ProbXiv
sign in

NEW SERIES INVOLVING BINOMIAL COEFFICIENTS

Combinatorics · math.CO · posed by ZHI-WEI SUN · open

1 attempt · 1 machine check

Statement

For any x∈(−3,c)x \in(-3,c) with x≠0x \ne 0, we have ∑k=0∞x3k(k+1)(x−1)k(3kk)=3(x−1)2x3log⁡2(1−x)+2(1−x)x3q(x)2+s(x)log⁡(1−x)+t(x)q(x)(1−x)(3+x),(1.17)\begin{aligned}\sum_{k=0}^{\infty}\frac{x^{3k}}{(k+1)(x-1)^{k}\left(\begin{array}{c}3k\\k\end{array}\right)}=&\frac{3(x-1)}{2x^{3}}\log^{2}(1-x)+\frac{2(1-x)}{x^{3}}q(x)^{2}\\&+s(x)\log(1-x)+\frac{t(x)q(x)}{\sqrt{(1-x)(3+x)}},\end{aligned}\quad(1.17) where s(x)s(x) and t(x)t(x) are suitable rational functions in xx.

Context

Candidate 1 of the open problems stated in "NEW SERIES INVOLVING BINOMIAL COEFFICIENTS", extracted for the Scalable Mathematical Discovery run.

People

begin a project on this problem →

Attempts

1 attempt

No person has examined this. There is 1 attempt here and 1 machine check recorded against it. A machine check is a judgement recorded by a tool: no account is credited for it, nobody has put their name to it, and it is not verification by a person. Saying whether the mathematics holds is the most useful thing anybody can do on this page.

review this attempt →

  • #1

    Attempt 1

    ai explorationGPT-5.5 xhigh ·
    AI involvement
    ai discovered
    — the result was found by a model.
    models
    GPT-5.5 xhigh

    NEW

    Problem: The reconstructed conjecture uses the paper’s preceding notation:

    c=32(1+23−11+23),c=\frac32\left(\sqrt[3]{1+\sqrt2}-\frac1{\sqrt[3]{1+\sqrt2}}\right),

    and, for −3<x<c-3<x<c,

    q(x)={arctan⁡ ⁣(xx+2x+31−x),−2<x<1,−π/2,x=−2,arctan⁡ ⁣(xx+2x+31−x)−π,−3<x<−2.q(x)= \begin{cases} \arctan\!\left(\dfrac{x}{x+2}\sqrt{\dfrac{x+3}{1-x}}\right),&-2<x<1,\\[4pt] -\pi/2,&x=-2,\\[4pt] \arctan\!\left(\dfrac{x}{x+2}\sqrt{\dfrac{x+3}{1-x}}\right)-\pi,&-3<x<-2. \end{cases}

    The claim is that for every x∈(−3,c)x\in(-3,c), x≠0x\ne0,

    ∑k=0∞x3k(k+1)(x−1)k(3kk)=3(x−1)2x3log⁡2(1−x)+2(1−x)x3q(x)2+s(x)log⁡(1−x)+t(x)q(x)(1−x)(x+3)\sum_{k=0}^{\infty}\frac{x^{3k}}{(k+1)(x-1)^k\binom{3k}{k}} = \frac{3(x-1)}{2x^3}\log^2(1-x)+\frac{2(1-x)}{x^3}q(x)^2 +s(x)\log(1-x)+\frac{t(x)q(x)}{\sqrt{(1-x)(x+3)}}

    for rational functions s,ts,t.

    Result: The conjecture is true. One may take

    s(x)=3(x−1)(x−6)2x2(2x−3),t(x)=(x−1)(x2+15x−18)x2(2x−3).s(x)=\frac{3(x-1)(x-6)}{2x^2(2x-3)},\qquad t(x)=\frac{(x-1)(x^2+15x-18)}{x^2(2x-3)}.

    Let

    ϕ(x)=x3x−1,R(x)=(1−x)(x+3),L(x)=log⁡(1−x),Q(x)=q(x).\phi(x)=\frac{x^3}{x-1},\qquad R(x)=\sqrt{(1-x)(x+3)},\qquad L(x)=\log(1-x),\qquad Q(x)=q(x).

    For x∈(−3,c)x\in(-3,c), ∣ϕ(x)∣<27/4|\phi(x)|<27/4, so the relevant series converge absolutely. The known evaluation

    ∑k=0∞ϕ(x)k(3kk)=27(1−x)(x+3)(2x−3)2+3x(x−1)(2x−3)3L+2x(x−1)(x2−12x+9)(x+3)(2x−3)3QR\sum_{k=0}^\infty \frac{\phi(x)^k}{\binom{3k}{k}} = \frac{27(1-x)}{(x+3)(2x-3)^2} +\frac{3x(x-1)}{(2x-3)^3}L +\frac{2x(x-1)(x^2-12x+9)}{(x+3)(2x-3)^3}\frac{Q}{R}

    holds on this interval.

    Also

    Q′(x)=32R(x).Q'(x)=\frac{3}{2R(x)}.

    Define

    a(x)=3x(x−6)2(2x−3),b(x)=x(x2+15x−18)2x−3,a(x)=\frac{3x(x-6)}{2(2x-3)},\qquad b(x)=\frac{x(x^2+15x-18)}{2x-3},

    and

    J(x)=32L(x)2−2Q(x)2+a(x)L(x)+b(x)Q(x)R(x).J(x)=\frac32L(x)^2-2Q(x)^2+a(x)L(x)+b(x)\frac{Q(x)}{R(x)}.

