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Permutations, moments, measures

Combinatorics · math.CO · posed by Natasha Blitvić, Einar Steingrímsson · open

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Statement

Interestingly, all the properties of Proposition 2 hold even for negative k, and it seems that for any k the A_k(n) eventually become positive for n sufficiently large, ...

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  • Permutations, moments, measures
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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: In the paper’s notation, Ak(n)A_k(n) is extended from positive integer kk to arbitrary integer kk by the identities in Proposition 2:

    Ak(0)=1,Ak(n)=nAk(n−1)+(k−1)n,A_k(0)=1,\qquad A_k(n)=nA_k(n-1)+(k-1)^n,

    equivalently

    ∑n≥0Ak(n)xnn!=e(k−1)x1−x.\sum_{n\ge0}A_k(n)\frac{x^n}{n!}=\frac{e^{(k-1)x}}{1-x}.

    The conjectural statement is:

    ∀k∈Z ∃N ∀n≥N,Ak(n)>0.\forall k\in\mathbb Z\ \exists N\ \forall n\ge N,\quad A_k(n)>0.

    The phrase “for any kk” is slightly ambiguous; the context concerns integer kk, including negative kk. We prove the stronger statement for every real kk.

    Result: For all k∈Rk\in\mathbb R,

    Ak(n)=n!∑j=0n(k−1)jj!.A_k(n)=n!\sum_{j=0}^n \frac{(k-1)^j}{j!}.

    Indeed,

    e(k−1)x1−x=(∑j≥0(k−1)jxjj!)(∑m≥0xm),\frac{e^{(k-1)x}}{1-x} =\left(\sum_{j\ge0}\frac{(k-1)^j x^j}{j!}\right)\left(\sum_{m\ge0}x^m\right),

    so the coefficient of xnx^n is ∑j=0n(k−1)j/j!\sum_{j=0}^n (k-1)^j/j!.

    Let

    Sn(k)=∑j=0n(k−1)jj!.S_n(k)=\sum_{j=0}^n \frac{(k-1)^j}{j!}.

    Then

    Sn(k)→ek−1>0.S_n(k)\to e^{k-1}>0.

    Hence there exists NN such that for all n≥Nn\ge N,

    Sn(k)>12ek−1>0.S_n(k)>\frac12 e^{k-1}>0.

    Since n!>0n!>0, it follows that

    Ak(n)=n!Sn(k)>0A_k(n)=n!S_n(k)>0

    for all n≥Nn\ge N.

    Thus the conjecture is true, even for all real kk. Degenerate cases are consistent: for example k=1k=1 gives A1(n)=n!>0A_1(n)=n!>0, while k=0k=0 gives derangement numbers, eventually positive despite A0(1)=0A_0(1)=0.

    Citation: The defining recurrence and exponential generating function are from Blitvić–Steingrímsson, “Permutations, moments, measures,” arXiv:2001.00280, Proposition 2. The eventual positivity follows from the elementary convergence of the exponential series.

  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 TYPE1

    PASS

    The proof attacks the correct statement. Using the cited exponential generating function,

    e(k−1)x1−x,\frac{e^{(k-1)x}}{1-x},

    coefficient extraction gives

    Ak(n)=n!∑j=0n(k−1)jj!.A_k(n)=n!\sum_{j=0}^n \frac{(k-1)^j}{j!}.

    For each fixed real (hence integer) kk, the partial sums converge to ek−1>0e^{k-1}>0, so they are eventually positive. Since n!>0n!>0, Ak(n)>0A_k(n)>0 eventually. This proves the conjecture as stated, with no apparent gap.

    Novelty assessment

    TYPE1

    Classification rationale: This is a one-line consequence of the exponential generating function already given in the original paper:

    Ak(n)=n!∑j=0n(k−1)jj!→n!ek−1A_k(n)=n!\sum_{j=0}^n \frac{(k-1)^j}{j!}\to n! e^{k-1}

    in sign after normalization. It is mathematically correct but entirely routine and not publishable as a standalone combinatorics result.

    Literature check: I found no independent paper or note explicitly stating the eventual positivity claim for negative/integer kk. The closest relevant follow-up is Fu–Han–Lin, “kk-arrangements, statistics and patterns,” which confirms other Blitvić–Steingrímsson conjectures about kk-arrangements, but not this eventual-positivity observation. The original Blitvić–Steingrímsson paper already contains the EGF/recurrence from which the claim follows immediately.

    Citation: Natasha Blitvić and Einar Steingrímsson, “Permutations, moments, measures,” arXiv:2001.00280, Proposition 2. Shishuo Fu, Guo-Niu Han, and Zhicong Lin, “kk-arrangements, statistics and patterns,” arXiv:2005.06354.

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