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For any $A\subseteq \mathbb{N}$ we let the (lower) density of $A$ be defined by $$d(A) = \liminf_{n\to\infty}\frac{|A\cap\{0,\ldots,n\}|}{n+1}.$$ If $\pi:\mathbb{N}\to\mathbb{N}$ is a permutation (bijection), let $\text{ex}(\pi) = \{n\in\mathbb{N}: \pi(n) > n\}$ be the set of excedances, and let $\text{negex}(\pi)=\{n\in\mathbb{N}: \pi(n) < n\}$ be the set of "negative excedances".

To me it seems inconceivable that there is a permutation $\pi:\mathbb{N}\to\mathbb{N}$ with $d\big(\text{ex}(\pi)\big) \neq d\big(\text{negex}(\pi)\big)$ -- but my intuition has let me down many times.

Is my intuition correct this time? ${}$

For any $A\subseteq \mathbb{N}$ we let the (lower) density of $A$ be defined by $$d(A) = \liminf_{n\to\infty}\frac{|A\cap\{0,\ldots,n\}|}{n+1}.$$ If $\pi:\mathbb{N}\to\mathbb{N}$ is a permutation (bijection), let $\text{ex}(\pi) = \{n\in\mathbb{N}: \pi(n) > n\}$ be the set of exceedances, and let $\text{negex}(\pi)=\{n\in\mathbb{N}: \pi(n) < n\}$ be the set of "negative exceedances".

To me it seems inconceivable that there is a permutation $\pi:\mathbb{N}\to\mathbb{N}$ with $d\big(\text{ex}(\pi)\big) \neq d\big(\text{negex}(\pi)\big)$ -- but my intuition has let me down many times.

Is my intuition correct this time? ${}$

For any $A\subseteq \mathbb{N}$ we let the (lower) density of $A$ be defined by $$d(A) = \lim\inf_{n\to\infty}\frac{|A\cap\{0,\ldots,n\}|}{n+1}.$$ If $\pi:\mathbb{N}\to\mathbb{N}$ is a permutation (bijection), let $\text{ex}(\pi) = \{n\in\mathbb{N}: \pi(n) > n\}$ be the set of excedances, and let $\text{negex}(\pi)=\{n\in\mathbb{N}: \pi(n) < n\}$ be the set of "negative excedances".

To me it seems inconceivable that there is a permutation $\pi:\mathbb{N}\to\mathbb{N}$ with $d\big(\text{ex}(\pi)\big) \neq d\big(\text{negex}(\pi)\big)$ -- but my intuition has let me down many times.

Is my intuition correct this time?

For any $A\subseteq \mathbb{N}$ we let the (lower) density of $A$ be defined by $$d(A) = \liminf_{n\to\infty}\frac{|A\cap\{0,\ldots,n\}|}{n+1}.$$ If $\pi:\mathbb{N}\to\mathbb{N}$ is a permutation (bijection), let $\text{ex}(\pi) = \{n\in\mathbb{N}: \pi(n) > n\}$ be the set of excedances, and let $\text{negex}(\pi)=\{n\in\mathbb{N}: \pi(n) < n\}$ be the set of "negative excedances".

To me it seems inconceivable that there is a permutation $\pi:\mathbb{N}\to\mathbb{N}$ with $d\big(\text{ex}(\pi)\big) \neq d\big(\text{negex}(\pi)\big)$ -- but my intuition has let me down many times.

Is my intuition correct this time? ${}$

For any $A\subseteq \mathbb{N}$ we let the (lower) density of $A$ be defined by $$d(A) = \lim\inf_{n\to\infty}\frac{|A\cap\{0,\ldots,n\}|}{n+1}.$$ If $\pi:\mathbb{N}\to\mathbb{N}$ is a permutation (bijection), let $\text{pos}(\pi) = \{n\in\mathbb{N}: \pi(n) > n\}$ and $\text{neg}(\pi)=\{n\in\mathbb{N}: \pi(n) < n\}$.

To me it seems inconceivable that there is a permutation $\pi:\mathbb{N}\to\mathbb{N}$ with $d\big(\text{pos}(\pi)\big) \neq d\big(\text{neg}(\pi)\big)$ -- but my intuition has let me down many times.

Is my intuition correct this time?

For any $A\subseteq \mathbb{N}$ we let the (lower) density of $A$ be defined by $$d(A) = \lim\inf_{n\to\infty}\frac{|A\cap\{0,\ldots,n\}|}{n+1}.$$ If $\pi:\mathbb{N}\to\mathbb{N}$ is a permutation (bijection), let $\text{ex}(\pi) = \{n\in\mathbb{N}: \pi(n) > n\}$ be the set of excedances, and let $\text{negex}(\pi)=\{n\in\mathbb{N}: \pi(n) < n\}$ be the set of "negative excedances".

To me it seems inconceivable that there is a permutation $\pi:\mathbb{N}\to\mathbb{N}$ with $d\big(\text{ex}(\pi)\big) \neq d\big(\text{negex}(\pi)\big)$ -- but my intuition has let me down many times.

Is my intuition correct this time?