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

                                           n
                                         -----
                                          \                  4
     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



                                   4           4
                               (k!)  ((n - k)!)
                        -------------------------------
                                    4                 4
                        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
-4 (4 n + 5) (4 n + 3) (n + 1) A(n) - 2 (2 n + 3) (3 n  + 9 n + 7) A(n + 1)

              3
     + (n + 2)  A(n + 2) = 0



                             and in Maple notation



-4*(4*n+5)*(4*n+3)*(n+1)*A(n)-2*(2*n+3)*(3*n^2+9*n+7)*A(n+1)+(n+2)^3*A(n+2) = 0


                       Subject to the initial condition



A(1) = 2, A(2) = 18




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



-4*(4*n+5)*(4*n+3)*(n+1)*B(n)-2*(2*n+3)*(3*n^2+9*n+7)*B(n+1)+(n+2)^3*B(n+2) = C
(n)


                          with the initial conditions



B(1) = 70, B(2) = 3075/8


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





-(15*n^2+70*n+82)*(n+1)^2/(n+3)/(n+2)/(15*n^2+40*n+27)*C(n)+C(n+1) = 0




                          with the initial conditions



C(1) = 205/3


   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, 23.8111111416450402133520813239



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



                               23.80033, 23.80567

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



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



         This ends this paper that took, 29.824, seconds to generate.