Let Ackermann function $A(m,n)$ be defined as
\begin{align} &A(0,n) = n+1,\\ &A(m+1,0) = A(m,1),\\ &A(m+1,n+1) = A(m, A(m+1,n)),~n,m\in \mathbb{N},\\ \end{align}
and the inverse Ackermann function $\alpha(n)$ be
$$ \alpha(n) = \min\lbrace m|A(m,m) \geq n\rbrace. $$
How could I find any closed form (if it exists) or some inequalities for $\alpha(kn)$, where $k \in \mathbb{N}$, in terms of $\alpha(n)$ and some function $f(k)$ (without $n$ involved)?
My problem: I would like to show that in Big O notation for sufficiently large $n$ the following will hold:
$$ \mathcal{O}\left(\alpha(kn\right)) \approx \mathcal{O}\left(\alpha(n\right)),~k\text{ - constant number bigger than 0} $$
or get some nicer inequalities for $\mathcal{O}\left(\alpha(kn\right))$.
I tried to start with function $d(n) = 2^{2^{\vdots^{2}}}$, where the height of the power tower is $n+2$ from there, with inequality $A(n,n) > d(n),~n\geq 4$, but it was dead end for me.
What is better approach to this problem?
The fact that $A(n,n)$ is convex and $A(n,n)\gt n$ already gives you the asymptotic $O(\alpha(kn))\equiv O(\alpha(n))$; $A(n,n)\gt \exp(n)$ is enough to give you that $\lim_{n\to\infty} \frac{\alpha(kn)}{\alpha(n)}=1$ for all $k$. (To see this, note that we get $\alpha(kn)\lt \alpha(k)+\alpha(n)$, so $\frac{\alpha(kn)}{\alpha(n)}\lt 1+\frac{\alpha(k)}{\alpha(n)}\to 1$.) From a practical standpoint, you should ignore $\alpha()$ entirely; for every $n$ smaller than the size of the universe, $\alpha(n)\lt 5$, so it can be treated as effectively constant.