(nm);
      {for(k=1,nm,bv=abs(6*b[k+1]/k);
      if(denominator(bv)==1&&ispseudoprime(bv),
      print([k,2*k-1])))}

      [caution: set a smaller value of nm, say
      nm=100, to test this code quickly, with
      a small core memory requirement]

      After running it overnight, I found this output:

      [13, 25]
      [17, 33]
      [19, 37]
      [37, 73]
      [59, 117]
      [337, 673]
      [2153, 4305]

      which means that

      zeta(-4305)/zeta(-1)=6*Bern(2*2153)/2153

      is probably prime.

      As it has 10,342 decimal digits,
      we will have to wait a long while for
      hardware or software to develop
      to the stage where a proof of primality
      is feasible.

      The number in question is filed in

      http://groups.yahoo.com/group/primenumbers/files/Prime%20Tables/
      zm4305.txt

      > zeta(-4305)/zeta(-1)=6*Bern(2*2153)/2153 is PrP

      Henri: Can you handle that in your database?

      David
    • Jon Perry
      Purely for amusement - don t burn too many brain cells thinking about it: When is zeta(a+ib)/zeta(c+id) prime? Jon Perry perry@globalnet.co.uk
      Message 2 of 9 , Sep 2, 2002
        Purely for amusement - don't burn too many brain cells thinking about it:

        When is zeta(a+ib)/zeta(c+id) prime?

        Jon Perry
        perry@...
        http://www.users.globalnet.co.uk/~perry/maths
        BrainBench MVP for HTML and JavaScript
        http://www.brainbench.com
      • mikeoakes2@aol.com
        In a message dated 02/09/02 11:50:29 GMT Daylight Time, ... Congrats on a really spendid puzzle, David, and on a big new prime and an even bigger PrP! I had
        Message 3 of 9 , Sep 3, 2002
          In a message dated 02/09/02 11:50:29 GMT Daylight Time,
          d.broadhurst@... writes:


          > I wrote:
          >
          > > zeta(-k)/zeta(-1) is prime for
          > > k=25,33,37,73,117,673,
          > > what, if anything, comes next?
          >
          >
          > zeta(-4305)/zeta(-1)=6*Bern(2*2153)/2153
          > is probably prime.
          >
          >
          Congrats on a really spendid puzzle, David, and on a big new prime and an
          even bigger PrP!

          I had tumbled to the fact that B[k+1]*12/(k+1) must be prime, which also
          entails B[k+1] = 1/6 mod 1, and was wondering how best to tackle the
          computation of huge B[n]'s, when your solution thundered into my in-tray.

          The problem with the most straightforward method, which uses the recurrence
          relation
          (n+2)B[n+1] = - sigma{k=0 to n}(n+2,k)B[k]
          where (n+2,k) = (n+2)!/(k!*(n+2-k)!)
          is that you have to store all the B[n]. This limits the max. index to about
          n=15000 on a m/c with 500Mb memory, doesn't it?

          The following URL
          http://numbers.computation.free.fr/Constants/Miscellaneous/bernoulli.html
          gives a mass of info on Bernoulli numbers.

          And
          http://www.cecm.sfu.ca/CAG/Projects/High_Precision_Numerics/H1.html
          gives a nice way of computing a single B[-k], by evalutating zeta(k+1) to
          "sufficient" precision, once B[-k] mod 1 is known exactly (which is an easier
          computation).

          So memory would be no longer a limitation, only time. And one could restrict
          attention to the subset for which the fractional part = 1/6.

          Why not go for the world's biggest prime?!

          Mike Oakes



          [Non-text portions of this message have been removed]
        • Bouk de Water
          I received message from Thomas Nicely about a new record prime gap: From: Thomas R. Nicely To: Selected parties Re: A new maximal prime
          Message 4 of 9 , Sep 3, 2002
            I received message from Thomas Nicely about a new record prime gap:

            From: Thomas R. Nicely <trnicely@...>
            To: Selected parties
            Re: A new maximal prime gap of 1184
            Date: 2002.09.03.0856 GMT

            (1) Dr. Bertil Nyman <bertil.nyman@...> has discovered a new first
            occurrence and maximal prime gap of measure 1184 following the prime
            43841547845541059.
            (2) Dr. Nyman first detected the gap 31 August 2002.
            It was confirmed as a first occurrence and maximal prime gap 2 September 2002,
            when his exhaustive scan of prime gaps was completed to 4.4445*10^16.
            (3) This is the first maximal gap found since the maximal gap of 1132 following
            the prime 1693182318746371, also discovered by Dr. Nyman, 24 January 1999, and
            confirmed as maximal 18 February 1999 by the work of Dr. Nyman and Thomas R.
            Nicely. The gap of 1132 constitutes the first occurrence of any gap of 1000 or
            greater.
            (4) The gap of 1184 has a merit, M=G/ln(P_1), of 30.90, second among known gaps
            only to the maximal gap of 1132, whose merit is 32.28. These are the only two
            gaps currently known (to me) with merits >= 30.
            (5) These gaps will be among the subjects of a paper (in preparation) on first
            occurrence and maximal prime gaps, authored by Dr. Nyman and Thomas R. Nicely.
            (6) Further information is available at Nicely´┐Żs website
            <http://www.trnicely.net>; see in particular the listing of all presently known
            first occurrence and maximal prime gaps at
            <http://www.trnicely.net/gaps/gaplist.html>.
            (7) Feel free to distribute this information as you wish.
            (8) I do not wish to spam your e-mail with unwanted information.
            I have attempted to restrict this mailing list to parties believed to have an
            interest in the message. If you do not wish to receive such messages
            henceforth, please so advise me, and your name will be removed from the
            distribution list.
            Regards,

