The Hankel determinants of the Catalan numbers are well known and can be written as $d(k,n)= \det \left( C_{k + i + j} \right)_{i,j = 0}^{n - 1}=\prod_{i=1}^{k-1}\frac{\binom{2n+2j}{j}}{\binom{2j}{j}}$ with $d(k,0)=1.$

Computations suggest that $$D_k(x)=\sum_{n\geq 0}d(k,n)x^n=\frac{A_{k}(x)}{(1-x)^{\binom{k}{2}+1}}$$ where $A_{k}(x)$ is a palindromic and unimodal polynomial of degree $\binom{k-1}{2}.$

Moreover it seems that $A_{k}(x)$ is gamma-nonnegative, i.e. a linear combination of polynomials $x^j(1+x)^{\binom{k-1}{2}-2j}$ with positive coefficients.

Is this known? Any idea how to prove this?


1 Answer 1


I'm upgrading my comments to an answer.

As I've mentioned in comments/answers to some of your previous MO questions (e.g. Number of bounded Dyck paths with negative length as Hankel determinants and Some nice polynomials related to Hankel determinants), this Hankel determinant of Catalan numbers counts fans of nested Dyck paths (see, e.g., Section 3.1.6 of https://arxiv.org/abs/1409.2562), which are the same as plane partitions of staircase shape with bounded entries, which were first counted by Proctor (see his "Odd symplectic groups" paper https://doi.org/10.1007/BF01404455).

At any rate, this interpretation means that your generating function $D_k(x)$ is the same as the generating function $\sum_{m \geq 0}\Omega_P(m)x^m$ where $\Omega_P(m)$ counts the number of order preserving maps $P\to \{0,1,\ldots,m\}$ for a certain poset $P$ (namely, the staircase partition shape poset; equivalently, the Type A root poset). The general theory of $P$-partitions, as developed by Stanley, thus says that $D_k(x) = \sum_{L} x^{\mathrm{des}(L)}/(1-x)^{\binom{k}{2}-1}$, where the sum is over linear extensions of $P$, when it is naturally labeled (see e.g. Theorem 3.15.8 of EC1). Thus your $A_k(x)$ is the $P$-version of an Eulerian number generating function, and a general result of Brändén (see https://doi.org/10.37236/1866 or https://arxiv.org/abs/1410.6601) says that such poset linear extension descent genearting functions are $\gamma$-nonnegative as long as the poset in question is graded, which it is in this case.

  • 3
    $\begingroup$ Great answer! I knew all ingredients but yet didn't make the connection before reading it... $\endgroup$ Aug 30, 2021 at 6:23
  • $\begingroup$ @Sam Hopkins: Thank you for this information. There is an analogous situation for the Hankel determinants $c(k,n)= \det \left( b_{k + i + j} \right)_{i,j = 0}^{n - 1}$ of $b(n)= \binom{n}{\lfloor{\frac{n}{2}}\rfloor}.$ Let $C_k(x)=\sum_{n\geq 0}c(2k,n)x^n=\frac{B_{k}(x)}{(1-x)^{k^2+1}}$ be the generating function of $c(2k,n)=\prod_{i=1}^n\prod_{j=0}^{k-1}\frac{k+i+j}{i+j}.$ Then it seems that $B_k(x)$ is also gamma positive. Is there an analogous interpretation? $\endgroup$ Aug 30, 2021 at 6:45
  • 1
    $\begingroup$ @JohannCigler: Yes, there is an analogous interpretation. This time $P$ will be the Type B root poset (alias "shifted double staircase shape"). The fact that this second Hankel determinant counts such bounded $P$-partitions is closely related to the fact that so-called "Dyck path prefixes" are counted by $\binom{n}{\lfloor n/2 \rfloor}$ (see e.g. arxiv.org/abs/1406.1709); briefly, we now consider fans of Dyck path prefixes. Incidentally, these kind of bounded $P$-partitions were again first counted by Proctor, in doi.org/10.2307/2045516. $\endgroup$ Aug 30, 2021 at 11:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.