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A sequence of numbers $a_{0}, a_{1}, \cdots, a_{n}, \cdots$ is said to be log-concave if for $1 \leq i \leq n-1, a_{i-1} a_{i+1} \leq a_{i}^{2} .$ consider the sequence $\left(\begin{array}{c}k \\ k\end{array}\right),\left(\begin{array}{c}k+1 \\ k\end{array}\right),\left(\begin{array}{c}k+2 \\ k\end{array}\right), \cdots, .$ Show that this sequence is log-concave. Also give a combinatorial proof.

I did solve the question using actual computation (That is by expansion of terms). Also I could find a combinatorial proof of a similar sequence,

namely sequence $\left(\begin{array}{l}n \\ 0\end{array}\right),\left(\begin{array}{l}n \\ 1\end{array}\right), \cdots,\left(\begin{array}{l}n \\ n\end{array}\right)$ is log-concave, by considering the pairs of subsets.

But I couldn't prove that $\left(\begin{array}{c}k \\ k\end{array}\right),\left(\begin{array}{c}k+1 \\ k\end{array}\right),\left(\begin{array}{c}k+2 \\ k\end{array}\right), \cdots, .$ is log concave by any combinatorial argument. (here $k$ is fixed). Any help is highly appreciated, thanks in advance!

RobPratt
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1 Answers1

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Thanks to Sharky Kesa for pointing out a huge error in the initial version answer. The answer is now correct, and has been upgraded from a hint to a full solution.


Let $[n]$ denote $\{1,\dots,n\}$, and for a set $S$, let $\binom{S}{k}$ denote the set of size $k$ subsets of $S$.

Given $k,n\in \mathbb N$, with $n\ge k$, you need to find an injection $$ f: \binom{[n]}{k}\times \binom{[n+2]}{k} \to \binom{[n+1]}{k}\times \binom{[n+1]}{k}. $$ Given an ordered pair of subsets $(A,B)$, with $A\in \binom{[n]}k$ and $B\in \binom{[n+2]}{k}$, we proceed in several cases.

  • If $n+2\notin B$, then $f(A,B)=(A,B)$.

  • If $n+2\in B$, but $n+1\notin B$, then $$f(A,B)=(B\setminus \{n+2\}\cup \{n+1\},\;A) .$$

  • Finally, suppose that $\{n+1,n+2\}\subseteq B$. This is the tricky case. To do this, suppose the existence of a map $$\varphi:\binom{[n]}{k}\times \binom{[n]}{k-2} \to \binom{[n]}{k-1}\times \binom{[n]}{k-1}$$such that $\varphi$ is injective. Let $\varphi(A, B\setminus \{n+1,n+2\})=(C,D)$. Then $$ f(A,B)=(C\cup \{n+1\},D\cup \{n+1\}). $$

Assuming we are able to construct $\varphi$, it should be clear that $f$ is an injection. The cases in the three bullet points all have distinct images. Let $(E,F)=f(A,B)$. In the first case, $(n+1)\notin E$. For the second bullet, $(n+1)\in E$, but $(n+1)\notin F$. For the third bullet, $(n+1)\in E$ and $(n+1)\in F$. Furthermore, it should be clear that $f$ restricted to the domains described in each bullet point is injective.

All that remains is to construct the map, $\varphi$. This was done at this other MSE answer: https://math.stackexchange.com/a/2881497/177399. A different presentation of the same injection, using lattice walks and the reflection principle, is given here: https://math.stackexchange.com/a/4742856/177399.

Mike Earnest
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  • Great Thanks! I guess this works, f((A,B))= \begin{cases} (A,B),& \text{if } n + 1 \not\in A \ (A/{n+1}, B \cup {n+2}), & \text{otherwise} \end{cases} – Mathaddict May 31 '21 at 02:56
  • I'm fairly sure this method is wrong. An injection would prove the other inequality, which is just wrong by considering $n = k = 1$. Furthermore, the map given in @Mathaddict doesn't preserve the fact that both subsets should be of size $k$. – Sharky Kesa Jul 26 '23 at 13:41
  • @SharkyKesa Thank you very much for pointing out that huge error on my part. This was a bit trickier than I initially envisioned, but my answer should now be correct. – Mike Earnest Jul 26 '23 at 15:55
  • @SharkyKesa Thank you for noticing the blunder in my thinking. Also Thank you MikeEarnest for updating your solution. It indeed seems correct to me now. – Mathaddict Jul 27 '23 at 14:26