Theorem:

        (Cojectured and alomost proved by Paul Heideman and Emilie Hogan,

      in undergraduate research advised by Jim Propp, at the U. of Wisc.)



                           Let K be a positive integer

      Define an integer sequence a(n) (n>=0) by the non-linear recurrence



                     a(n - 1) a(n - 2 K) + a(n - K) + a(n - K - 1)
              a(n) = ---------------------------------------------
                                    a(n - 2 K - 1)

             subject to the initial conditions a(i)=1 for 0<=i<=2K

                 Then lo-and-behold all the terms are integers



                  Completion of the proof by Shalosh B. Ekhad

               Let b(n) be the sequence annihilated by the linear

                 recurrence equation with constant coefficients

                      2                            2
-b(n) + (4 + 8 K + 2 K ) b(n + K) + (-4 - 8 K - 2 K ) b(n + 2 K) + b(n + 3 K)

     = 0

                     With the following initial conditions

                         For n in the discrete interval

                                  [1, 1 + 2 K]

                            it equals the polynomial

                                       1

                         For n in the discrete interval

                               [2 + 2 K, 1 + 3 K]

                            it equals the polynomial

                                 -1 + 2 n - 4 K

                         For n in the discrete interval

                               [2 + 3 K, 1 + 4 K]

                            it equals the polynomial

                                  2                      2
                     1 - 2 n + 2 n  + 8 K - 12 n K + 18 K

                         For n in the discrete interval

                               [2 + 4 K, 1 + 5 K]

                            it equals the polynomial

                          2                      2        2       3
            -3 + 2 n + 2 n  - 16 K - 4 n K - 18 K  + 4 n K  - 16 K

                         For n in the discrete interval

                               [2 + 5 K, 1 + 6 K]

                            it equals the polynomial

                       4      2       2        2  2         3        3
3 - 10 n + 60 K + 100 K  + 8 n  + 16 n  K + 4 n  K  - 40 n K  + 424 K

              2        2
     - 164 n K  + 296 K  - 96 n K

            Note that b(n) is well-defined and manifestly an integer

                                 (by induction)

                             I claim that b(n)=a(n).

        All we have to prove is that b(n) satisfies the same non-linear

              recurrence as a(n), with the same initial conditions

        The initial conditions are right, of course, and Paul and Emilie

                          have already proved that if

    b(n) b(n - 2 K - 1) - b(n - 1) b(n - 2 K) - b(n - K) - b(n - K - 1) = 0

               is true for n<=6K, then it is true for ever after.

                   So it only remains to prove this for n<=6K

                       But this is a routine verification

       using the above explicit expressions for b(n), and is indeed true.

                                       QED

               The whole thing took, 37.542, seconds of CPU time