            Thomas R. Nicely


            __________________________________________________
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          • djbroadhurst
            ... Here is a numerical workaround; for large p, it computes twice as many terms in zeta(2*p) as needed, so the _relative_ error in ... =6*|Bern(2*p)|/p
            Message 5 of 9 , Sep 3, 2002
              Mike Oakes wrote:

              > The problem with the most straightforward method,
              > which uses the recurrence relation
              > (n+2)B[n+1] = - sigma{k=0 to n}(n+2,k)B[k]
              > where (n+2,k) = (n+2)!/(k!*(n+2-k)!)
              > is that you have to store all the B[n].

              Here is a numerical workaround;
              for large p, it computes twice as many terms in
              zeta(2*p) as needed, so the _relative_ error in

              |zeta(1-2*p)/zeta(-1)|
              =6*|Bern(2*p)|/p
              =24*(2*p-1)!/(2*Pi)^(2*p)*zeta(2*p)

              is O(1/2^(2*p))

              But you have to be careful in choosing the precision,
              which should be comfortably more than

              log(abs(zeta(1-2*p)))/log(10) decimal digits

              where

              |zeta(1-2*p)|=O((p/Pi/exp(1))^(2*p))

              Example, with p=337:

              -----

              \p1200

              {fn(p)=local(a,m,r);m=2*ceil(p/Pi/exp(1));
              a=24*(2*p-1)!/(2*Pi)^(2*p)*sum(k=1,m,1./k^(2*p));
              r=round(a);if(abs(r-a)<1/10^10&&ispseudoprime(r),print(r))}

              fn(337)

              ------

              OK, it's crude and slow, but it sure solves the storage problem.

              PS: Of course one should not use Pari's "ispseudoprime".
              Better to output the integer to Pfgw.

              David
            • djbroadhurst
              Whoops! I said ... which was just the opposite of what I meant. The _absolute_ error is O(1/2^(2*p)) if you take ... terms in zeta(2*p), with p Pi*exp(1)
              Message 6 of 9 , Sep 3, 2002
                Whoops! I said

                > so the _relative_ error

                which was just the opposite of what I meant.

                The _absolute_ error is O(1/2^(2*p)) if you take

                > m=2*ceil(p/Pi/exp(1))

                terms in zeta(2*p), with p >> Pi*exp(1)

                David
              • mikeoakes2@aol.com
                David s puzzle continues to haunt me... von Staudt s theorem [Hardy & Wright, pp. 90 et seq.] says:- B[2*k] = - sigma{p:(p-1)|2*k} (1/p) MOD 1 or, in words:-
                Message 7 of 9 , Sep 8, 2002
                  David's puzzle continues to haunt me...

                  von Staudt's theorem [Hardy & Wright, pp. 90 et seq.] says:-
                  B[2*k] = - sigma{p:(p-1)|2*k} (1/p) MOD 1
                  or, in words:-
                  the fractional part of B[2*k] is the negative of the fractional part of the
                  sum (1/2+1/3+1/5+1/7+...),
                  the sum being taken over all primes p such that (p-1) divides (2*k).

                  Any straightforward approach to solving David's puzzle calls for a search
                  over those B[2*k] with denominator 6.
                  The sum on the r.h.s. of the theorem must therefore be _only_ over the primes
                  2 and 3, giving B[2*k] = -(1/2+1/3) = -5/6 = 1/6 MOD 1.

                  The search can be restricted to a subset of the even numbers (2*k) by using
                  the following result, which may be new - at any rate, I have not come across
                  it in a fairly extensive literature/web search on Bernoulli numbers.

                  Theorem: If k has any Sophie-Germain prime factor f, then the denominator of
                  B[2*k] is > 6.
                  Proof: Suppose f is a Sophie-Germain prime factor of k, 2 <= f <= k.
                  Then g = 2*f+1 is a prime, 5 <= g <= (2*k+1); and (g-1) = 2*f, which divides
                  (2*k), so the 1/g must be included in the sum, so the denominator of B[2*k]
                  must be divisible by the prime g >= 5, and so must be > 6.
                  QED

                  In particular, taking f=2 shows that we can restrict attention to 2*k = 2 MOD
                  4, i.e. k odd; and taking f=3 shows that 2*k = 2 or 4 MOD 6.

                  This test reduces the number of B[2*k] which need to be evaluated by about
                  50%, at least for the "small" k (<= 500) which I have looked at.

                  Note that the converse of this theorem is false. The first countexample
                  occurs (I believe) for k=119=7*17, as neither 7 nor 17 is a Sophie Germain,
                  yet B[238] has denominator 1434=6*239.

                  Mike Oakes


                  [Non-text portions of this message have been removed]
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