2
The Apery Constant Inspired by the coefficient of , t ,

     in the Zudilin-Straub transform of,

      n
    -----
     \
      )   binomial(n, k) binomial(n + k, 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 ,

     2
    t ,  in the following quantity



                                             2
                        RF(1 + t, n + k) (k!)  (n - k)!
                    ---------------------------------------
                                2
                    RF(1 + t, k)  RF(1 - t, n - k) (n + k)!



                  1/2
                 2    n
   and k equals, ------, that in floating-point is, 0.70710678118654752440 n
                   2



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



           (n + 1) A(n) + (-6 n - 9) A(n + 1) + (n + 2) A(n + 2) = 0



                             and in Maple notation



(n+1)*A(n)+(-6*n-9)*A(n+1)+(n+2)*A(n+2) = 0


                       Subject to the initial condition



A(1) = 3, A(2) = 13




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



(n+1)*B(n)+(-6*n-9)*B(n+1)+(n+2)*B(n+2) = C(n)


                          with the initial conditions



B(1) = 3, B(2) = 69/4


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





-2*(n+1)*(21881459371394930680052*n^3+185707271237033180603765*n^2+
462510605143354924832742*n+287346091083506801360925)*(3+2*n)^2/(n+5)/(
5195974367286240732095*n^3+50480267836210758755222*n^2+151875453836769500089538
*n+140567016521152468356999)/(6+n)^2*C(n)+(179441922580159316447104*n^6+
1569792273476603943975468*n^5-523587211358108796729190*n^4-\
47648870347684265979593089*n^3-195503664853490031926413804*n^2-\
317904077383696722576867585*n-187718197628604115039897476)/(n+5)/(
5195974367286240732095*n^3+50480267836210758755222*n^2+151875453836769500089538
*n+140567016521152468356999)/(6+n)^2*C(n+1)+1/2*(52473635287185179020092*n^6+
3311190153395584865202363*n^5+46619044862356341290874091*n^4+
287580723084879249399079559*n^3+895521339568144994051674661*n^2+
1377550071595020209833995042*n+828013724834069863848126744)/(n+5)/(
5195974367286240732095*n^3+50480267836210758755222*n^2+151875453836769500089538
*n+140567016521152468356999)/(6+n)^2*C(n+2)-1/2*(71646079239757402497658*n^5+
1707977555794646280584937*n^4+14778302332559848724947974*n^3+
59169020325727004755830659*n^2+110500568361712478106966612*n+
76970000311622216619797952)/(5195974367286240732095*n^3+50480267836210758755222
*n^2+151875453836769500089538*n+140567016521152468356999)/(6+n)^2*C(n+3)+C(n+4)
= 0




                          with the initial conditions



C(1) = 22, C(2) = 1037/12, C(3) = 10483/30, C(4) = 42247/30


   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, 1.64493406684822643647241516665



                                                    2
                                                  Pi
                           This is most probably, ---
                                                   6



#The direct computation of the approximation was terminated.