4
The Apery Constant Inspired by the coefficient of , t ,

                                           n
                                         -----
                                          \                  3
     in the Zudilin-Straub transform of,   )   binomial(n, k)
                                          /
                                         -----
                                         k = 0



                              By Shalosh B. Ekhad



      Let RF(t,k), as usual, be the raising factorial, t*(t+1)...(t+k-1)



We are interested in the limit, as n goes to infinity, of the coefficient of ,

     4
    t ,  in the following quantity



                                   3           3
                               (k!)  ((n - k)!)
                        -------------------------------
                                    3                 3
                        RF(1 + t, k)  RF(1 - t, n - k)



     and k equals, n/2, that in floating-point is, 0.50000000000000000000 n



Let's call this constant c (it can also be described directly in terms of ha\
    rmonic-type expressions). Using this definition the convergence is extre\
    mely slow.



    We will show how to compute it much faster, with exponential error-rate



     Let A(n) be the sequence of integers satisfying the linear recurrence



              2             2                                2
    -8 (n + 1)  A(n) + (-7 n  - 21 n - 16) A(n + 1) + (n + 2)  A(n + 2) = 0



                             and in Maple notation



-8*(n+1)^2*A(n)+(-7*n^2-21*n-16)*A(n+1)+(n+2)^2*A(n+2) = 0


                       Subject to the initial condition



A(1) = 2, A(2) = 10




    Let  B(n) be the sequence satisfying the INHOMOGENEOUS recurrence i.e.



-8*(n+1)^2*B(n)+(-7*n^2-21*n-16)*B(n+1)+(n+2)^2*B(n+2) = C(n)


                          with the initial conditions



B(1) = 30, B(2) = 1083/8


             where the right side, C(n), satisfies the recurrence





-(2253*n+7454)*(n+1)*(n+2)^2/(n+4)/(1049*n+2346)/(5+n)^2*C(n)+(n+3)*(n+2)*(5555
*n^2+30307*n+37590)/(n+4)/(1049*n+2346)/(5+n)^2*C(n+1)-(n+3)*(4351*n^2+25938*n+
35616)/(1049*n+2346)/(5+n)^2*C(n+2)+C(n+3) = 0




                          with the initial conditions



C(1) = 154/3, C(2) = 1057/24, C(3) = 387/10


   Note that it is very fast to compute many terms of C(n) and hence of B(n)



             The ratios B(n)/A(n) tend very fast to the constant c



            Here it is to 30 digits, 13.7996212298170119418290471309



To illustrate how fast the convergence is, and also as check, let's compute \
    the n=5000 and n=10000 values of the above sequence, whose limit is our \
    C



                               13.79374, 13.79669

     as you can see the convergence is extremely slow using the definition



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         This ends this paper that took, 21.570, seconds to generate.