IRL_ABooklet

Solution 1. We first prove inductively that the following equations hold for all $i\in \mathbb{N}$: $$\begin{array}{rl}f(1\cdot 4^{i-1})& =2\cdot 4^{i-1},\\ f(2\cdot 4^{i-1})& =3\cdot 4^{i-1},\\ f(3\cdot 4^{i-1})& =4\cdot 4^{i-1}=4^i.\end{array}$$ The base case follows easily from the string of inequalities $$1<f(1)<f(f(1))<f(f(f(1)))=4.$$ The inductive step then follows from the assumed identity for $f(f(f(n)))$.

Let $A=1\cdot 4^{i-1}$ for some $i\in \mathbb{N}$, let $S$ be the set of integers between $A$ and $2A$ inclusive, and let $T$ be the set of integers between $2A$ and $3A$ inclusive. Since $f$ maps the endpoints of $S$ to the endpoints of $T$, the strictly increasing condition for $f$ implies that $f(S)\subset T$, and $S$ and $f(S)$ have the same cardinality. But $S$ and $T$ clearly have the same cardinality, so we must have $f(S)=T$, and monotonicity now implies that $f(A+k)=2A+k$ for all $1\le k\le 4^{i-1}$. In the same way, we see that $f(2A+k)=3A+k$ for all $1\le k\le 4^{i-1}$. Finally, $f(3A+k)=f(f(f(A+k)))=4A+4k$ for all $1\le k\le 4^{i-1}$. We now have a formula for $f(n)$ for all $n$. Writing $2022=1024+998=4^5+998$, we see that $$f(2022)=2\cdot 1024+998=3046.$$

Solution 2. Applying $f$ to $f(f(f(n)))=4n$ and then using the same equation with $n$ replaced by $f(n)$ we get $$f(4n)=f(f(f(f(n))))=4f(n)$$ for all $n\in \mathbb{N}$, and hence $$f(4^{n}n)=4f(4^{n-1}n)=\cdots =4^{n}f(n).$$ We know $f(1)=2$, $f(2)=3$, $f(3)=4$ since the function is increasing with $$1<f(1)<f(f(1))<f(f(f(1)))=4.$$ Hence $f(4^i)=4^{i}f(1)=2\cdot 4^i$ and $f(2\cdot 4^i)=4^{i}f(2)=3\cdot 4^i$ so $$4^i\le \sum _{x=4^i}^{2\cdot 4^i-1}[f(x+1)-f(x)]=f(2\cdot 4^i)-f(4^i)=4^i.$$ Thus $f(x+1)-f(x)=1$ for each $x$ with $4^i\le x\le 2\cdot 4^i-1$. It follows that $f(4^i+k)=f(4^i)+k=2\cdot 4^i+k$ for $1\le k\le 4^i$ and hence $f(2022)=2\cdot 1024+998=3046$ as above.