Magnitude of $f_3(n)$ compared to power towers of tens In the fast growing hierarchy , the sequence $f_2(n)$ is defined as $$f_2(n)=n\cdot 2^n$$
The number $f_3(n)$ is defined by $$f_3(n)=f_2^{\ n}(n)$$
For example, to calculate $f_3(5)$, we have to apply the operator $n\cdot 2^n$ five times with start value $5$.
Denote $$T(n):=10\uparrow 10 \uparrow \cdots \uparrow 10 \uparrow 10$$ with $n$ tens, so a power tower of tens with height $n$.
With the help of the computer, I found out that $f_{30}<T(31)$ , but $f_{31}>T(32)$, so $31$ is the smallest number $n$ with $f_3(n)>T(n+1)$

  
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*Can this value also be found without electronic help by bounding the function $f_3(n)$ ?
  
*Can I also find the smallest number $n$ with $f_3(n)>T(n+k)$ for $k=2,3,4,\cdots$ without brute force ?
  


Interestingly, 10 is not the best base to choose, in fact, 2 is much more optimal, if you allow the top exponent to be altered. As shown in this post of mine, we have the following bounds:
$$n\operatorname{Tet}(2^n,n,1)\le f_3(n)\le\operatorname{Tet}(2\sqrt[n]n,n,n)$$
where
$$\operatorname{Tet}(a,b,c)=\underbrace{a\widehat{}a\widehat{}\dots\widehat{}a\widehat{}}_bc=\begin{cases}c,&b=0\\a\widehat{}\operatorname{Tet}(a,b-1,c),&b>0\end{cases}$$
Thus, it follows that
$$\operatorname{Tet}(2,n,n)\le f_3(n)<^\star\operatorname{Tet}(2+\varepsilon,n,n)$$
where $<^\star$ represents eventual dominance and $\varepsilon>0$.
As for base conversion, you can ignore most of the 2's, only considering the top 2's and converting them to base 10.

From this method, we may find bounds for $h(k)=\min\{n|f_3(n)>T(n+k)\}$ by considering the smallest $n$ such that $n\operatorname{Tet}(2^n,j,1)>T(j+k)$ for varying values of $j$.
For $j=1$, we get that which you already noticed,
$$h(k)\approx\log_2(T(k+1))\\h(k)\approx\log_2(T(k+1))-\log_2(h(k))$$
which readily provides a recursive approach for approximating $h(k)$. For example, at $k=2$, we get an initial approximation of
$$h(2)\approx\log_2(T(3))\approx33219280949\\h(2)\approx\log_2(T(3))-\log_2(33219280949)\approx33219280914$$
which is fairly accurate. This is the fixed-point approximation to $n2^n=n\operatorname{Tet}(2^n,1,1)=T(3)$.
