Strict unimodality of bipartite partitions

Question

For non-negative integers $k$ and $l$ let $p(k,l)$ denote the number of vector partitions of $(k,l)$. In other words, $p(k,l)$ is the number of ways of writing $$ (k,l) = (k_1,l_1)+\dotsb + (k_r,l_r), $$ where $(k_i,l_i)$ are non-zero vectors with non-negative integer coordinates, the order of the summands being irrelevant.

Some values of $p(k,l)$, $k,l\geq 0$, are given in the following table:

[    1     1     2     3     5     7    11    15    22    30]
[    1     2     4     7    12    19    30    45    67    97]
[    2     4     9    16    29    47    77   118   181   267]
[    3     7    16    31    57    97   162   257   401   608]
[    5    12    29    57   109   189   323   522   831  1279]
[    7    19    47    97   189   339   589   975  1576  2472]
[   11    30    77   162   323   589  1043  1752  2876  4571]
[   15    45   118   257   522   975  1752  2998  4987  8043]
[   22    67   181   401   831  1576  2876  4987  8406 13715]
[   30    97   267   608  1279  2472  4571  8043 13715 22652]

The main theorem of an article by Kim and Hahn is that whenever $k\geq l\geq 1$, then $p(k+1,l-1)\leq p(k,l)$.

Question

Is it true that $p(k+1,l-1)<p(k,l)$ for all $k\geq l\geq 1$ except $k=l=1$?

Answer 1

I can prove that $p(k+1, l-1) < p(k,l)$ for all $k \ge l > 1$ except $k=l$ odd. Modifying the notation from the articles, let $p_j(k,l)$ be the number of partitions of $(k,l)$ into exactly $j$ parts, so that $p(k,l) = \sum_{j=1}^\infty p_j(k,l)$. The argument below shows that $p_2(k+1,l-1) < p_2(k,l)$ except when $k=l$ is odd. Kim & Hahn's Theorem 7 is equivalent to $p_j(k+1,l-1) \le p_j(k,l)$ for every $j$ (and our $p_j$), so the strict inequality for $j=2$ gives $p(k+1, l-1) < p(k,l)$.

The Landman, Brown, Portier paper that Kim & Hahn cite includes, in the Theorem 2 proof, the observation that $p_2(k,l) = \lfloor \frac{kl+k+l}{2} \rfloor$. Without using the greatest integer function, \begin{equation} p_2(k,l) = \begin{cases} \frac{kl + k + l}{2} & \text{ if both $k$ and $l$ are even}, \\ \frac{kl + k + l-1}{2} & \text{ if one or both of $k$ and $l$ are odd}. \end{cases} \end{equation} Working out the analogous expression for $p_2(k+1,l-1)$ in terms of $k$ and $l$ gives \begin{equation} p_2(k+1,l-1) = \begin{cases} \frac{kl + 2l -1}{2} & \text{ if both $k$ and $l$ are odd}, \\ \frac{kl + 2l-2}{2} & \text{ if one or both of $k$ and $l$ are even}. \end{cases} \end{equation} There are three cases to check for equality. For $k$ and $l$ both even, the expressions for $p_2(k,l)$ and $p_2(k+1,l-1)$ are equal when $k = l - 2$, but we assumed $k \ge l$. The case one of $k$, $l$ even and the other odd leads to $k = l-1$, another impossibility. The only issue occurs when $k$ and $l$ are both odd, as $p_2(k,l) = p_2(k+1,l-1)$ reduces to $k=l$.

I believe your conjecture is true also for $k=l$ odd, but the difference does not arise in 2 part partitions. Later in Landman et al. Theorem 2, they work out some of the $j=3$ case which involves Burnside's lemma and various possibilities modulo 6.