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{\bf Two Hankel Determinants dear to Volodia Retakh }
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\centerline
{\it Shalosh B. EKHAD and Doron ZEILBERGER}
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\qquad\qquad{\it Dedicated to Vladimir Retakh (b. May 20, 1948) on his 70th birthday}
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Let $B_s:=\frac{(2s)!}{s!^2}$, then
$$
\det \, (B_{r+i+j})_{0 \leq i,j \leq n-1} \, =
$$
$$
 \left( {\frac { \left( 2\,r \right) !}{ \left( r! \right) ^{2}}} \right) ^{n}\prod _{n_{{1}}=1}^{n} \left( \prod _{n_{{2}}=2}^{n_{{1}}}4\,{\frac { \left( n_{{2}}-1 \right)  \left( 2\,n_{{2}}-3
 \right)  \left( 2\,n_{{2}}-3+2\,r \right)  \left( n_{{2}}-2+r \right) }{ \left( 2\,n_{{2}}-2+r \right)  \left( 2\,n_{{2}}-4+r \right)  \left( 2\,n_{{2}}-3+r \right) ^{2}}} \right) \quad .
$$

Let $C_s:=\frac{(2s)!}{s!(s+1)!}$, then
$$
\det \, (C_{r+i+j})_{0 \leq i,j \leq n-1} \, =
$$
$$
 \left( {\frac { \left( 2\,r \right) !}{r!\, \left( r+1 \right) !}} \right) ^{n}\prod _{n_{{1}}=1}^{n} \left( \prod _{n_{{2}}=2}^{n_{{1}}}4\,{\frac { \left( n_{{2}}-1 \right)  \left( 2\,n_{{2}}-1
 \right)  \left( n_{{2}}-1+r \right)  \left( 2\,n_{{2}}-3+2\,r \right) }{ \left( 2\,n_{{2}}-1+r \right)  \left( 2\,n_{{2}}-3+r \right)  \left( 2\,n_{{2}}-2+r \right) ^{2}}} \right) 
$$


{\bf Proofs}

{\bf First Proof of both determinants} (by SBE). Go to the Maple package

{\tt http://sites.math.rutgers.edu/\~{}zeilberg/tokhniot/CLD} \quad ,

and type

{\tt EvalHpaper((2*r)!/(r!*r!),n,r,4);} \quad ,

and

{\tt EvalHpaper((2*r)!/(r!*(r+1)!),n,r,4);} \quad ,

respectively, getting, {\it ab initio}, the above expressions, together with  proofs, using Dodgson's rule and induction. See [Z].



{\bf Sketch of the Second Proof of the First determinant} (by DZ)

We use the fact that
$$
\frac{(2r)!}{r!^2} = ConstantTerm_x (\frac{1}{x} + 2 + x)^r \quad .
$$
Let $f(x)=\frac{1}{x}+2+x$. The determinant is the Constant term, in $x_1, \dots , x_n$, of
$$
\det \, ( f(x_i)^{r+i+j-2} )_{1 \leq i, j \leq n} \quad .
$$
$$
= \, ConstantTerm_{x_1, \dots , x_n} \prod_{i=1}^{n} f(x_i)^r \, 
\det \, ( f(x_i)^{i+j-2} )_{1 \leq i, j \leq n} \quad .
$$

By Vandermonde (also dear to Volodia Retakh!), this equals
$$
ConstantTerm_{x_1, \dots , x_n} \prod_{i=1}^{n} f(x_i)^r \,  \prod_{i=1}^{n} f(x_i)^{i-1} \, 
\prod_{1 \leq i<j \leq n} (f(x_i)-f(x_j)) \quad.
$$
Symmetrizing, and using Vandermonde one more time, this equals
$$
\frac{1}{n!} \,CT_{x_1, \dots , x_n} \prod_{i=1}^{n} f(x_i)^r \,
\prod_{1 \leq i<j \leq n} (f(x_i)-f(x_j))^2 \quad.
$$
Now, {\it lo and behold}, this is a special case $k_1=r$,  $k_2=1$, $k_3=0$,
of the $BC_n$ Macdonald Constant-Term  Identity (see, e.g., [FW], top of p. 501).


