Associated Hermite matching-generating-function conjecture

From papers

Let $n_1,n_2,\dots,n_k$ be positive integers. For each $i$ , let $[n_i]=\{1,2,\dots,n_i\}$ , and arrange the sets in the disjoint union below in weakly increasing order by size. Let $H_{n_i}(x;c)$ be the associated Hermite polynomials and let $\L_c$ denote the corresponding integral functional. A matching is inhomogeneous when it contains the inhomogeneous edges specified by the model, and the rightmost-choice moment weighting assigns weight $c$ to every edge with no right crossing.

Associated Hermite matching-generating-function conjecture. The integral

\L_c \left( \prod_{i=1}^k H_{n_i}(x;c) \right)

is the generating function for inhomogeneous matchings on $\bigsqcup_{i=1}^k [n_i]$ , with the vertex sets arranged in weakly increasing order by size and the edges weighted using the rightmost-choice moment weighting.

The conjecture is motivated by computational evidence after the other orderings and moment or polynomial weightings fail in small examples. The supplied material does not establish a proof or a disproof, so its resolution remains open.

Progress summary

Open

No public source found here proves or disproves the conjecture, so it remains open.

The conjecture asserts that $\mathcal{L}_c\!\left(\prod_{i=1}^k H_{n_i}(x;c)\right)$ counts the specified inhomogeneous matchings when the block sizes are weakly increasing. It appears as Conjecture 2.1 in a 2007 paper, motivated by small computational tests.

Known results

The 2007 paper gives weighted-matching interpretations for the associated Hermite polynomials and two weighted-complete-matching interpretations for their moments.
It proves orthogonality: $\mathcal{L}_c(H_nH_m)=0$ for $n\ne m$ and $\mathcal{L}_c(H_n^2)=(c+1)^n$ .
It discusses linearization for products of two associated Hermite polynomials, but records no combinatorial interpretation resolving the multi-factor conjecture.

Current status (as of August 2026): The conjecture remains unsettled; the related moment, matching, and orthogonality results are known, but no independently sourced proof or counterexample to the stated multi-factor identity was found.

Sources

ar5iv.labs.arxiv.org

Sources & referencesView supporting material

Primary source

Dan Drake, “The combinatorics of associated Hermite polynomials”, arXiv:0709.0987 (2008).

1 week ago

The conjecture is false. Take six blocks of weakly increasing sizes

(n_1,n_2,n_3,n_4,n_5,n_6)=(1,1,1,1,3,3).

The associated-Hermite recurrence

H_{r+1}(x;c)=xH_r(x;c)-(r-1+c)H_{r-1}(x;c)

gives $H_1=x$ and $H_3=x^3-(2c+1)x$ . Let $\mu_j=\mathcal L_c(x^j)$ . The Dyck-path moment rule gives

\begin{aligned} \mu_6&=3c+7c^2+5c^3,\\ \mu_8&=15c+39c^2+37c^3+14c^4,\\ \mu_{10}&=105c+296c^2+326c^3+176c^4+42c^5. \end{aligned}

These follow directly from $D_{0,0}=1$ and the transitions

D_{t+1,h+1}\mathrel{+}=D_{t,h},\qquad D_{t+1,h-1}\mathrel{+}=(h-1+c)D_{t,h}\quad(h>0),

with $\mu_t=D_{t,0}$ . Therefore

\begin{aligned} \mathcal L_c(H_1^4H_3^2) &=\mu_{10}-2(2c+1)\mu_8+(2c+1)^2\mu_6\\ &=78c+177c^2+141c^3+48c^4+6c^5. \end{aligned}

For an independently checkable enumeration of the proposed matching side, write the ordered block-label word as

b=(1,2,3,4,5,5,5,6,6,6).

Process vertices from left to right and let $L=(\ell_1,\ldots,\ell_h)$ be the ordered list of block labels of currently open edges. If $F_p(L)$ is the weighted generating function for completions starting at vertex $p$ , then

F_p(L)=F_{p+1}(L,b_p) +\sum_{\substack{1\le j\le h\\ell_j\ne b_p}} c^{[j=h]}F_{p+1}(\ell_1,\ldots,\widehat{\ell_j},\ldots,\ell_h),

with $F_{11}(\varnothing)=1$ and $F_{11}(L)=0$ for $L\ne\varnothing$ . The restriction $\ell_j\ne b_p$ enforces inhomogeneity. An edge has no right crossing exactly when its opening vertex is the most recently opened edge still present at its closing vertex, giving the factor $c^{[j=h]}$ .

The recurrence yields

F_1(\varnothing)=84c+180c^2+135c^3+45c^4+6c^5.

Thus the 450 admissible matchings split into $84,180,135,45,6$ according to their number of edges without a right crossing. Consequently,

\mathcal L_c(H_1^4H_3^2)-F_1(\varnothing) =3c(c-1)(c+1)(c+2).

For the admissible value $c=2$ , the two sides are respectively $2952$ and $2880$ . Therefore the conjectured identity fails already on ten vertices.

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Shivam Patel · 1 day ago