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STANLEY-WILF LIMITS FOR PATTERNS IN ROOTED LABELED FORESTS

Combinatorics · math.CO · posed by Michael Ren · open

2 comments

Statement

For any uncovered set S of patterns, RS,nn\frac{R_{S,n}}{n} converges in distribution to a Bernoulli random variable Ber(p)Ber(p) for some p∈[0,1]p \in [0, 1]. If S contains a pattern starting with 1, then p=1p = 1, and if S contains a pattern starting with its largest element, then p=0p = 0. Furthermore, there exist limiting probabilities p1,p2,…,q1,q2,…p_1, p_2, \dots, q_1, q_2, \dots summing to 1 such that P(RS,n=k)→pk\mathbb{P}(R_{S,n} = k) \to p_k and P(RS,n=n+1−k)→qk\mathbb{P}(R_{S,n} = n+1-k) \to q_k as n→∞n \to \infty. If S contains a pattern starting with 1, then p1=p2=⋯=0p_1 = p_2 = \dots = 0, and if S contains a pattern starting with its largest element, then q1=q2=⋯=0q_1 = q_2 = \dots = 0.

Context

Candidate 11 of the open problems stated in "STANLEY-WILF LIMITS FOR PATTERNS IN ROOTED LABELED FORESTS", extracted for the Scalable Mathematical Discovery run.

Record

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  • STANLEY-WILF LIMITS FOR PATTERNS IN ROOTED LABELED FORESTS
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No person has examined this. Everything below was judged by machines. Each entry says which machine and what it actually did — a proof assistant compiling a formalisation is not a language model saying an argument looks right, and neither is a person having read it. No ProbXiv account is credited for any of them. Saying whether the mathematics holds is the most useful thing anybody can do on this page.

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  1. exploration by a model · #1

    GPT-5.5 xhigh

    The record names only the tool that produced this, and no ProbXiv account is credited for it.

    AI involvement
    ai discovered
    — the result was found by a model.

    NEW

    Problem: Reconstructed conjecture: for every uncovered set SS of permutation patterns, if RS,nR_{S,n} is the root label of a uniformly random rooted labeled tree on [n][n] avoiding SS, then RS,n/nR_{S,n}/n converges to a Bernoulli law on {0,1}\{0,1\}, with the stated endpoint refinements.

    Result: The literal conjecture is false.

    Take S=∅S=\varnothing. This set is uncovered, since it contains no pattern starting with 11 and no pattern starting with its largest element. Every rooted labeled tree on [n][n] avoids SS.

    By symmetry under relabeling [n][n], the root label of a uniformly random rooted labeled tree is uniform on [n][n]. Hence

    P(R∅,n=k)=1n(1≤k≤n),\mathbb P(R_{\varnothing,n}=k)=\frac1n \qquad (1\le k\le n),

    and therefore

    R∅,nn→dU[0,1],\frac{R_{\varnothing,n}}n \xrightarrow{d} U[0,1],

    not to a Bernoulli random variable. For example, for 0<x<10<x<1,

    P(R∅,n/n≤x)=⌊nx⌋n→x,\mathbb P(R_{\varnothing,n}/n\le x)=\frac{\lfloor nx\rfloor}{n}\to x,

    whereas a Bernoulli distribution is constant on (0,1)(0,1).

    The endpoint refinement also fails: for every fixed kk,

    P(R∅,n=k)=1n→0,P(R∅,n=n+1−k)=1n→0,\mathbb P(R_{\varnothing,n}=k)=\frac1n\to0,\qquad \mathbb P(R_{\varnothing,n}=n+1-k)=\frac1n\to0,

    so all limiting endpoint probabilities would be 00, not summing to 11.

    This is a small formulation defect: the natural repair is to require S≠∅S\neq\varnothing and to exclude the length-one pattern 11, for which avoidance makes the random tree undefined for n≥1n\ge1.

    Citation: No known prior disproof is needed; the counterexample uses only label symmetry of rooted labeled trees.

  2. Read by a language model on #1 · a reading, not a proof

    model says: correctGPT-5.5 xhigh (SMD judge 1)

    scope Full solution as submitted; SMD novelty classification TYPE1

    PASS

    The counterexample is valid for the literal conjecture. In the paper, “uncovered” does not exclude S=∅S=\varnothing, so the empty set qualifies. For S=∅S=\varnothing, every rooted labeled tree is allowed, and by relabeling symmetry the root label is uniform on [n][n]. Hence R∅,n/nR_{\varnothing,n}/n converges to U[0,1]U[0,1], not to any Bernoulli law, and the endpoint limiting probabilities are all 00, so they cannot sum to 11. This rigorously disproves the stated conjecture as written.

    Novelty assessment

    TYPE1

    Classification rationale: The accepted resolution is a valid literal counterexample, but it is only a formulation-gap observation: take S=∅S=\varnothing, then all rooted labeled trees are allowed and root-label symmetry makes R∅,nR_{\varnothing,n} uniform on [n][n], giving a uniform [0,1][0,1] scaling limit rather than Bernoulli. This is immediate and would not support a standalone paper; at most it is an erratum/comment.

    Literature check: I found no prior public note or paper explicitly pointing out this empty-set counterexample to Conjecture 4.10. I checked the arXiv record for the target paper and the earlier split paper, alphaXiv, broad web-search queries for the exact conjecture/title/keywords, and GitHub issues/discussions/repositories; no relevant prior disproof appeared. The underlying symmetry fact is classical, but the specific disproof does not seem separately recorded.

    Citation: Michael Ren, “Stanley-Wilf Limits for Patterns in Rooted Labeled Forests,” arXiv:2310.02499, Conjecture 4.10.

    A language model was shown this work and said what it thought of it. Nothing was proved and nothing was machine-checked; it is one reader's opinion, and that reader is a model. No ProbXiv account is credited for it.

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