User blog:Simply Beautiful Art/Even tighter bounds on f 3 in the fast growing hierarchy

Brief refresher on the fast growing hierarchy for finite subscripts:

$$f_k(n)=\begin{cases}n+1,&k=0\\f_{k-1}^n(n),&k>0\end{cases}$$

where we use function iteration notation.

I then define tetration with a modified top exponent as follows:

$$a\uparrow\uparrow^bc=a^{a^{\dots a^b}}=\begin{cases}b,&c=0\\a\uparrow(a\uparrow\uparrow^b(c-1)),&c>0\end{cases}$$

where $$a\uparrow b=a^b$$.

I shall then prove the following bounds for all $$n>0$$:

$$n(2^n\uparrow\uparrow^1n)\le f_3(n)\le(2\sqrt[n]n)\uparrow\uparrow^nn$$

We do this by applying induction over $$j$$ in the following:

$$n(2^j\uparrow\uparrow^1k)\le f_2^j(n)\le(2\sqrt[n]n)\uparrow\uparrow^nj$$

Trivially it is true for $$j=1$$. Assume it holds for some arbitrary $$j>0$$, and we shall prove it holds for $$j+1$$.

$$\begin{align}n(2^n\uparrow\uparrow^1(k+1))&=n(2\uparrow(2^n\uparrow\uparrow^1k))\\&\le(2^n\uparrow\uparrow^1k)(2\uparrow(2^n\uparrow\uparrow^1k))\\&=f_2(2^n\uparrow\uparrow^1k)\\&\le f_2(f_2^k(n))\\&=f_2^{k+1}(n)\\&=f_2(f_2^k(n))\\&\le f_2((2\sqrt[n]n)\uparrow\uparrow^nk)\\&=((2\sqrt[n]n)\uparrow\uparrow^nk)(2\uparrow((2\sqrt[n]n)\uparrow\uparrow^nk))\\&=2\uparrow(((2\sqrt[n]n)\uparrow\uparrow^nk)+\log_2((2\sqrt[n]n)\uparrow\uparrow^nk))\\&=2\uparrow\left(((2\sqrt[n]n)\uparrow\uparrow^nk)\left(1+\frac{\log_2((2\sqrt[n]n)\uparrow\uparrow^nk)}{(2\sqrt[n]n)\uparrow\uparrow^nk}\right)\right)\\&\le2\uparrow(((2\sqrt[n]n)\uparrow\uparrow^nk)\left(1+\frac{\log_2(n)}n\right)\right)=(2\sqrt[n]n)\uparrow((2\sqrt[n]n)\uparrow\uparrow^nk)\\&=(2\sqrt[n]n)\uparrow\uparrow^n(k+1)\end{align}$$

Q.E.D.