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Re: Lehmer sequence puzzle

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  • Maximilian Hasler
    ... Just exclude oo explicitly... (What is the largest finite value...). An hint through an experimental approach: u(R,Q,N=9)={
    Message 1 of 33 , May 7, 2009
      > > For integer Q > 0 and integer R
      > > with both R and R-4*Q non vanishing,
      > > let u(R,Q) be the number of integers n > 0 for which
      > > (x^n+y^n)/(x+y) is a unit, with
      > > x = sqrt(R)/2 + sqrt(R-4*Q)/2,
      > > y = sqrt(R)/2 - sqrt(R-4*Q)/2.
      > > What is the maximum value of u(R,Q) when R and Q
      > > range over all the integers allowed above?
      > >
      > > The solution remains very simple: there is no such maximum!

      > I posed the puzzle because I have found, experimentally, examples where u(R,Q) is 1, 2 and even (a great surprise to me!) 3.
      >
      > I want to know if there are any examples >3 (and <oo).
      >
      > Help me, please, to both formulate the puzzle securely and solve it!

      Just exclude oo explicitly... (What is the largest finite value...).

      An hint through an experimental approach:

      u(R,Q,N=9)={ my(x=sqrt(R)/2+sqrt(R-4*Q)/2,y=sqrt(R)/2-sqrt(R-4*Q)/2);sum(n=1,N,round((x^n+y^n)/(x+y))==1)};
      ? matrix(20,20,R,Q,u(R,Q))
      %415 =
      [3 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1]
      [4 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1]
      [2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1]
      [9 1 2 1 1 1 2 1 1 1 1 1 1 1 1 1 1 1 1 1]
      [2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1]
      [1 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1]
      [1 3 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1]
      [1 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1]
      [1 1 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1]
      [1 1 3 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1]
      [1 1 2 2 1 1 1 2 1 1 1 1 1 1 1 1 1 1 1 1]
      [1 1 1 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1]
      [1 1 1 3 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1]
      [1 1 1 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1]
      [1 1 1 1 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1]
      [1 1 1 1 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1]
      [1 1 1 1 1 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1]
      [1 1 1 1 1 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1]
      [1 1 1 1 1 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1]
      [1 1 1 1 1 1 2 2 1 1 1 1 1 1 1 1 1 1 1 1]
      The numbers in column 1, rows 1..4, grow if you give a larger 3rd argument (as would (Q,R)=(0,1)), but not the others.

      Maximilian, per proxy SEAPS
      (Society for Experimental Approaches to Puzzle Solving)
    • David Broadhurst
      ... Yes, Mike, that last form is by far the neatest, if one removes the unnecessary lucas , which should be well understood, when one speaks of U or V . 1)
      Message 33 of 33 , May 10, 2009
        --- In primenumbers@yahoogroups.com, "Mike Oakes"
        <mikeoakes2@...> wrote [with unfailing courtesy]:

        > lucasU(338,25,11584)-5*lucasU(338,25,11583)
        > I guess Chris would expect the last of these as being the most
        > "canonical"? And I concur, it being also the shortest.

        Yes, Mike, that last form is by far the neatest,
        if one removes the unnecessary "lucas", which should
        be well understood, when one speaks of "U" or "V".

        1) François Édouard Anatole Lucas defined the integer sequence
        U(P,Q,n) = P*U(P,Q,n-1) - Q*U(P,Q,n-2),
        with U(P,Q,0) = 0 and U(P,Q,1) = 1,
        in terms of elementary arithmetic.

        2) I like to imagine that Derrick Norman Lehmer (1867-1938) saw that
        U(P,Q^2,2*k+1) = U(P,Q^2,k+1)^2 - (Q*U(P,Q^2,n))^2
        and told his son: "Go look at the factors for your Ph.D."

        3) Certainly, Derrick Henry Lehmer (1905-1991) did study
        U(P,Q^2,k+1) - Q*U(P,Q^2,k)
        being very well aware of the super-Lucasian extension to powers
        of algebraic numbers of degree 4, in the wider complex plane.

        4) Mike Oakes has found that the Lehmer number
        U(P,Q^2,k+1) - Q*U(P,Q^2,k)
        is probably prime, when P = 338, Q = 5, k = 11583.

        5) David Broadhurst is able to characterize this
        circumstance without using a dirty 4-letter word :-)

        Thank ye, kindly, Sir, for this interesting thread!

        David, pp SSSR
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