User blog:Simply Beautiful Art/Slightly better upper bound to f 3 in the fgh

See also.

$$\operatorname{Tet}(a,b,c)=\underbrace{a\widehat{~}a\widehat{~}\dots\widehat{~}a\widehat{~}}_cb=\begin{cases}b,&c=0\\a\widehat{~}\operatorname{Tet}(a,b,c-1),&c>0\end{cases}$$

I shall then prove the upper bound

$$f_3(n)\le\operatorname{Tet}\left(2\widehat{~}\left(1+{n+\log_2(n)\over n2^n}\right),\frac1{\frac1{n+\log_2(n)}+\frac1{n2^n}},n\right)$$

by proving

$$f_2^j(n)\le\operatorname{Tet}\left(2\widehat{~}\left(1+{n+\log_2(n)\over n2^n}\right),\frac1{\frac1{n+\log_2(n)}+\frac1{n2^n}},j\right)$$

Which holds trivially for $$j=1$$. Assume it holds for some arbitrary $$k\ge1$$.

$$\begin{align}f_2^{k+1}(n)&=f_2(f_2^k(n))\\&=f_2^k(n)2\widehat{~}f_2^k(n)\\&=2\widehat{~}(f_2^k(n)+\log_2(f_2^k(n)))\\&=2\widehat{~}\left(f_2^k(n)\left(1+{\log_2(f_2^k(n))\over f_2^k(n)}\right)\right)\\&\le2\widehat{~}\left(f_2^k(n)\left(1+{\log_2(f_2(n))\over f_2(n)}\right)\right)\\&=2\widehat{~}\left(f_2^k(n)\left(1+{n+\log_2(n)\over n2^n}\right)\right)\\&=\left(2\widehat{~}\left(1+{n+\log_2(n)\over n2^n}\right)\right)\widehat{~}f_2^k(n)\\&\le\left(2\widehat{~}\left(1+{n+\log_2(n)\over n2^n}\right)\right)\widehat{~}\operatorname{Tet}\left(2\widehat{~}\left(1+{n+\log_2(n)\over n2^n}\right),\frac1{\frac1{n+\log_2(n)}+\frac1{n2^n}},k\right)\\&=\operatorname{Tet}\left(2\widehat{~}\left(1+{n+\log_2(n)\over n2^n}\right),\frac1{\frac1{n+\log_2(n)}+\frac1{n2^n}},k+1\right)\end{align}$$

Q.E.D.