Binomial Ward numbers and central Stirling numbers conjecture

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.

References

Primary source

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

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

A reader-provided complete-proof claim does not match the stated formulas, so the conjecture has not been settled.

Aleks Žigon Tankosić introduced both identities in 2025 as experimentally supported conjectures relating binomial Ward-number row sums to central unsigned Stirling numbers.

August 2025 paper

Tankosić’s Section 5 states the two identities as Conjecture 5.4 and gives recurrences and generating information for the relevant binomial Ward numbers, but supplies no proof or disproof.

Posted attempt

A reader-provided argument claims a complete bijective proof, identifying Ward numbers with permutations or set partitions having no singleton components. However, its final identities contain binomial weights, whereas the conjecture’s displayed sums do not; the claimed proof is therefore not an independently verified resolution of the stated problem.

Current status (as of August 2026): The paper’s two identities remain unproved and undisproved in the retrieved evidence; a posted complete-proof claim is unverified and appears to address different, weighted identities.

Sources

arxiv.org

ProofThis solution needs a summarySee full solutionHide full solution

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.