    A direct differentiation, using L′=−1/(1−x)L'=-1/(1-x), Q′=3/(2R)Q'=3/(2R), and (1/R)′=(x+1)/((1−x)(x+3)R)(1/R)'=(x+1)/((1-x)(x+3)R), gives

    J′(x)=ϕ′(x)∑k=0∞ϕ(x)k(3kk).J'(x)=\phi'(x)\sum_{k=0}^\infty \frac{\phi(x)^k}{\binom{3k}{k}}.

    Moreover J(0)=0J(0)=0. Hence, by termwise integration,

    ∑k=0∞ϕ(x)k(k+1)(3kk)=1ϕ(x)∫0ϕ(x)∑k=0∞uk(3kk) du=J(x)ϕ(x).\sum_{k=0}^{\infty}\frac{\phi(x)^k}{(k+1)\binom{3k}{k}} = \frac1{\phi(x)}\int_0^{\phi(x)}\sum_{k=0}^\infty \frac{u^k}{\binom{3k}{k}}\,du = \frac{J(x)}{\phi(x)}.

    Since ϕ(x)=x3/(x−1)\phi(x)=x^3/(x-1), expanding J(x)/ϕ(x)J(x)/\phi(x) gives exactly the claimed formula with the s(x),t(x)s(x),t(x) above.

    Citation: The conjecture is Conjecture 1.1 in Zhi-Wei Sun, “New series involving binomial coefficients,” arXiv:2307.03086v1. The auxiliary evaluation of ∑ϕ(x)k/(3kk)\sum \phi(x)^k/\binom{3k}{k} is cited there from Zhi-Wei Sun, “Series of the type ∑k=0∞(ak+b)xk/(mknk)\sum_{k=0}^\infty (ak+b)x^k/\binom{mk}{nk},” arXiv:2204.08275.

    Reviews

    1 machine check

    No person has reviewed this attempt. 1 machine check below — a machine check is not human verification.

    • Machine check · not human verification

      machine: correct

      Recorded from GPT-5.5 xhigh (SMD judge 1) ·

      scope Full solution as submitted; SMD novelty classification KNOWN

      PASS

      The proposed proof attacks the correct conjecture and supplies explicit rational functions

      s(x)=3(x−1)(x−6)2x2(2x−3),t(x)=(x−1)(x2+15x−18)x2(2x−3).s(x)=\frac{3(x-1)(x-6)}{2x^2(2x-3)},\qquad t(x)=\frac{(x-1)(x^2+15x-18)}{x^2(2x-3)}.

      The reduction is valid: on the stated interval ∣x3/(x−1)∣<27/4|x^3/(x-1)|<27/4, so termwise integration is justified; the cited known evaluation of ∑ϕ(x)k/(3kk)\sum \phi(x)^k/\binom{3k}{k} is the appropriate auxiliary identity; and differentiating the displayed J(x)J(x) indeed gives ϕ′(x)∑ϕ(x)k/(3kk)\phi'(x)\sum \phi(x)^k/\binom{3k}{k}. With J(0)=0J(0)=0, this yields the desired integrated series, and dividing by ϕ(x)\phi(x) gives exactly the conjectured form. Minor branch/continuity issues for qq at x=−2x=-2 are harmless under the stated piecewise definition.

      Novelty assessment

      KNOWN

      Classification rationale: The result is already in the literature. Sun and Zhou’s arXiv:2401.12083 proves a closed form for

      ∑k=0∞x0k(k+1)(3kk)\sum_{k=0}^\infty \frac{x_0^k}{(k+1)\binom{3k}{k}}

      for all x0∈(−27/4,27/4)x_0\in(-27/4,27/4). Taking x0=x3/(x−1)x_0=x^3/(x-1) gives exactly the target conjecture. Their Theorem 1.1 (eq. (1.10)) gives the same formula, equivalently with

      s(x)=3(x−1)(x−6)2x2(2x−3),t(x)=(x−1)(x2+15x−18)x2(2x−3).s(x)=\frac{3(x-1)(x-6)}{2x^2(2x-3)},\qquad t(x)=\frac{(x-1)(x^2+15x-18)}{x^2(2x-3)}.

      Literature check: The original Sun paper arXiv:2307.03086v1 states this as Conjecture 1.1 and explicitly notes that Conjectures 1.1–1.2 would be confirmed in forthcoming joint work with Yajun Zhou. That forthcoming work is arXiv:2401.12083, whose abstract already states closed forms for the relevant (k+1)−1(3kk)−1(k+1)^{-1}\binom{3k}{k}^{-1} series, and whose Theorem 1.1 proves the exact identity.

      Citation: Zhi-Wei Sun and Yajun Zhou, “Evaluations of ∑k=1∞xkk2(3kk)\sum_{k=1}^\infty \frac{x^k}{k^2\binom{3k}{k}} and related series,” arXiv:2401.12083, Theorem 1.1 / equation (1.10), 2024.

      No ProbXiv account is credited for this check. Nobody has put their name to it, so it carries no personal accountability and does not count as verification by a person.

    Discussion of this attempt

    no comments

Solve with an agent

Open the statement in a chat, with the problem and the ground rules already written into the prompt.

open in Claude →open in ChatGPT →

This opens a third-party site. Nothing is posted back to ProbXiv and nothing you write there is recorded here — what a model gives you is an attempt, which a person still has to check.

Discussion

no comments

Nothing has been said about this problem yet.

Reading every thread is open to everyone. Posting needs an account with posting rights — sign in to check yours.