# How can I show, that $N\uparrow\uparrow N$ is not “much larger” than $N$ for very large $N\$?

Here :

https://sites.google.com/site/largenumbers/home/3-2/knuth

Saibian demonstrates that for very large numbers $N$, $N\uparrow\uparrow N$ is only "slightly larger" than $N$.

I would like to demonstrate it for the number

$$N=4\uparrow^4 4=4\uparrow^3 4\uparrow^3 4\uparrow^3 4$$

I want to bound $N\uparrow\uparrow N$ from above. I think $4\uparrow^5 4$ would be an upper bound, but even if this is the case, I would like to find a better upper bound.

For which $k$ do we have $4\uparrow\uparrow\uparrow k>N\uparrow\uparrow N\$ ?

The value $k$ should be near $4\uparrow^3 4\uparrow^3 4$. This would show that $N\uparrow\uparrow N$ is "not much larger" than $N$.

• @Deedlit another small exercise with up-arrows :) – Peter Jun 5 '16 at 18:25

## 1 Answer

Letting $M = 4\uparrow^3 4 \uparrow^3 4$, we have

$$4\uparrow^3(M+1) = 4 \uparrow\uparrow (4 \uparrow^3 M) = 4\uparrow\uparrow N < N \uparrow\uparrow N$$

but

$$4 \uparrow^3 (M+2) = 4\uparrow\uparrow (4 \uparrow\uparrow (4 \uparrow^3 M)) = 4\uparrow\uparrow (4 \uparrow\uparrow N) > 4 \uparrow\uparrow (2N) > (4 \uparrow\uparrow N) \uparrow\uparrow N > N \uparrow\uparrow N$$

• Wow, wonderful answer! I did not expect that $(4\uparrow^3 4\uparrow^3 4)+2$ would already be sufficient. – Peter Jun 6 '16 at 8:40
• Your second line can be replaced by an immediate application of Saibian's theorem: $4\uparrow^3(M+2)>(4\uparrow^3M)\uparrow^32=N\uparrow\uparrow N$. – r.e.s. Jun 14 '16 at 2:44
• Also, it might be worth noting that the above proof, mutatis mutandis, shows that if $n=b\uparrow^p m$ with $b\ge 2,\ p\ge 3,\ m\ge 1$, then $b\uparrow^p(m+1)\le n\uparrow^{p-1}n<b\uparrow^p(m+2)$. – r.e.s. Jun 14 '16 at 2:46