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Let $\newcommand{\ome}{\omega_+}\ome=\omega\setminus\{0\}$ denote the set of positive integers. For $a,b\in \ome$ we let $$||a-b|| = |(a\setminus b)\cup(b\setminus a)|$$ be the absolute difference of $a,b$.

If $\varphi:\ome\to\ome$ is a bijection, we let the absolute derivative of $\varphi$, denoted by $\varphi':\ome\to\ome$, be defined by $\varphi'(n)=||\varphi(n)-\varphi(n+1)||$ for all $n\in\ome$.

We also write $\varphi^{(1)}=\varphi'$, and inductively for $n\in\ome$ we let $\varphi^{(n+1)} =\big(\varphi^{(n)}\big)'$.

Question. Is there a bijection $\varphi:\ome\to\ome$ such that $\varphi':\ome\to\ome$ is a bijection again? And if yes, can we continue this to higher absolute derivatives $\varphi^{(k)}$ for $k>1$?

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    $\begingroup$ I'm a bit confused by your notations. In your formula for $||a - b||$, you write set operations, so you're viewing $a$ and $b$ as von Neumann ordinals, right? But isn't your $||a-b||$ just $|a-b|$ where $|\cdot|$ is the standard absolute value on $\mathbb{Z}$? What's the point of ordinals here? $\endgroup$ Commented Jul 9 at 20:43
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    $\begingroup$ As someone who enjoys your contributions Dominic, and someone who loves the ordinals, when I saw that $\omega_+$ meant the positive naturals I said 'oh god why'. I think some of the kickback you're getting is related to this notational choice. $\endgroup$ Commented Jul 9 at 21:47
  • $\begingroup$ Or is it preparing us to generalizing it to higher ordinals? $\endgroup$ Commented Jul 9 at 22:05
  • $\begingroup$ What does $a\b$ mean? $\endgroup$ Commented Jul 10 at 0:17
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    $\begingroup$ @Claude, so, Jean is right – the whole thing is just a complicated way of writing $|b-a|$. $\endgroup$ Commented Jul 10 at 6:49

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If I understood the notations correctly (see my comment), you are looking for a bijection $φ : ℕ^* → ℕ^*$ (where $ℕ^* = \{1, 2, …\}$) such that $φ' : ℕ^* → ℕ^*, n ↦ |φ(n+1)-φ(n)|$ is also a bijection.

Such a bijection can be built by stages. Assume $φ(0), …, φ(n-1)$ are already chosen. Let $x ∈ ℕ^*$ be the smallest not among $φ(0), …, φ(n-1)$ and let $x' ∈ ℕ^*$ be the smallest not among $φ'(0), …, φ'(n-2)$. Now pick $φ(n)$ large enough so that $φ(n)$ is not among $φ(0), …, φ(n-1)$, and so that $φ'(n-1) = |φ(n)-φ(n-1)|$ as well as $|φ(n)-x|$ are not among $φ'(0), …, φ'(n-2)$, and set $φ(n+1) = x$. Then pick $φ(n+2)$ large enough so that $φ(n+2)$ as well as $φ(n+2)+y$ are not among $φ(0), …, φ(n+1)$ and $|φ(n+2)-φ(n+1)|$ is not among $φ'(0), …, φ'(n)$, and set $φ(n+3) = φ(n+2) + x'$. The construction ensures that $φ$ and $φ'$ are injective and surjective.

Adapting the construction to work diagonally ensures that $φ, φ', φ'', …$ are all bijections. You can incrementally add values to make sure that the smallest value not hit by $φ$ gets hit, then the smallest value for $φ$ followed by $φ'$, then $φ$ followed by $φ'$ followed by $φ''$, etc. At each step, just go high enough to preserve injectivity.

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