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On the Central Antecedents of Integer(and Other) Sequences

Combinatorics · math.CO · posed by Paul Barry · open

1 attempt · 1 machine check

Statement

We have [xn]gm(x)=[xn](−x(1−W(−x))mW(−x))n,[x^{n}]g_{m}(x)=[x^{n}]\left(\frac{-x(1-W(-x))^{m}}{W(-x)}\right)^{n}, where gm(x)g_{m}(x) is the generating function of the central coefficients (m−2)n+1(m−1)n+1(mn)\frac{(m-2)n+1}{(m-1)n+1}\left(\begin{array}{c}mn\end{array}\right) of the Riordan array (c(x),xc(x)m−2)(c(x),xc(x)^{m-2}) .

Context

Candidate 1 of the open problems stated in "On the Central Antecedents of Integer(and Other) Sequences", extracted for the Scalable Mathematical Discovery run.

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1 attempt

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  • #1

    Attempt 1

    ai explorationGPT-5.5 xhigh ·
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    Problem: Reconstructed statement: let c(x)=(1−1−4x)/(2x)c(x)=(1-\sqrt{1-4x})/(2x), and for an integer m≥2m\ge2 let gm(x)g_m(x) be the generating function of the central coefficients r2n,nr_{2n,n} of the Riordan array

    (c(x), xc(x)m−2).(c(x),\,x c(x)^{m-2}).

    Thus

    [xn]gm(x)=r2n,n=[xn]c(x)(m−2)n+1=(m−2)n+1(m−1)n+1(mnn).[x^n]g_m(x)=r_{2n,n} =[x^n]c(x)^{(m-2)n+1} =\frac{(m-2)n+1}{(m-1)n+1}\binom{mn}{n}.

    The conjecture asserts that, for every n≥0n\ge0,

    [xn]gm(x)=[xn](−x(1−W(−x))mW(−x))n,[x^n]g_m(x)= [x^n]\left(\frac{-x(1-W(-x))^m}{W(-x)}\right)^n,

    where WW is the formal Lambert WW-series, W(z)eW(z)=zW(z)e^{W(z)}=z. The missing lower entry in the binomial in the prompt is forced by the Riordan central coefficient computation.

    Result: The conjecture is true.

    Set

    T(x)=−W(−x).T(x)=-W(-x).

    Then T=xeTT=x e^T, and

    −x(1−W(−x))mW(−x)=x(1+T)mT=e−T(1+T)m.\frac{-x(1-W(-x))^m}{W(-x)} =\frac{x(1+T)^m}{T} =e^{-T}(1+T)^m.

    For n=0n=0, both sides equal 11. For n≥1n\ge1, Lagrange inversion applied to T=xeTT=x e^T gives

    [xn](e−T(1+T)m)n=1n[un−1]ddu(e−nu(1+u)mn)enu.[x^n]\left(e^{-T}(1+T)^m\right)^n =\frac1n[u^{n-1}] \frac{d}{du}\left(e^{-nu}(1+u)^{mn}\right)e^{nu}.

    Since

    ddu(e−nu(1+u)mn)=ne−nu(1+u)mn−1(m−1−u),\frac{d}{du}\left(e^{-nu}(1+u)^{mn}\right) =n e^{-nu}(1+u)^{mn-1}(m-1-u),

    we obtain

    [xn](−x(1−W(−x))mW(−x))n=[un−1](1+u)mn−1(m−1−u).[x^n]\left(\frac{-x(1-W(-x))^m}{W(-x)}\right)^n =[u^{n-1}](1+u)^{mn-1}(m-1-u).

    Hence

    [un−1](1+u)mn−1(m−1−u)=(m−1)(mn−1n−1)−(mn−1n−2)=(m−2)n+1(m−1)n+1(mnn).\begin{aligned} &[u^{n-1}](1+u)^{mn-1}(m-1-u)\\ &=(m-1)\binom{mn-1}{n-1}-\binom{mn-1}{n-2}\\ &=\frac{(m-2)n+1}{(m-1)n+1}\binom{mn}{n}. \end{aligned}

    This is exactly the stated central coefficient sequence, so

    [xn]gm(x)=[xn](−x(1−W(−x))mW(−x))n.[x^n]g_m(x)= [x^n]\left(\frac{-x(1-W(-x))^m}{W(-x)}\right)^n.

    Verification audit: the apparent singularity at x=0x=0 is removable because W(−x)=−x+O(x2)W(-x)=-x+O(x^2); Lagrange inversion applies since T=xeTT=x e^T with e0=1e^0=1; the case n=0n=0 was handled separately; no hypothesis beyond the reconstructed m≥2m\ge2 Riordan-array setting was used.

    Citation: No exact prior resolution of Conjecture 14 was identified. The proof uses the standard Lagrange inversion theorem; see, e.g., R. P. Stanley, Enumerative Combinatorics, Vol. 2, Cambridge University Press, §5.4.

    Reviews

    1 machine check

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    • Machine check · not human verification

      machine: correct

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

      scope Full solution as submitted; SMD novelty classification TYPE1

      PASS

      The proof attacks the stated coefficient identity and is mathematically sound. Setting T=−W(−x)T=-W(-x) gives T=xeTT=xe^T and rewrites the antecedent as e−T(1+T)me^{-T}(1+T)^m. Lagrange inversion then yields

      [xn](e−T(1+T)m)n=[un−1](1+u)mn−1(m−1−u),[x^n](e^{-T}(1+T)^m)^n=[u^{n-1}](1+u)^{mn-1}(m-1-u),

      which simplifies to the stated central coefficient

      (m−2)n+1(m−1)n+1(mnn).\frac{(m-2)n+1}{(m-1)n+1}\binom{mn}{n}.

      The n=0n=0 case and removable singularity at x=0x=0 are handled. No fatal gaps are present.

      Novelty assessment

      TYPE1

      Classification rationale: The accepted resolution appears to be genuinely new as an explicit proof of Barry’s Conjecture 14, but it is a very short and routine application of Lagrange-Bürmann inversion to T=−W(−x)T=-W(-x), T=xeTT=xe^T. The coefficient family itself is already standard/recorded as ballot or two-row standard Young tableau numbers. This would not support a standalone combinatorics paper; at most it is suitable as a short note, erratum-style observation, or OEIS/JIS comment.

      Literature check: I found no prior explicit resolution of Barry’s Conjecture 14 or the exact Lambert-WW coefficient identity for arbitrary mm. The original JIS article states the result as a conjecture. Searches of the web, arXiv, OEIS, and metadata/indexing sources for combinations of “central antecedents,” “Conjecture 14,” “W(−x)W(-x),” “(1−W(−x))m(1-W(-x))^m,” and “A214776 Lambert W” did not locate a later proof. OEIS A214776 and its columns record the same closed-form coefficient family and Young-tableau/ballot interpretations, but not the Lambert-WW antecedent identity.

      Citation: Paul Barry, “On the Central Antecedents of Integer (and Other) Sequences,” J. Integer Sequences 23 (2020), Article 20.8.3. See also OEIS A214776 for the known coefficient array.

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