Candidate counterexample · combinatorics
Conjecture 6.6 of Combinatorial study of the q‑Catalan triangle and its generalizations (Wirdane, arXiv:2605.14682v1, 14 May 2026) is false as stated. The smallest counterexample uses five permutations; the failure extends to an infinite family.
Fix a positive integer μ. The paper partitions the inversions of a permutation π ∈ Sn according to the residue class of the larger entry.
invi(π) = |{ (a,b) : 1 ≤ a < b ≤ n, πa > πb, πa ≡ i (mod μ) }|
Cn,k(q) = ∑π ∈ S′n,k(312) ∏i=1μ qiinvi(π)
Here S′n,k(312) is the paper's family of 312‑avoiding permutations counted by the Catalan‑triangle entry Cn,k; for k = n it is all of Sn(312). The cyclotomic specialisation sets every variable to the same root of unity:
“For ω = e2iπ/μ, the cyclotomic analogues are Cn,k(μ) = Cn,k(ω, …, ω), related to the cyclic sieving phenomenon.”
If μ | (n+1): Cn,n(μ) = 1μn+1(n+1)/μ
The multivariate definition looks richer than it is at the cyclotomic point. This is the paper's own Corollary 6.5, which records that setting all variables equal recovers the univariate triangle: qi = q ⇒ Cn,k(q) = Cn,k(q).
Lemma 1. For every permutation π ∈ Sn and every μ ≥ 1,
∏i=1μ ωinvi(π) = ωinv(π).
Proof. Every entry πa lies in exactly one residue class modulo μ, so each inversion (a,b) is counted by exactly one invi. Hence the classes partition the inversion set and ∑i invi(π) = inv(π). Since every factor carries the same base ω, the product telescopes to ω∑i invi(π). □
Consequently the conjecture is a statement about a single‑variable polynomial evaluated at a root of unity:
Cn,n(μ) = Cn,n(ω), where Cn,n(q) = ∑π ∈ Sn(312) qinv(π).
Take μ = 2 and n = k = 3. The divisibility hypothesis holds since 2 | 4, and ω = eiπ = −1.
For k = n the index set is all of S3(312) — the six permutations of length three minus the single pattern occurrence 312, leaving five. By Lemma 1 each carries weight (−1)inv(π).
| π | inv(π) | weight (−1)inv |
|---|---|---|
123 | 0 | +1 |
132 | 1 | −1 |
213 | 1 | −1 |
231 | 2 | +1 |
321 | 3 | −1 |
| total | −1 |
So C3,3(2) = 1 − 1 − 1 + 1 − 1 = −1, whereas Conjecture 6.6 predicts
1242 = 62 = 3.
Theorem. Conjecture 6.6 is false as stated: at (μ, n, k) = (2, 3, 3) the left side equals −1 and the right side equals 3. □
The value does not depend on the permutation model. The paper's own triangle recurrence (Definition 2.2, and Theorem 6.3 with all qi = q) is
Cn,0(q) = q&binom;(n,2), Cn,k(q) = Cn,k−1(q) + qn−k−1 Cn−1,k(q).
Running it to n = 3 gives C3,3(q) = 1 + 2q + q2 + q3, matching the paper's own Example 5.5 at p = 1. Evaluating,
C3,3(−1) = 1 − 2 + 1 − 1 = −1.
A third route — the Fürlinger–Hofbauer recurrence Cn(q) = ∑j=0n−1 qj Cj(q) Cn−1−j(q) — produces the same polynomial. All three agree on −1.
The failure is not isolated to n = 3. Write an = Cn,n(−1) and A(z) = ∑n≥0 an zn. Setting q = −1 in the Fürlinger–Hofbauer recurrence and splitting the convolution by the parity‑twisting factor (−1)j gives
A(z) = 1 + z A(−z) A(z).
Substituting z → −z yields A(−z) = 1 − z A(z) A(−z). Adding the two equations, the product terms cancel:
A(z) + A(−z) = 2.
Put F(z) = A(z) − 1, so A(−z) = 1 − F(z). The first equation becomes 1 + F = 1 + z(1 − F)(1 + F), that is
F(z) = z (1 − F(z)2).
Writing F(z) = z H(z2) turns this into H = 1 − yH2 with y = z2, which is the Catalan equation with y negated. Hence H(y) = ∑m≥0 (−1)m Catm ym and
C2m,2m(−1) = 0 (m ≥ 1), C2m+1,2m+1(−1) = (−1)m Catm.
For μ = 2 the hypothesis 2 | (n+1) selects exactly the odd n = 2m+1, where the conjectured value simplifies to
122m+2m+1 = (2m+1) Catm.
