The uniform rectangle-creation rate conjecture for the finite k-chain

The uniform rectangle-creation conjecture holds for every (k); the argument also proves the associated limit-shape conjecture.

Put $K=k+1$ and $R_i=(i^{K-i})$, $1\le i\le k$. We prove simultaneously that every stationary rectangle-creation rate is

$$\rho_i=\binom{k+2}{3}^{-1}$$

and that the limit shape is $D_{k+1}$.

The rectangle property and unique rectangle decomposition give

$$s^{(k)}_{\lambda\cup R_i}=s_{R_i}s^{(k)}_\lambda,\qquad \mathbb Q[h_1,\ldots,h_k] =\bigoplus_{\lambda\in\mathcal R_k} \mathbb Q[s_{R_1},\ldots,s_{R_k}]s^{(k)}_\lambda.$$

Indeed, divide the multiplicity of each part $i$ by $K-i$; the remainder gives a unique $\lambda\in\mathcal R_k$. Let $A(r)$ denote multiplication by $h_1$ in this free basis, with $r_i=s_{R_i}$. Each nonrectangle transition contributes $1$, and each transition creating $R_i$ contributes $r_i$.

At exponential specialization $p_1=1,\ p_j=0$ for $j>1$, write

$$c_\lambda=\frac{d_\lambda^{(k)}}{|\lambda|!},\qquad r_i^0=\frac{f^{R_i}}{|R_i|!}.$$

The source's transition matrix is

$$P=\operatorname{diag}(c)^{-1} A(r^0)\operatorname{diag}(c).$$

For each $j=2,\ldots,k$, vary $p_j=t$ while keeping $p_1=1$. The specialized Pieri identity gives

$$A(r(t))c(t)=c(t).$$

Positivity near $t=0$ and irreducibility show that the Perron eigenvalue remains exactly $1$. Differentiating against its stationary left eigenvector gives

$$\sum_{i=1}^k\rho_iD_{i,j}=0,\qquad D_{i,j} =\left.\partial_{p_j}\log s_{R_i}\right|_{\operatorname{ex}} =\frac1j\operatorname{Ch}_j(R_i). \tag{1}$$

Here $\operatorname{Ch}_j$ is the normalized symmetric-group character of a $j$-cycle.

Stanley's rectangular-character residue formula gives

$$\operatorname{Ch}_j((K-q)\times q) =-\frac1j[x^{-1}] \frac{(x)_j(x-K)_j}{(x-q)_j},$$

with falling factorials and expansion at infinity. For $j\ge2$,

$$\sum_{q=1}^{K-1}\frac1{(x-q)_j} =\frac1{j-1} \left(\frac1{(x-K)_{j-1}} -\frac1{(x-1)_{j-1}}\right).$$

Multiplication by $(x)_j(x-K)_j$ makes both terms polynomials, so their $x^{-1}$ coefficients vanish. Hence

$$\sum_{i=1}^kD_{i,j}=0 \qquad(2\le j\le k). \tag{2}$$

Moreover, Stanley's permutation-factorization formula and its Narayana top-degree term imply that

$$Q_j(q)= \frac{\operatorname{Ch}_j((K-q)\times q)} {q(K-q)}$$

is a polynomial of exact degree $j-1$, with nonzero leading coefficient $\operatorname{Cat}_j$. Thus $Q_1,\ldots,Q_k$ form a polynomial basis; evaluating at $q=1,\ldots,k$ gives the nonzero Vandermonde determinant

$$\det[D_{i,j}]_{1\le i,j\le k} = \frac1{k!} \left(\prod_{i=1}^ki(K-i)\right) \left(\prod_{j=1}^k\operatorname{Cat}_j\right) \prod_{1\le i<r\le k}(r-i)>0.$$

Therefore columns $j=2,\ldots,k$ have rank $k-1$. By (2) their left kernel is exactly the span of $(1,\ldots,1)$. Equation (1) forces

$$\rho_1=\cdots=\rho_k=\rho.$$

Finally, each step adds one box, while a rectangle transition removes $|R_i|=i(K-i)$ boxes from the residual state. Stationarity of residual size gives

$$0=1-\sum_{i=1}^ki(K-i)\rho_i =1-\binom{k+2}{3}\rho.$$

Consequently

$$\boxed{\rho_i=\binom{k+2}{3}^{-1}\quad(1\le i\le k).}$$

For $k=1$, the same conclusion follows directly from this size-drift equation. The finite-state ergodic theorem gives equal asymptotic multiplicities of all rectangles; the geometric implication immediately following Conjecture 6 in the source then identifies the limiting boundary as

$$\boxed{C_k=D_{k+1}.}$$

Sources: S. Linusson and A. Özdemir, The $k$-Plancherel Measure and a Finite Markov Chain, arXiv:2512.24346, Conjectures 5 and 6; R. Stanley, Irreducible Symmetric Group Characters of Rectangular Shape, arXiv:math/0109093, equations (8)–(9) and Theorem 1.