Loading ...
Sorry, an error occurred while loading the content.

Re: Prime chains x-->Ax+B

Expand Messages
  • mikeoakes2
    ... Thanks for those refinements, David. As you know, my 4 came from the fact that the 3rd and 4th terms of the sequence are known to be odd; and I didn t
    Message 1 of 143 , Dec 1, 2010
    • 0 Attachment
      --- In primenumbers@yahoogroups.com, "djbroadhurst" <d.broadhurst@...> wrote:
      >
      > --- In primenumbers@yahoogroups.com,
      > "mikeoakes2" <mikeoakes2@> wrote:
      >
      > > For 1<=t<=10000, there are only the 3 we know of, namely:
      > > t=1,2,6.
      >
      > Mike was asking that
      > N = 3^t+4
      > M = (N-2)*(N-3)+3
      > be simultaneously prime. The Poisson mean for
      > such cases with t > x >> 1 may be estimated as
      > C/(2*log(3)^2)*sum(t>x,1/t^2) =~ C/(2*log(3)^2)/x
      > where C is a constant of order unity.
      > To determine C we would need to take account of the discriminant
      > D = -11 of the quadratic M = (N-2)*(N-3)+3. For example,
      > M is always coprime to 2, 3, 7, 13, 17, 19, 29 ...
      >
      > > Predicted number, from PNT, is:
      > > 2*zeta(2)/log(3)^2=(Pi/log(3))^2/3=2.72577237357
      > > which agrees nicely with the experimental value of 3.
      > > Exercise for the reader: derive this formula.
      >
      > To "derive" the formula we would need to make 2 rather poor choices:
      > 1) set C = 4, thus ignoring the discriminant, above,
      > 2) set x = 0, thus applying the PNT all the way down to N = 7.
      >
      > More meaningfully, I estimate that C =~ 5.5 and hence that
      > the probabilty of a 4th case, with t > 10^4, is about
      > 5.5/(2*log(3)^2)/10^4 =~ 0.00023.

      Thanks for those refinements, David.
      As you know, my "4" came from the fact that the 3rd and 4th terms of the sequence are known to be odd; and I didn't get round to including those other factors which increase the probability of primehood by a amall multiple, analogous to what happens for AP-k's.

      I have an analogous bound, considerably less than 1, for the expected number of plupluperfect maximal power chains (in my sense) of length L=6 (of which exactly one example is known).

      As L increases beyond 6, the expectation is << 1, as there are more and more powers of log(2), log(3), log(5) etc. in the denominator of the expression, and also a factor (L-2)! there. I reckon the series converges.

      It's relatively unusual in [prime] number theory to find a small finite set of objects, so I'm happy to proffer the following.

      Definition: A "power chain" of length L is a chain
      p[n+1] = A*p[n] + B, such that p[n] is prime,
      for n = 1 to L, with p[2] > p[1].

      Definition: A "maximal power chain" is a power chain with
      integers (A,B) for which there is a proof that no chain
      of greater length exists for that A.

      Defininition: L2(A) = (largest prime divisor of (A-1)) - 1.

      Conjecture: there are only 4 maximal power chains of length >= L2(A)+2.

      $$lots for anyone who can find a 5th one (not:-)

      Mike
    • djbroadhurst
      ... Suppose that we want to start with a square and get a square. Then we must solve the Diophantine equation y^2 = a*x^2 + b For any pair (a,b), Dario will
      Message 143 of 143 , Jan 7, 2011
      • 0 Attachment
        --- In primenumbers@yahoogroups.com,
        Kevin Acres <research@...> wrote:

        > x=a*x+b either is a square or has a square as a major factor.

        Suppose that we want to start with a square and get a square.
        Then we must solve the Diophantine equation
        y^2 = a*x^2 + b

        For any pair (a,b), Dario will tell us all the solutions:
        http://www.alpertron.com.ar/QUAD.HTM

        David
      Your message has been successfully submitted and would be delivered to recipients shortly.