The first factorization discovered (and "proved") in James Sellers' paper

           Domino Tilings and Products of Fibonacci and Pell Numbers

published in        Journal of Integer Sequences, Vol. 5 (2002), Article 02.\
    1.2, is:

          Theorem: Let F(n) be the sequence defined by the recurrence

             F(n) = 2 F(n - 1) + 7 F(n - 2) + 2 F(n - 3) - F(n - 4)

                       subject to the initial conditions

                   F(0) = 2, F(1) = 10, F(2) = 36, F(3) = 145



  Let G(n) be the sequence defined on non-negative integers by the recurrence

                           G(n) = G(n - 1) + G(n - 2)

                       subject to the initial conditions

                               G(0) = 1, G(1) = 2



               Let H(n) be the sequence defined by the recurrence

                          H(n) = 2 H(n - 1) + H(n - 2)

                       subject to the initial conditions

                               H(0) = 2, H(1) = 5

          Then the following is true for every non-negative integer n

                                F(n) = G(n) H(n)



           Proof: Routine! (since everything in sight is C-finite) .

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      The second factorization discovered (and "proved") in that paper is:

          Theorem: Let F(n) be the sequence defined by the recurrence

           F(n) = 15 F(n - 1) - 32 F(n - 2) + 15 F(n - 3) - F(n - 4)

                       subject to the initial conditions

                   F(0) = 1, F(1) = 8, F(2) = 95, F(3) = 1183



  Let G(n) be the sequence defined on non-negative integers by the recurrence

                          G(n) = 3 G(n - 1) - G(n - 2)

                       subject to the initial conditions

                               G(0) = 1, G(1) = 2



               Let H(n) be the sequence defined by the recurrence

                          H(n) = 5 H(n - 1) - H(n - 2)

                       subject to the initial conditions

                               H(0) = 1, H(1) = 4

          Then the following is true for every non-negative integer n

                                F(n) = G(n) H(n)



           Proof: Routine! (since everything in sight is C-finite) .

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The third factorization discovered (and whose "proof" was left to the reader\
    , since it was too "complicated") in that paper is:

          Theorem: Let F(n) be the sequence defined by the recurrence

F(n) = F(n - 1) + 76 F(n - 2) + 69 F(n - 3) - 921 F(n - 4) - 584 F(n - 5)

     + 4019 F(n - 6) + 829 F(n - 7) - 7012 F(n - 8) + 829 F(n - 9)

     + 4019 F(n - 10) - 584 F(n - 11) - 921 F(n - 12) + 69 F(n - 13)

     + 76 F(n - 14) + F(n - 15) - F(n - 16)

                       subject to the initial conditions

F(0) = 1, F(1) = 1, F(2) = 34, F(3) = 153, F(4) = 2245, F(5) = 14824,

    F(6) = 167089, F(7) = 1292697, F(8) = 12988816, F(9) = 108435745,

    F(10) = 1031151241, F(11) = 8940739824, F(12) = 82741005829,

    F(13) = 731164253833, F(14) = 6675498237130, F(15) = 59554200469113



  Let G(n) be the sequence defined on non-negative integers by the recurrence

                           G(n) = G(n - 1) + G(n - 2)

                       subject to the initial conditions

                               G(0) = 1, G(1) = 1



               Let H(n) be the sequence defined by the recurrence

H(n) = H(n - 1) + 25 H(n - 2) + 11 H(n - 3) - 47 H(n - 4) - 11 H(n - 5)

     + 25 H(n - 6) - H(n - 7) - H(n - 8)

                       subject to the initial conditions

H(0) = 1, H(1) = 1, H(2) = 17, H(3) = 51, H(4) = 449, H(5) = 1853, H(6) = 12853,

    H(7) = 61557

          Then the following is true for every non-negative integer n

                                F(n) = G(n) H(n)



           Proof: Routine! (since everything in sight is C-finite) .

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