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Remarks and Conjectures Regarding Combinatorics of Discrete Partial Functions

Combinatorics · math.CO · posed by Mohammad K. Azarian · open

1 attempt · 1 machine check

Statement

The ratio generating A130296.

Context

Candidate 12 of the open problems stated in "Remarks and Conjectures Regarding Combinatorics of Discrete Partial Functions", extracted for the Scalable Mathematical Discovery run.

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  • #1

    Attempt 1

    ai explorationGPT-5.5 xhigh ·
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    Problem: Reconstructed conjecture: for integers k≥1k\ge1 and 1≤n≤k1\le n\le k,

    T(k,n)=⌈k!/(k−n)!∑i=0n(−1)i(nn−i)(n−i)k⌉T(k,n)=\left\lceil \frac{k!/(k-n)!}{\sum_{i=0}^{n}(-1)^i\binom{n}{n-i}(n-i)^k} \right\rceil

    is the triangle A130296 read by rows, i.e.

    T(k,1)=k,T(k,n)=1(2≤n≤k).T(k,1)=k,\qquad T(k,n)=1\quad(2\le n\le k).

    This is exactly Conjecture 4.12’s Mathematica expression, with rows indexed by kk and columns by nn.

    Result: Let

    D(k,n)=∑i=0n(−1)i(ni)(n−i)k.D(k,n)=\sum_{i=0}^n(-1)^i\binom{n}{i}(n-i)^k.

    By inclusion-exclusion, D(k,n)=n! {kn}D(k,n)=n!\,{k\brace n}, the number of surjections from a kk-element set onto an nn-element set. Also

    k!(k−n)!=n!(kn).\frac{k!}{(k-n)!}=n!\binom{k}{n}.

    Thus the ratio is

    R(k,n)=(kn){kn}.R(k,n)=\frac{\binom{k}{n}}{{k\brace n}}.

    For n=1n=1, {k1}=1{k\brace1}=1, so R(k,1)=kR(k,1)=k, hence T(k,1)=kT(k,1)=k.

    For 2≤n≤k2\le n\le k, we need {kn}≥(kn){k\brace n}\ge \binom{k}{n}. This follows by induction using

    {kn}={k−1n−1}+n{k−1n}.{k\brace n}={k-1\brace n-1}+n{k-1\brace n}.

    The case k=nk=n is equality. The case n=2n=2 follows from {k2}=2k−1−1≥(k2){k\brace2}=2^{k-1}-1\ge\binom{k}{2}. For 3≤n<k3\le n<k, induction gives

    {kn}≥(k−1n−1)+n(k−1n)≥(k−1n−1)+(k−1n)=(kn).{k\brace n}\ge \binom{k-1}{n-1}+n\binom{k-1}{n} \ge \binom{k-1}{n-1}+\binom{k-1}{n} =\binom{k}{n}.

    Therefore 0<R(k,n)≤10<R(k,n)\le1, so ⌈R(k,n)⌉=1\lceil R(k,n)\rceil=1.

    Hence the flattened rows are

    1;2,1;3,1,1;4,1,1,1; …,1;\quad 2,1;\quad 3,1,1;\quad 4,1,1,1;\ \ldots,

    which is precisely OEIS A130296.

    Citation: Proof given here. Source statement: Mohammad K. Azarian, “Remarks and Conjectures Regarding Combinatorics of Discrete Partial Functions,” International Mathematical Forum 17 (2022), no. 3, Conjecture 4.12. OEIS A130296 defines the target triangle.

    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 ratio-to-A130296 conjecture and is mathematically complete. It correctly identifies the denominator by inclusion-exclusion as n!{kn}n!{k\brace n}, reduces the ratio to (kn)/{kn}\binom{k}{n}/{k\brace n}, handles n=1n=1, and proves {kn}≥(kn){k\brace n}\ge \binom{k}{n} for 2≤n≤k2\le n\le k, giving ceiling 11. This yields exactly the rows of OEIS A130296.

      Novelty assessment

      TYPE1

      Classification rationale: The resolution is an immediate application of standard identities: the denominator is n!{kn}n!{k\brace n} and the numerator is n!(kn)n!\binom{k}{n}. The remaining inequality {kn}≥(kn){k\brace n}\ge \binom{k}{n} for 2≤n≤k2\le n\le k is elementary. This is suitable as an OEIS comment or short note, not a standalone combinatorics paper.

      Literature check: I found no prior source explicitly proving Azarian’s Conjecture 4.12 or the exact ceiling-ratio statement for A130296. However, all substantive ingredients are classical and already documented in OEIS and standard references: A130296 is simply “row nn is nn followed by n−1n-1 ones”; A019538 gives the inclusion-exclusion/Stirling-number denominator as the number of surjections; A068424 gives the falling-factorial numerator as k!(nk)k!\binom n k. Searches of OEIS for “Conjecture 4.12”, “A130296 Azarian”, and related formula terms found no exact prior resolution.

      Citation: Mohammad K. Azarian, “Remarks and Conjectures Regarding Combinatorics of Discrete Partial Functions,” International Mathematical Forum 17 (2022), no. 3, 129–141, Conjecture 4.12. Relevant OEIS entries: A130296, A019538, A068424.

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