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Combinatorial Aspects of Total Positivity

Combinatorics · math.CO · posed by Lauren Kiyomi Williams · open

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Statement

The coefficient of qq in E^k,n(q)\hat{E}_{k,n}(q) is (nk+1)(nk−2)\binom{n}{k+1}\binom{n}{k-2}.

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  • Combinatorial Aspects of Total Positivity
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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
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    NEW

    Problem: For n≥1n\ge1 and 1≤k≤n1\le k\le n, Williams’s E^k,n(q)\widehat E_{k,n}(q) is the renormalized qq-Eulerian polynomial defined by

    E^(q,x,y):=∑n,kE^k,n(q)xnyk=11−b0x−λ1x21−b1x−λ2x2⋯,\widehat E(q,x,y):=\sum_{n,k}\widehat E_{k,n}(q)x^ny^k = \cfrac{1}{1-b_0x-\cfrac{\lambda_1x^2}{1-b_1x-\cfrac{\lambda_2x^2}{\cdots}}},

    where [m]q=1+q+⋯+qm−1[m]_q=1+q+\cdots+q^{m-1}, bh=y[h+1]q+[h]qb_h=y[h+1]_q+[h]_q, and λh=y[h]q2\lambda_h=y[h]_q^2. This is Corteel’s continued-fraction form for Williams’s polynomials. The conjecture is

    [q]E^k,n(q)=(nk+1)(nk−2),[q]\widehat E_{k,n}(q)=\binom n{k+1}\binom n{k-2},

    with the usual convention (nr)=0\binom nr=0 outside 0≤r≤n0\le r\le n.

    Result: The conjecture is true.

    Let FhF_h denote the tail of the continued fraction beginning at height hh:

    Fh=11−bhx−λh+1x2Fh+1.F_h=\frac{1}{1-b_hx-\lambda_{h+1}x^2F_{h+1}}.

    At q=0q=0, for h≥1h\ge1 all tails are equal to a series MM satisfying

    M=11−(1+y)x−yx2M.M=\frac1{1-(1+y)x-yx^2M}.

    Put Ah=[q]FhA_h=[q]F_h. Since

    [m]q={0,m=0,1,m=1,1+q+O(q2),m≥2,[m]_q=\begin{cases} 0,&m=0,\\ 1,&m=1,\\ 1+q+O(q^2),&m\ge2, \end{cases}

    one obtains, for h≥2h\ge2, a common value AA satisfying

    A=M2((1+y)x+2yx2M+yx2A),A=M^2\bigl((1+y)x+2yx^2M+yx^2A\bigr),

    and

    A1=M2(yx+2yx2M+yx2A).A_1=M^2\bigl(yx+2yx^2M+yx^2A\bigr).

    Solving gives

    A1=M2xy(1+xM)21−yx2M2.A_1=\frac{M^2xy(1+xM)^2}{1-yx^2M^2}.

    Also F0(0)=M/(1+xM)F_0(0)=M/(1+xM), and b0,λ1b_0,\lambda_1 have zero qq-coefficient, hence

    [q]E^(q,x,y)=A0=F0(0)2 yx2A1=x3y2M41−yx2M2.[q]\widehat E(q,x,y)=A_0=F_0(0)^2\,yx^2A_1 =\frac{x^3y^2M^4}{1-yx^2M^2}.

    Now set T=xMT=xM. Then

    T=x(1+T)(1+yT).T=x(1+T)(1+yT).

    Let ϕ(t)=(1+t)(1+yt)\phi(t)=(1+t)(1+yt). Since T=xϕ(T)T=x\phi(T),

    [q]E^(q,x,y)=y2T3ϕ(T)1−yT2=y2T31−xϕ′(T).[q]\widehat E(q,x,y) =\frac{y^2T^3\phi(T)}{1-yT^2} =\frac{y^2T^3}{1-x\phi'(T)}.

    By Lagrange inversion,

    [xn]T31−xϕ′(T)=[tn−3]ϕ(t)n.[x^n]\frac{T^3}{1-x\phi'(T)} =[t^{n-3}]\phi(t)^n.

    Therefore

    [xnyk][q]E^(q,x,y)=[tn−3yk−2](1+t)n(1+yt)n.[x^ny^k][q]\widehat E(q,x,y) =[t^{n-3}y^{k-2}](1+t)^n(1+yt)^n.

    Extracting yk−2y^{k-2} from (1+yt)n(1+yt)^n contributes

    (nk−2)tk−2,\binom n{k-2}t^{k-2},

    so the remaining tt-power is n−3−(k−2)=n−k−1n-3-(k-2)=n-k-1, giving

    [tn−k−1](1+t)n=(nn−k−1)=(nk+1).[t^{n-k-1}](1+t)^n=\binom n{n-k-1}=\binom n{k+1}.

    Thus

    [q]E^k,n(q)=(nk+1)(nk−2).[q]\widehat E_{k,n}(q)=\binom n{k+1}\binom n{k-2}.

    Citation: Williams posed the conjecture in “Combinatorial Aspects of Total Positivity,” Conjecture 2.5.10. The continued fraction used above is due to Sylvie Corteel, “Crossings and alignments of permutations,” arXiv:math/0505031.

  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 addresses the exact conjecture and is mathematically sound. The first-order expansion of Corteel’s continued fraction at q=0q=0, the tail recurrences, the simplification to

    [q]E^(q,x,y)=y2T31−xϕ′(T)[q]\widehat E(q,x,y)=\frac{y^2T^3}{1-x\phi'(T)}

    with T=x(1+T)(1+yT)T=x(1+T)(1+yT), and the Lagrange inversion extraction are all correct and give

    [q]E^k,n(q)=(nk+1)(nk−2).[q]\widehat E_{k,n}(q)=\binom n{k+1}\binom n{k-2}.

    No fatal gap or indexing mismatch is apparent.

    Novelty assessment

    TYPE1

    Classification rationale: The explicit coefficient formula appears not to be recorded, but it is a very small consequence of Corteel’s known continued fraction. The accepted proof is a routine first-order expansion at q=0q=0 plus standard Lagrange inversion. Thus it resolves a specific thesis conjecture, but is unlikely to support a standalone journal paper without additional results or a broader generalization.

    Literature check: I found no source explicitly proving

    [q]E^k,n(q)=(nk+1)(nk−2).[q]\widehat E_{k,n}(q)=\binom n{k+1}\binom n{k-2}.

    Searches for “Conjecture 2.5.10,” the exact binomial product, E^k,n(q)\widehat E_{k,n}(q), “coefficient of qq,” and related terms around Williams’s qq-Eulerian numbers led only to the original thesis, Corteel’s continued-fraction paper, and related work on permutation tableaux/unimodality. Corteel’s continued fraction essentially contains the needed information, but the binomial coefficient extraction itself does not appear to have been stated.

    Citation: Lauren K. Williams, Combinatorial Aspects of Total Positivity, PhD thesis, Conjecture 2.5.10.
    Sylvie Corteel, “Crossings and alignments of permutations,” arXiv:math/0505031.
    Petter Brändén, “Actions on permutations and unimodality of descent polynomials,” European J. Combin. 29 (2008), 514–531.

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