Binomial Ward numbers and central Stirling numbers conjecture

From papers

Let $n$ and $k$ be integers with $n\geq k$ , and let $\left\uparrow \begin{matrix} n \\ k \end{matrix} \right\downarrow^{\circ}$ and $\left\updownarrow \begin{matrix} n \\ k \end{matrix} \right\updownarrow^{\circ}$ denote the binomial Ward numbers of the first and second kinds, respectively. Let $\genfrac [ ] {0pt} {0} {2n} {n}$ and $\genfrac \{ \} {0pt} {0} {2n} {n}$ denote the central unsigned Stirling numbers of the first and second kinds, respectively.

Central Stirling numbers conjecture. The binomial Ward numbers satisfy

\sum_{k=0}^{n} \left\uparrow \begin{matrix} n \\ k \end{matrix} \right\downarrow^{\circ} = \genfrac [ ] {0pt} {0} {2n} {n}

and

\sum_{k=0}^{n} \left\updownarrow \begin{matrix} n \\ k \end{matrix} \right\updownarrow^{\circ} = \genfrac \{ \} {0pt} {0} {2n} {n}.

These relations are presented as conjectures based on experimental evidence and connect binomial Ward numbers with central Stirling numbers of both kinds.

Progress summary

Open

The conjecture remains open, with the available paper giving experimental evidence but no proof or counterexample.

The conjecture concerns two identities linking binomial Ward numbers with central Stirling numbers. Aleks Žigon Tankosić presents them as conjectures in a 2025 paper, based on experimental evidence rather than a proof.

Known results

The retrieved literature records the identities as conjectural; no classical proof or partial theorem specific to them was found.

August 2025 paper

Tankosić’s paper states both identities and labels them conjectures based on experimental evidence. The retrieved record contains no claimed proof, counterexample, verification, or subsequent resolution.

Current status (as of August 2026): Both identities remain open conjectures; no proof, counterexample, or verified resolution was found in the retrieved sources.

Sources

Sources & referencesView supporting material

Primary source

Aleks Žigon Tankosič, “Recurrence Relations for Some Integer Sequences Related to Ward Numbers”, arXiv:2508.04754 (2025).

1 week ago

Both identities follow from a common bijection, which also identifies every individual summand.

Write $W_1(n,k)$ and $W_2(n,k)$ for the first- and second-kind Ward numbers, with $W_i(0,0)=1$ and the defining recurrences

W_1(n,k)=(n+k-1)\bigl(W_1(n-1,k)+W_1(n-1,k-1)\bigr),

W_2(n,k)=kW_2(n-1,k)+(n+k-1)W_2(n-1,k-1).

Let $C(m,k)$ count permutations of $m$ labeled elements with exactly $k$ cycles, all of length at least two. Inspect the cycle containing the largest element. If its length exceeds two, remove that element, leaving $m-1$ possible insertion positions; if its length is two, choose its partner in $m-1$ ways. Hence

C(m,k)=(m-1)C(m-1,k)+(m-1)C(m-2,k-1).

Putting $m=n+k$ gives the first Ward recurrence, so

W_1(n,k)=C(n+k,k).

Similarly, let $T(m,k)$ count partitions of $m$ labeled elements into $k$ blocks of size at least two. The largest element either joins one of $k$ existing blocks or forms a two-element block with one of $m-1$ partners. Therefore

T(m,k)=kT(m-1,k)+(m-1)T(m-2,k-1),

and consequently

W_2(n,k)=T(n+k,k).

Now classify permutations of $[2n]$ having exactly $n$ cycles by their number $n-k$ of fixed points. Choose the fixed points in

\binom{2n}{n-k}=\binom{2n}{n+k}

ways; the remaining $n+k$ elements form $k$ nonsingleton cycles in $W_1(n,k)$ ways. Thus

\sum_{k=0}^{n}\binom{2n}{n+k}W_1(n,k) =\left[{2n\atop n}\right].

Exactly the same argument classifies partitions of $[2n]$ into $n$ blocks by their $n-k$ singleton blocks. The remaining $n+k$ elements form $k$ nonsingleton blocks in $W_2(n,k)$ ways, yielding

\sum_{k=0}^{n}\binom{2n}{n+k}W_2(n,k) =\left\{{2n\atop n}\right\}.

This proves both conjectured identities for every $n\ge0$ , with the stronger refinement that each summand counts objects with exactly $n-k$ singleton components.

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Shivam Patel · 22 hours ago