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
                  -----                   4      3    2
                   \            j     -3 t  - 2 t  - t  + 1
                    )     Z[j] t  = --------------------------
                   /                  5    4    3    2
                  -----             -t  - t  - t  - t  - t + 1
                  j = 0



                                      Let



            n - 1
           --------'
          '  |  |          Z[i + 4]    (Z[i + 3] + Z[i + 4])
F[n](x) =    |  |    (1 + x         + x
             |  |
             |  |
            i = 0

        (Z[i + 2] + Z[i + 3] + Z[i + 4])
     + x

        (Z[i + 1] + Z[i + 2] + Z[i + 3] + Z[i + 4])
     + x

        (Z[i] + Z[i + 1] + Z[i + 2] + Z[i + 3] + Z[i + 4])
     + 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, 6, 36, 216, 1296, 7776, 46656, 279936, 1679616, 10077696, 60466176,

    362797056, 2176782336, 13060694016, 78364164096, 470184984576,

    2821109907456, 16926659444736, 101559956668416, 609359740010496,

    3656158440062976, 21936950640377856, 131621703842267136, 789730223053602816,

    4738381338321616896, 28430288029929701376, 170581728179578208256,

    1023490369077469249536, 6140942214464815497216, 36845653286788892983296,

    221073919720733357899776]



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



                          This took, 0.007, seconds.



                               Theorem Number, 2





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



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



                 infinity
                  -----                   4      3    2
                   \            j     -3 t  - 2 t  - t  + 1
                    )     Z[j] t  = --------------------------
                   /                  5    4    3    2
                  -----             -t  - t  - t  - t  - t + 1
                  j = 0



                                      Let



            n - 1
           --------'
          '  |  |          Z[i + 4]    (Z[i + 3] + Z[i + 4])
F[n](x) =    |  |    (1 + x         + x
             |  |
             |  |
            i = 0

        (Z[i + 2] + Z[i + 3] + Z[i + 4])
     + x

        (Z[i + 1] + Z[i + 2] + Z[i + 3] + Z[i + 4])
     + x

        (Z[i] + Z[i + 1] + Z[i + 2] + Z[i + 3] + Z[i + 4])
     + x                                                  )



                                    Write:



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



                                     Let :



                                   infinity
                                    -----
                                     \             2
                            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, 12751



                   and v is a certain vector of length, 12751



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, 6, 118, 2332, 43040, 776182, 14096160, 258863060, 4759479794, 87307373274,

    1599583984610, 29306925519860, 537073862300162, 9844882125355244,

    180455814972656336, 3307371523093715084, 60614841687014981498,

    1110886864454334276802, 20359841497444717922446, 373148858318572700506456,

    6838860279949142328789470, 125337543981460140154985380,

    2297083930693040153209146546, 42099186031615964836943478146,

    771563281389973674073809026598, 14140623933328946509238287954584,

    259157975853975433204646533550570, 4749635199606267509604698368205096,

    87047411387014608684874079375400224, 1595334298122534586383857294138576980,

    29237983733568975046238512191348334982]



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



                         This took, 4681.837, seconds.





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     This concludes this article that took, 21811.677, seconds to produce.