# Holonomic generating function

Let $\lambda$ denote a hook of size $d$ and $c(\Box)$ the content of $\Box \in \lambda$. Let $\text{Hooks}(d)$ be the set of hooks with $d$ boxes. Define \begin{align} B(d)&= \frac{1}{d!h^{d-1}} \sum_{\lambda\in \text{Hooks}(d)} (-1)^{ht(\lambda)-1} \, \dim \lambda \,\color{red}{\prod_{\Box \in \lambda}(1-c(\Box)h)^m}\\ &= \frac{1}{dd!{h^{d-1}}}\sum_{\ell=1}^{d} (-1)^{(\ell+1)}\binom{d-1}{\ell-1}\prod_{i=1}^{d}(1-(\ell-i)h)^m \end{align} So $B(d)$ is a polynomial in $h$

I have noticed the following recursion for $B(d)$ for $m=2$. \begin{align} d(d+1)B(d)=(d-1)(4d-2)B(d-1)+h^2(d-1)^2(d-2)^2B(d-2) \end{align}

I want to prove existence of recursion of $B(d)$ for general $m$ so in the last post Rational generating function and recursion

I wanted to show that the generating function $F(x,h)=\sum B_d (h)x^d$ is rational. Now I realize that it might not necessarily be rational. My claim of recursion would be true if I could show the generating function to be holonomic.

Recently I am studying generating function with the polynomial coefficient. So, for example, $a_n=x^n$ and the generating function is $F(x,y)=\sum a_n y^n$. Is this a rational generating function or a holonomic generating function? Is there are a method to show it?

The polynomial ring $C[x]$ has two more interesting basis $x^{\underline{n}}:=x(x-1)\cdots(x-(n-1))$ and $x^{\overline{n}}:=x(x+1)\cdots(x+n-1)$

How can I show the generating function of $F'(x,y)=\sum x^{\underline{n}}y^n$ is holonomic? Same question for $F'(x,y)=\sum x^{\overline{n}}y^n$.

The reason I am interested to show the generating function $F'(x,y)$ is holonomic is that I feel $B(d)$ would be some linear combination of $h^{\overline{d}}$ or $h^{\underline{d}}$. It will help me to prove that generating function of $B(d)$ is holonomic. Please guide me if I make some sense or give me some direction of progress.

• I get $B(1)=1, B(2)=h, B(3)=h^2(5+h^2)/3$ and the recursion does not hold for $d=3$. But if I modify it to $d(d+1)B(d)/h^2=(d-1)(4d-2)B(d-1)/h+h^2(d-1)^2(d-2)^2B(d-2)$ then it works. – Somos Sep 2 '17 at 16:13
• The OEIS sequence A177267 seems to be the triangle of coefficients of powers of $h^2$ in $dB(d)/h^{d-1}$. – Somos Sep 2 '17 at 17:15
• For this last remark, this is explained at the end of my answer to the previous question (mathoverflow.net/questions/279369/…). Indeed, the coefficients of the polynomial count some numbers similar to the genus described in the OEIS sequence that you gave. – Synia Sep 2 '17 at 19:15
• Sorry I missed a factor of $h^{d-1}$ in the denominator of $B(d)$ – GGT Sep 3 '17 at 1:16