{\bf Sketch of the Second Proof of the Second determinant} (by DZ)

We use the fact that
$$
\frac{(2r)!}{r!(r+1)!} = ConstantTerm_x (1-x)(\frac{1}{x} + 2 + x)^r \quad .
$$
The determinant is the constant term, in $x_1, \dots , x_n$, of
$$
\prod_{i=1}^{n} (1-x_i) \, \det \, ( f(x_i)^{r+i+j-2} )_{1 \leq i, j \leq n} \quad .
$$
This equals
$$
ConstanTerm_{x_1, \dots , x_n} \prod_{i=1}^{n} (1-x_i) \, \prod_{i=1}^{n} f(x_i)^r \, 
\det \, ( f(x_i)^{i+j-2} )_{1 \leq i, j \leq n} \quad .
$$
By Vandermonde, this equals
$$
ConstantTerm_{x_1, \dots , x_n}   \prod_{i=1}^{n} (1-x_i) \prod_{i=1}^{n} f(x_i)^r \,  \prod_{i=1}^{n} f(x_i)^{i-1} \, \,
\prod_{1 \leq i<j \leq n} (f(x_i)-f(x_j)) \quad.
$$
Symmetrizing this equals
$$
\frac{1}{n!} \,ConstantTerm_{x_1, \dots , x_n}  \prod_{i=1}^{n} (1-x_i) \, \prod_{i=1}^{n} f(x_i)^r \,
\prod_{1 \leq i<j \leq n} (f(x_i)-f(x_j))^2 \quad.
$$
Now, {\it lo and behold}, this is a special case $k_1=r-\frac{1}{2}$, $k_2=\frac{1}{2}$, $k_3=1$,
of the $BC_n$ Macdonald Constant-Term  Identity mentioned above  ([FW], top of p. 501).

\vfill\break

{\bf Challenges for Volodia}

In the beautiful article [BR], the authors prove a non-commutative analog, using quasi-determinants, of
the special cases $r=0$ and $r=1$ of the second determinant (where the determinant happens to be $1$, in the
commutative case). What is the non-commutative analog, for general $r$, of the second determinant?
The first determinant?


\bigskip

{\bf References}

[BR] Arkady Berenstein and Vladimir Retakh, {\it Noncommutative Catalan Numbers}, \hfill\break
{\tt https://arxiv.org/abs/1708.03316} \quad .

[FW] Peter J. Forrester and S. Ole Warnaar , {\it The importance of the Selberg integral},
Bull. Amer. Math. Soc. {\bf 45} (2008), 489-534 \hfill\break
{\tt http://www.ams.org/journals/bull/2008-45-04/S0273-0979-08-01221-4/} \quad .

[Z] Doron Zeilberger,  {\it Lieber Opa Paul, Ich Bin Auch Ein Experimental Scientist!},
 Adv. Applied Math. {\bf 31} (2003), 532-543, \hfill\break
{\tt http://sites.math.rutgers.edu/\~{}zeilberg/mamarim/mamarimhtml/opa.html } \quad .

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\hrule
\bigskip
Shalosh B. Ekhad, c/o D. Zeilberger, Department of Mathematics, Rutgers University (New Brunswick), Hill Center-Busch Campus, 110 Frelinghuysen
Rd., Piscataway, NJ 08854-8019, USA. \hfill\break
Email: {\tt ShaloshBEkhad at gmail dot com}   \quad .
\bigskip
Doron Zeilberger, Department of Mathematics, Rutgers University (New Brunswick), Hill Center-Busch Campus, 110 Frelinghuysen
Rd., Piscataway, NJ 08854-8019, USA. \hfill\break
Email: {\tt DoronZeil at gmail  dot com}   \quad .
\bigskip
First Written: May 20, 2018. This version: July 5, 2018.
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