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Clustering of Consecutive Numbers in Permutations Avoiding a Pattern and in Separable Permutations

Probability & statistics · math.CO · math.PR · posed by Ross G. Pinsky · open

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Statement

lim⁡l→∞lim sup⁡n→∞Pnav(τ)(Al;kn(n))=0\lim_{l \to \infty}\limsup_{n \to \infty}P_{n}^{\mathrm{av}(\tau)}(A_{l;k_{n}}^{(n)})=0, for all τ∈∪m=2∞Sm\tau \in \cup_{m=2}^{\infty}S_{m} and for all {kn}\{k_{n}\}.

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  • Clustering of Consecutive Numbers in Permutations Avoiding a Pattern and in Separable Permutations
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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
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    NEW

    Problem: Reconstructed conjecture: for every pattern τ∈⋃m≥2Sm\tau\in\bigcup_{m\ge2}S_m and every admissible sequence knk_n with 1≤kn≤n−l+11\le k_n\le n-l+1 for each fixed ll and all large nn,

    lim⁡l→∞lim sup⁡n→∞Pnav⁡(τ) ⁣(Al;kn(n))=0,\lim_{l\to\infty}\limsup_{n\to\infty} P_n^{\operatorname{av}(\tau)}\!\left(A^{(n)}_{l;k_n}\right)=0,

    where Al;k(n)A^{(n)}_{l;k} is the event that the value interval {k,k+1,…,k+l−1}\{k,k+1,\dots,k+l-1\} occupies some ll consecutive positions in the permutation.

    Result: The conjecture is false as stated.

    Take τ=12\tau=12 and kn=1k_n=1. A permutation avoids 1212 iff it is strictly decreasing, so

    Sn(12)={n,n−1,…,1}.S_n(12)=\{n,n-1,\dots,1\}.

    In this unique permutation, for every fixed l≥2l\ge2 and every n≥l+1n\ge l+1, the values {1,2,…,l}\{1,2,\dots,l\} occur in the last ll positions:

    n,n−1,…,l+1,l,l−1,…,1.n,n-1,\dots,l+1,\boxed{l,l-1,\dots,1}.

    Hence Al;1(n)A^{(n)}_{l;1} occurs with probability 11 under Pnav⁡(12)P_n^{\operatorname{av}(12)}. Therefore

    lim sup⁡n→∞Pnav⁡(12)(Al;1(n))=1\limsup_{n\to\infty}P_n^{\operatorname{av}(12)}(A^{(n)}_{l;1})=1

    for every ll, and consequently

    lim⁡l→∞lim sup⁡n→∞Pnav⁡(12)(Al;1(n))=1≠0.\lim_{l\to\infty}\limsup_{n\to\infty} P_n^{\operatorname{av}(12)}(A^{(n)}_{l;1})=1\ne0.

    Thus the literal conjecture fails. The same defect occurs for τ=21\tau=21. A natural minimal repair would be to exclude the two length-22 monotone patterns, i.e. require ∣τ∣≥3|\tau|\ge3.

    Citation: No external disproof is needed; the counterexample follows immediately from the definitions in Pinsky, “Clustering of consecutive numbers in permutations avoiding a pattern and in separable permutations,” arXiv:2109.09370.

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

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

    scope Full solution as submitted; SMD novelty classification KNOWN

    PASS

    The proposed disproof is valid for the literal conjecture supplied. Since τ=12∈S2\tau=12\in S_2, the class Sn(12)S_n(12) consists only of the decreasing permutation. For kn=1k_n=1, the set {1,…,l}\{1,\dots,l\} always occupies the last ll consecutive positions, so the probability is 11 for every fixed ll and all large nn. Thus the stated limit is 11, not 00. This rigorously disproves the conjecture as written.

    Novelty assessment

    KNOWN

    Classification rationale: The counterexample is valid, but it is not genuinely new: Pinsky’s own paper contains a stronger general lower bound that already covers τ=12\tau=12 and τ=21\tau=21. Since L(12)=L(21)=1L(12)=L(21)=1, Corollary 1(iii) forces the relevant limiting probabilities to be 11, contradicting the later conjecture as stated.

    Literature check: The current arXiv version of Pinsky, arXiv:2109.09370, states the conjecture with τ∈⋃m≥2Sm\tau\in\bigcup_{m\ge2}S_m. In the same paper, Theorem 1(iii) and Corollary 1(iii) apply to patterns not containing tightly at least one of 12,2112,21, which includes τ=12\tau=12 and τ=21\tau=21. Thus the paper itself already contains the stronger ingredient implying this length-two counterexample. Follow-up arXiv:2211.12090 concerns length-three/simple-pattern cases and is not needed.

    Citation: Ross G. Pinsky, “Clustering of consecutive numbers in permutations avoiding a pattern and in separable permutations,” arXiv:2109.09370, Theorem 1(iii), Corollary 1(iii), and the Conjecture in §1.

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