Generating function for the squared Catalan numbers

catalan-numberselliptic integralsgenerating-functionshadamard-producthypergeometric function

The squares of the Catalan numbers: 1, 1, 4, 25, 196, 1764… are given in OEIS A001246. In the OEIS entry two ordinary generating functions for the series are given in terms of elliptic integrals/hypergeometric functions:

$$-1/4x+(16x-1)\textrm{EllipticK}(4\sqrt{x})/2x\pi + \textrm{EllipticE}(4\sqrt{x})/\pi x,$$ and
$${}_3F_2( (1, 1/2, 1/2); (2, 2); 16x) = (-1 + {}_2F_1( (-1/2, -1/2); (1); 16x))/(4x). $$

Unfortunately I cannot find a derivation for either of these expressions. The squared Catalan numbers are given by the Hadamard product of the Catalan number generating funcion with itself, so one approach would be to use the integral form of the Hadamard product, but I am not sure how to evaluate this integral, and Mathematica also struggles (even though it would seems the first expression was probably derived using Mathematica given the notation?)

Were these expressions derived by evaluating such an integral, or perhaps appealing to the recursion relations satisfied by the Catalan numbers like the proof for the g.f. for the Catalan numbers themselves?

Best Answer

As a starting point, use the definition of a generalized hypergeometric series

$$ {}_pF_q( a, b; x) := \sum_{k=0}^\infty \frac{(a_1)_k\cdots(a_p)_k} {(b_1)_k\cdots(b_q)_k}\cdot\frac{x^k}{k!} \tag1 $$

and verify the desired equality of the ordinary generating function of the square of the Catalan numbers with a generalized hypergeometric series

$$ {}_3F_2( (1, 1/2, 1/2); (2, 2); 16x) = \sum_{k=0}^\infty \, (C_n)^2 x^k \tag2 $$

using the Wolfram Mathematica code

HypergeometricPFQn[a_List, b_List, z_, n_] := (
    z^n/n! Product[Pochhammer[ap, n], {ap, a}] /
    Product[Pochhammer[bq, k], {bq, b}] );
HypergeometricPFQn[ {1, 1/2, 1/2}, {2, 2}, 16*x, n]/16^n ==
    x^n/2^(4n) CatalanNumber[n]^2 //FunctionExpand
(* True *)

which really only depends on the gamma function identity

$$ C_n = \frac{4^n \Gamma(n+1/2)}{\pi^{1/2}\Gamma(n+2)}. \tag3 $$

Next, use the OEIS to identify the generating function of OEIS A002894 as

$$ \frac{\text{EllipticK}(m)}{\pi/2} = \sum_{n=0}^\infty {2n\choose n}^2x^n \tag4 $$

and the generating function of OEIS A000891 as

$$ \frac1{4x} - \frac{\text{EllipticE}(m)}{2\pi x} = \sum_{n=0}^\infty \,(2n+1)(C_n)^2 x^n \tag5 $$

where, as usual,

$$ m = k^2 = 16x \tag6 $$

denotes the parameter. Now use the binomial coefficient identity

$$ (4n+3)(C_n)^2=4{2n \choose n}^2-\frac14 {2n+2 \choose n+1}^2 \tag7 $$

to verify that

$$ \frac1{4x} + \frac{\text{EllipticK}(m)(m-1)}{2\pi x} = \sum_{n=0}^\infty (4n+3)(C_n)^2 x^n. \tag8 $$

Combine equation $(5)$ and $(7)$ to get

$$ -\frac1{4x} + \frac{\text{EllipticK}(m)(m-1)}{2x\pi} + \frac{\text{EllipticE}(m)}{\pi x} = \sum_{n=0}^\infty (C_n)^2 x^n. \tag9 $$

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