So the conjecture compares (−1)m Catm against (2m+1) Catm. Since Catm ≥ 1, equality would force (−1)m = 2m+1, which holds only at m = 0. Every m ≥ 1 therefore gives a counterexample.
Computed directly from the paper's triangle recurrence:
| n | Cn,n(2) actual | conjectured | verdict |
|---|---|---|---|
| 1 | 1 | 1 | holds (m = 0) |
| 3 | −1 | 3 | fails |
| 5 | 2 | 10 | fails |
| 7 | −5 | 35 | fails |
| 9 | 14 | 126 | fails |
| 11 | −42 | 462 | fails |
n = 1 is the unique case where the conjecture is correct.
Immediately after the conjecture the paper asserts, for μ = 2,
Cn,k(2) = (−1)⌊n/2⌋ Cn,k (verified for n ≤ 5).
This is also false, and it fails inside the very range the paper claims to have checked. At n = k = 3 the right side is (−1)1 · 5 = −5 while the left side is −1. The first failure is earlier still, at n = 2. Across n ≤ 5 there are 14 mismatches:
| n | k | Cn,k(2) | (−1)⌊n/2⌋Cn,k | verdict |
|---|---|---|---|---|
| 2 | 1 | 0 | −2 | mismatch |
| 2 | 2 | 0 | −2 | mismatch |
| 3 | 1 | −1 | −3 | mismatch |
| 3 | 2 | −1 | −5 | mismatch |
| 3 | 3 | −1 | −5 | mismatch |
| 4 | 1 | 0 | 4 | mismatch |
| 4 | 2 | 1 | 9 | mismatch |
| 4 | 3 | 0 | 14 | mismatch |
| 4 | 4 | 0 | 14 | mismatch |
| 5 | 1 | 1 | 5 | mismatch |
| 5 | 2 | 2 | 14 | mismatch |
| 5 | 3 | 2 | 28 | mismatch |
| 5 | 4 | 2 | 42 | mismatch |
| 5 | 5 | 2 | 42 | mismatch |
Only the k = 0 column and the whole of n = 1 agree. The stated verification for n ≤ 5 cannot have been performed.
Since 1 | (n+1) always, μ = 1 is formally in scope. There ω = 1, so Cn,n(1) = Cn,n = Catn, while the conjecture predicts &binom;(n+1, n+1) = 1. Already n = 2 gives 2 ≠ 1. A referee would presumably read μ ≥ 2 as intended — but the paper discusses μ = 1 explicitly in Remark 6.4, so the omission is further evidence the statement was never tested.
The verifier uses the standard library and integer arithmetic only. It computes the value by two independent routes — brute‑force enumeration of 312‑avoiding permutations, and the paper's triangle recurrence — and asserts they agree with each other and disagree with the conjecture.
$ python3 verify_cyclotomic_q_catalan_counterexample.py
C_{3,3}(q) coefficients, low to high degree: [1, 2, 1, 1]
C_{3,3}(-1) from recurrence: -1
312-avoiding permutations and their (-1)^inv weights:
(1, 2, 3): inv=0, weight=1
(1, 3, 2): inv=1, weight=-1
(2, 1, 3): inv=1, weight=-1
(2, 3, 1): inv=2, weight=1
(3, 2, 1): inv=3, weight=-1
Sum from permutation definition: -1
Conjecture 6.6 predicts: 3
COUNTEREXAMPLE VERIFIED: -1 != 3An honest accounting of what this is worth.
What is established. Conjecture 6.6 and Remark 6.7 of arXiv:2605.14682v1 are false exactly as printed. The certificate is finite, checkable by hand in a few minutes, and confirmed by three independent computational routes.
What is not claimed. That the failure is deep or surprising. Once Lemma 1 is noticed — and the paper itself states it as Corollary 6.5 — the conjecture reduces to evaluating the Carlitz q‑Catalan polynomial at q = −1. That evaluation, 0 at even index and ±Catm at odd, is classical and catalogued. The “infinite family” above is a known evaluation applied to a new false claim, not new mathematics.
Context on the target. The preprint is a ten‑page single‑author note whose proofs are deferred to an unpublished reference in a projected series, and several of whose statements are marked “verified for n ≤ 5” — including Remark 6.7, which fails at n = 2. It has not been refereed and had attracted no scrutiny, which is the ordinary reason such a statement survives. Refuting it is a correction, not a breakthrough.
Priority. No published correction, erratum, or counterexample was found in targeted searches for the identifier, the conjecture number, or the value C3,3(2), and the preprint remains at v1 with no revision. That is not proof of priority: the author may have been notified privately, and the intended statement may differ from the printed one. The defensible claim is the narrow one — Conjecture 6.6 is false as stated in arXiv:2605.14682v1.