Recent developments in the theory of Stirling numbers

NEW

Problem: The displayed formula is evidently a typo: $K_{n+1}\in\{K_n,K_{n+1}\}$ is tautological. The standard Stirling-mode property is

Interpreting “Whitney and $r$-Whitney numbers” as Whitney numbers of the second kind, let

and

with $m>0,\ r\ge0$. The ordinary Whitney numbers are $W_m(n,k)=W_{m,1}(n,k)$. For each $n$, let $K_n$ be the least index at which the row $k\mapsto W_{m,r}(n,k)$ is maximal.

Result: Yes. For all $m>0,\ r\ge0$,

Proof. Put $a_{n,k}=W_{m,r}(n,k)$, $b_k=mk+r$, and $B_k=\prod_{i=1}^k b_i$. Define

Writing $\alpha=r/m$, the recurrence gives

If $\alpha>0$, the operator

sends polynomials with all zeros in $(-1,0)$ to polynomials with all zeros in $(-1,0)$: indeed, for $q=T_\alpha p$,

changes sign once in each interval between consecutive zeros of $p$, also once between $-1$ and the first zero and once between the last zero and $0$. Thus $C_n$ has only real zeros in $[-1,0]$; the case $\alpha=0$ follows by continuity.

By Newton’s inequalities, $(c_{n,k})_k$ is log-concave:

Since $c_{n,k}=B_k a_{n,k}$, this implies

Hence the ratios $R_{n,k}=a_{n,k}/a_{n,k-1}$ decrease with $k$, so every row is unimodal.

Now

and the preceding inequality gives the interlacing

If $K_n$ is the least mode, then

The interlacing therefore gives $R_{n+1,k}>1$ for $k\le K_n$, so $K_{n+1}\ge K_n$, and also

so $K_{n+1}\le K_n+1$. Therefore

The same argument works for the greatest maximizing index, so the conclusion is independent of the usual tie convention.

Citation: No external resolution is used; the proof above establishes the repaired conjecture. Newton’s inequalities are the only standard invoked fact, and their hypotheses are verified by the real-rootedness argument.