# Can $10\uparrow^n m<2\uparrow^n (m+2)$ be formally proven?

See here :

http://googology.wikia.com/wiki/Arrow_notation

for the definition of the up-arrow function.

Can $10\uparrow^n m<2\uparrow^n (m+2)$ be formally proven for all $m\ge 1$ and $n\ge 3$ ?

With Saibians theorem we get $$10\uparrow^n m<(2\uparrow^n 3)\uparrow ^n m<2\uparrow^n (m+3)$$ Also the claim is trivially true for $m=1$, but I did not manage to complete the induction step.

• Not to sound rude or anything, but does my answer suffice? – Simply Beautiful Art Aug 4 '18 at 21:44
• Ah, sorry. I've edited to include that now :-) – Simply Beautiful Art Aug 6 '18 at 14:59
• I will have to go through the details, but apparently, you have proven the claim :) – Peter Aug 6 '18 at 15:02

For $n=3$, we have:

For $m=1$, we have $10\uparrow^31=10$ and $2\uparrow^33=65536$.

Assume $10\uparrow^3m<2\uparrow^3(m+2)-3$ holds for some $m\ge1$. Then we have

\begin{align}10\uparrow^3(m+1)&=10\uparrow^210\uparrow^3m\\&<10\uparrow^2(2\uparrow^3(m+2)-3)\\&<(2\uparrow^23)\uparrow^2(2\uparrow^3(m+2)-3)-3\tag0\\&<2\uparrow^22\uparrow^3(m+2)-3\\&=2\uparrow^3(m+3)-3\end{align}

Hence $10\uparrow^3m<2\uparrow^3(m+2)-3<2\uparrow^3(m+2)$.

$(0)$ is due to the fact that up-arrows are strictly increasing in all arguments.

Assume it holds for some $n\ge3$ and all $m\ge1$. Then we get

\begin{align}10\uparrow^{n+1}m&=(10\uparrow^n)^{m-1}10\\&<(2\uparrow^n[2+)^{m-1}10]\tag1\\&<(2\uparrow^n)^{m-1}13\\&<(2\uparrow^n)^{m-1}2\uparrow^33\\&<(2\uparrow^n)^{m-1}2\uparrow^n2\uparrow^n2\\&=(2\uparrow^n)^{m+1}2\\&=2\uparrow^{n+1}(m+2)\end{align}

$(1)$ is the result of

\begin{align}2+2\uparrow^nm&<3+2\uparrow^nm\\&<2\uparrow^{n-1}2\uparrow^nm\\&=2\uparrow^n(m+1)\end{align}

Can $10\uparrow^n m<2\uparrow^n (m+2)$ be formally proven for all $m\ge 1$ and $n\ge 2$ ?

No, because $10\uparrow^2 2 > 2\uparrow^2 (2+2)$.

• Sorry, I will edit my post. – Peter Jan 15 '18 at 18:17