The denominator conjecture for pop-stacked permutation generating functions

From papers

Let Fk(x)F_k(x) denote the generating function for pop-stacked permutations with precisely kk ascending runs. For each positive integer kk, define the product

Dk(x)=i=1k(1ix)ki+1.D_k(x)=\prod_{i=1}^k(1-ix)^{k-i+1}.

Denominator conjecture. For all kk, the rational generating function Fk(x)F_k(x) can be written as

Fk(x)=Nk(x)Dk(x)=Nk(x)/i=1k(1ix)ki+1,F_k(x)=\frac{N_k(x)}{D_k(x)}=N_k(x)\Big/\prod_{i=1}^k(1-ix)^{k-i+1},

where Nk(x)N_k(x) is a polynomial of degree k(k+1)/2k(k+1)/2, equal to the degree of the denominator Dk(x)D_k(x).

Progress summary

Open

The conjecture proposes an exact pattern for these generating functions, but no public proof or counterexample was found.

Claesson’s denominator conjecture asserts that the generating function for pop-stacked permutations with exactly kk ascending runs has denominator i=1k(1ix)ki+1\prod_{i=1}^k(1-ix)^{k-i+1} and a numerator of degree k(k+1)/2k(k+1)/2. It remains stated as a conjecture in the available literature.

Known results

  • For each fixed kk, the generating function is rational, via a finite-automaton argument.
  • Rational fits were computed from the first 300300 terms for every k24k\le 24.
  • Those fits were verified exactly for k6k\le 6.
  • Explicit generating functions and the conjectured pattern are known in small cases.

Current status (as of August 2026): Rationality for each fixed kk and exact verification through k6k\le 6 are known, but the stated denominator formula and numerator-degree claim remain unproved, with no reported counterexample or verification.

Sources
Sources & referencesView supporting material

Primary source

Anders Claesson, Bjarki Ágúst Guðmundsson and Jay Pantone, “Counting pop-stacked permutations in polynomial time”, arXiv:1908.08910 (2019).

Solutions 1

Proof

Let F_k(x) enumerate pop-stacked permutations with exactly k ascending runs, and put D_k(x)=∏{i=1}^k(1−ix)^{k−i+1}, K=k(k+1)/2. We prove that F_k=N_k/D_k with deg N_k=K and, more strongly, lim{x→∞}F_k(x)=−1, [x^K]N_k(x)=(−1)^{K+1}∏_{j=1}^k j!.

By the source's overlapping-ballot characterization, such a permutation is equivalent to a surjective word w∈[k]^n in which the occurrence intervals of every adjacent pair of letters i,i+1 overlap. Failure of overlap has two disjoint orientations: every i precedes every i+1, or the converse. For a subset S of the k−1 adjacent edges and a choice ε of their orientations, let P(S,ε) be the resulting poset. Every such orientation is acyclic because the edge graph is a path. Inclusion–exclusion gives F_k(x)=Σ_{S⊆[k−1]}(−1)^{|S|} Σ_{ε∈{+,-}^S}G_{P(S,ε)}(x), where G_P enumerates surjective words respecting the precedence relations of P.

For any poset P on k elements, classify a compatible word by the order σ=(σ₁,…,σ_k) in which its distinct letters first occur. This order is a linear extension. Put I_j={σ₁,…,σ_j} and m_j=|Max_P(I_j)|. Between the first appearances of σ_j and σ_{j+1}, the permissible repeated letters are exactly Max_P(I_j); the same holds after the last first appearance. Therefore G_P(x)=x^k Σ_{σ∈Lin(P)} ∏_{j=1}^k(1−m_jx)^{-1}. Because 1≤m_j≤j, a factor 1−ix appears at most k−i+1 times in each denominator. Thus every denominator divides D_k, proving N_k=D_kF_k∈Z[x] and deg N_k≤K.

Furthermore, lim_{x→∞}G_P(x)=(−1)^k Σ_{σ∈Lin(P)}∏{j=1}^k1/|Max_P(I_j)|=(−1)^k. Indeed, delete a uniformly chosen maximal element at each step from P: the product is exactly the probability of the corresponding reverse linear extension, and these probabilities sum to one. Consequently lim{x→∞}F_k(x) =(−1)^k Σ_{S⊆[k−1]}(−1)^{|S|}2^{|S|} =(−1)^k(1−2)^{k−1} =−1. The numerator therefore has degree exactly K. Since the leading coefficient of D_k is (−1)^K∏{i=1}^k i^{k−i+1}=(−1)^K∏{j=1}^k j!, the stated numerator leading coefficient follows. This proves the complete conjecture for every k≥1.

A further conjectured numerator invariant also follows exactly. For k≥2, N_k(1)=2(−1)^{K−1}∏{j=1}^{k−1}j!. Indeed, a summand G_P has a pole of order k at x=1 precisely when m_j=|Max_P(I_j)|=1 for every j. Then every prefix ideal has a unique maximal element, which forces P to be a total chain. Among the oriented subforests of the path on [k], exactly two posets are total chains: the full path oriented consistently in either direction. Each has one linear extension and inclusion–exclusion sign (−1)^{k−1}. Consequently N_k(1)=2(−1)^{k−1}∏{i=2}^k(1−i)^{k−i+1} =2(−1)^{K−1}∏_{j=1}^{k−1}j!. For k=1, separately, N_1(1)=1. Thus the same argument proves the full denominator conjecture together with the exact numerator degree, leading coefficient, and coefficient sum.

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Shivam Patel · · edited