The [Kimberling sequence](http://oeis.org/A007063) is a recursively defined "shuffling sequence" (pictorial description [here](https://codegolf.stackexchange.com/questions/67542/the-kimberling-sequence)). Let $k:\mathbb{N}\to \mathbb{N}$ be the Kimberling sequence. Does $k$ map members of $\mathbb{N}$ arbitrarily far away, or more formally: given $N\in\mathbb{N}$ is there $m\in\mathbb{N}$ such that $|k(m)-m|>N$?



Does the Kimberling sequence map numbers "arbitrarily far away"?
The answer is yes. Indeed, as noted at [A007063](http://oeis.org/A007063), 
$$k(\theta\_j)=3\theta\_j-(j+1),
$$
where 
$$\theta\_j:=\sum\_{i=0}^{j-1}2^{\lfloor i/3\rfloor}\ge2^{\lfloor(j-1)/3\rfloor}. 
$$
So,
$$k(\theta\_j)-\theta\_j=2\theta\_j-(j+1)\underset{j\to\infty}\longrightarrow\infty,
$$
as desired. 


