Rational generating functions for the Certain Stanley-Stern Sums



                              By Shalosh B. Ekhad



                               Theorem Number, 1





                       ---------------------------------



          Let Z[n] be the integer sequence whose generating function is



                    infinity
                     -----                    3    2
                      \            j      -2 t  - t  + 1
                       )     Z[j] t  = ---------------------
                      /                  4    3    2
                     -----             -t  - t  - t  - t + 1
                     j = 0



                                      Let



            n - 1
           --------'
          '  |  |          Z[i + 3]    (Z[i + 2] + Z[i + 3])
F[n](x) =    |  |    (1 + x         + x
             |  |
             |  |
            i = 0

        (Z[i + 1] + Z[i + 2] + Z[i + 3])
     + x

        (Z[i] + Z[i + 1] + Z[i + 2] + Z[i + 3])
     + x                                       )



                                    Write:



                                   infinity
                                    -----
                                     \               i
                         F[n](x) =    )     a(n, i) x
                                     /
                                    -----
                                    i = 0



                                     Let :



                                   infinity
                                    -----
                                     \
                            H(n) =    )     a(n, k)
                                     /
                                    -----
                                    k = 0



                                      Then



                                infinity
                                 -----
                                  \            n
                                   )     H(n) t
                                  /
                                 -----
                                 n = 0

                            equals (I-Mt)^(-1) v [1]



               where M is a certain square matrix of dimension, 1



                     and v is a certain vector of length, 1



that are too big to display. At any rate we can use them to find the first, 31,

    terms  starting at n=0, for the sake of the OEIS.





[1, 5, 25, 125, 625, 3125, 15625, 78125, 390625, 1953125, 9765625, 48828125,

    244140625, 1220703125, 6103515625, 30517578125, 152587890625, 762939453125,

    3814697265625, 19073486328125, 95367431640625, 476837158203125,

    2384185791015625, 11920928955078125, 59604644775390625, 298023223876953125,

    1490116119384765625, 7450580596923828125, 37252902984619140625,

    186264514923095703125, 931322574615478515625]



                         -----------------------------



                          This took, 0.006, seconds.



                               Theorem Number, 2





                       ---------------------------------



          Let Z[n] be the integer sequence whose generating function is



                    infinity
                     -----                    3    2
                      \            j      -2 t  - t  + 1
                       )     Z[j] t  = ---------------------
                      /                  4    3    2
                     -----             -t  - t  - t  - t + 1
                     j = 0



                                      Let



            n - 1
           --------'
          '  |  |          Z[i + 3]    (Z[i + 2] + Z[i + 3])
F[n](x) =    |  |    (1 + x         + x
             |  |
             |  |
            i = 0

        (Z[i + 1] + Z[i + 2] + Z[i + 3])
     + x

        (Z[i] + Z[i + 1] + Z[i + 2] + Z[i + 3])
     + x                                       )



                                    Write:



                                   infinity
                                    -----
                                     \               i
                         F[n](x) =    )     a(n, i) x
                                     /
                                    -----
                                    i = 0



                                     Let :



                             infinity
                              -----
                               \
                      H(n) =    )     a(n, k) a(n, k + 1)
                               /
                              -----
                              k = 0



                                      Then



                                infinity
                                 -----
                                  \            n
                                   )     H(n) t
                                  /
                                 -----
                                 n = 0

                            equals (I-Mt)^(-1) v [1]



             where M is a certain square matrix of dimension, 1490



                   and v is a certain vector of length, 1490



that are too big to display. At any rate we can use them to find the first, 31,

    terms  starting at n=0, for the sake of the OEIS.





[0, 4, 63, 868, 11233, 145036, 1896034, 24702594, 320939693, 4168509507,

    54161394007, 703690375378, 9141431570591, 118745454279821, 1542455665520992,

    20035967844104422, 260258398681159859, 3380598151181562105,

    43911614329670982982, 570381048569505299346, 7408847766971480584012,

    96235493526207790122176, 1250026616015747916526358,

    16236897964945427549673250, 210904998373441401920840498,

    2739495053377290644149593887, 35583941330076637030005311418,

    462208066131520527815955631320, 6003727735438413886765517794080,

    77983806894072688210608805413357, 1012949635313850740084133122142035]



                         -----------------------------



                         This took, 188.849, seconds.





                   -----------------------------------------



      This concludes this article that took, 239.528, seconds to produce.