derivation of generating function for the reciprocal central binomial coefficients

According to the article, the ordinary generating function for (2nn)-1 is


To see this, let Cn=(2nn)-1, and C(x)=n0Cnxn its ordinary generating function. Then

Cn+1 =(2n+2n+1)-1=(n+1)!(n+1)!(2n+2)!



so that


A little algebra gives


so that


and, collecting terms,


We now have a first-order linear ODE to solve. Put it in the form


and we must now integrate the coefficient of C(x). Expand by partial fractions and integrate to get


Thus the solution to the equation is

C(x) =x(4-x)3/2(k+(4-x)3/2x-2x(4-x)𝑑x)

To determine the constant k, note that we should have C(x)|x=0=12; looking at limx0C(x) we see that for k=π this equation holds. Thus


We show below that the following is an identity:


Assuming that result, substitute x2-1 for z and simplify to get


so that


and then

C(x) =44-x+x(4-x)3/2(4arcsin(x2))

as desired.

Finally, to prove the identity, first expand the right-hand using the formula for sin(a+b), and then apply the half-angle formulas:

sin(π4+12arcsin(z)) =22(cos(12arcsin(z))+sin(12arcsin(z)))

Now square this expression to get


Thus the identity holds for 0z1; an almost identical computation using -z in of z shows that it also holds for -1z0.

Title derivation of generating functionMathworldPlanetmath for the reciprocal central binomial coefficientsMathworldPlanetmath
Canonical name DerivationOfGeneratingFunctionForTheReciprocalCentralBinomialCoefficients
Date of creation 2013-03-22 19:04:58
Last modified on 2013-03-22 19:04:58
Owner rm50 (10146)
Last modified by rm50 (10146)
Numerical id 4
Author rm50 (10146)
Entry type Result
Classification msc 05A10
Classification msc 05A15
Classification msc 05A19
Classification msc 11B65