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For every $n\geq1$, there is some $f(n)$ such that for the Stirling numbers of the first kind we have:

$c(n,0)<c(n,1)<...<c(n,f(n)-1)\leq c(n,f(n))>c(n,f(n)+1)>...>c(n,n)$

Moreover, $f(n)=f(n-1)$ or $f(n)=f(n-1)+1$.

Mike Earnest
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    Hints: Prove this by induction on $n$. Given that the $n^{th}$ row of the Stirling triangle increases up to a point and decreases after, you need to prove the same is true for the $(n+1)^{st}$ row. This is equivalent to showing that differences $c(n+1,k+1)-c(n+1,k)$ are positive for $k$ up to a point, and negative for $k$ past that point. You can use the relation $c(n+1,k+1)=c(n,k)+n\cdot c(n,k+1)$ to relate the differences for the $(n+1)^{st}$ row to the differences for the $n^{th}$ row. – Mike Earnest May 08 '22 at 17:00
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    Apparently, $f(n)\sim \ln n$: https://math.stackexchange.com/a/1156802 – Mike Earnest May 08 '22 at 17:04
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    Thanks for the suggestion. Is there a way to do this without induction? Is there a combinatorial proof? –  May 09 '22 at 07:02

1 Answers1

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We will prove, by induction on $n$, the existence of an integer $f(n)$ such that these three conditions hold:

  • for each $k\in \{1,2,\dots,f(n)-1\}$, we have $c(n,k)-c(n,k-1)>0$,

  • $c(n,f(n))-c(n,f(n)-1)\ge 0$,

  • for each $ k\in \{f(n)+1,f(n)+2,\dots,n\}$, we have $c(n,k)-c(n,k-1)<0$.

The base case is obvious, so onto the induction step. Assume that $f(n)$ exists. We want to prove $f(n+1)$ exists, so we consider the differences $c(n+1,k)-c(n+1,k-1)$. Note $$ \begin{align} c(n+1,k)-c(n+1,k-1) &= [c(n,k-1)+n\,c(n,k)]-[c(n,k-2)+n\,c(n,k-1)] \\ &= [\color{blue}{c(n,k-1)-c(n,k-2)}]+n\cdot[\color{green}{c(n,k)-c(n,k-1)}] \end{align} $$ Conclude as follows:

  • For all $k\le f(n)$, the induction hypothesis implies that the green difference is nonnegative, and the blue difference is positive. It follows that $c(n+1,k)-c(n+1,k-1)>0$ for $k\le f(n)$.

  • For all $k\ge f(n)+2$, the induction hypothesis implies that the blue and green differences are both negative. It follows that $c(n+1,k)-c(n+1,k-1)<0$ for $k\ge f(n)+2$.

  • The only difference left to consider is when $k=f(n)+1$.

    • If $c(n+1,f(n)+1)-c(n+1,f(n))<0$, we have proved $f(n+1)=f(n)$.

    • If $c(n+1,f(n)+1)-c(n+1,f(n))\ge 0$, we have proved $f(n+1)=f(n)+1$.

Either way, we have proved $f(n+1)$ exists, and that it equals either $f(n)$ or $f(n)+1$.

Mike Earnest
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