# Which arrangement produces the largest number?

I learnt that the power tower $2\uparrow3\uparrow4\uparrow...\uparrow n$ is larger than any power tower with a different order of the numbers $2,3,4,...,n$.

Is this also true for conway-chains and for bowers array notation ?

Are $$2\rightarrow 3\rightarrow 4\rightarrow...\rightarrow n$$

and {$2,3,4,...,n$} also larger than any other number of this form with different order of the numbers ?

No for both.

For example, for $n = 4$ we have $2 \rightarrow 3 \rightarrow 4 = 2 \rightarrow 4 \rightarrow 3 = 2 \uparrow \uparrow 65536$, whereas $3 \rightarrow 2 \rightarrow 4 = 3 \uparrow \uparrow 3^{27}$.

We can show by induction that $2 \rightarrow 3 \rightarrow n < 3 \rightarrow 2 \rightarrow n$, as

$$2 \rightarrow 3 \rightarrow 1 = 8 < 9 = 3 \rightarrow 2 \rightarrow 1$$

and assuming $2 \rightarrow 3 \rightarrow n = 2 \rightarrow 4 \rightarrow (n-1) < 3 \rightarrow 2 \rightarrow n = 3 \rightarrow 3 \rightarrow (n-1)$, we have

$$2 \rightarrow 3 \rightarrow (n+1) = 2 \rightarrow (2 \rightarrow 2 \rightarrow (n+1)) \rightarrow n = 2 \rightarrow 4 \rightarrow n$$ $$= 2 \rightarrow (2 \rightarrow 4 \rightarrow (n-1)) \rightarrow (n-1) < 2 \rightarrow (3 \rightarrow 3 \rightarrow (n-1)) \rightarrow (n-1)$$ $$< 3 \rightarrow (3 \rightarrow 3 \rightarrow (n-1)) \rightarrow (n-1) = 3 \rightarrow 3 \rightarrow n = 3\rightarrow 2 \rightarrow (n+1).$$

It follows that $2 \rightarrow 3 \rightarrow X < 3 \rightarrow 2 \rightarrow X$ for any chain $X$, since when you evaluate the chains you will get the same expressions, except one will have chains starting with $3 \rightarrow 2$ and one will have chains starting with $2 \rightarrow 3$. Whenever the expressions reduce a chain to a 3-chain, we will have $3 \rightarrow 2 \rightarrow n$ evaluate to a higher value than $2 \rightarrow 3 \rightarrow n$, so in the end $3 \rightarrow 2 \rightarrow X$ will evaluate to a higher value than $2 \rightarrow 3 \rightarrow X$. In particular, $2 \rightarrow 3 \rightarrow \cdots \rightarrow n < 3 \rightarrow 2 \rightarrow \cdots \rightarrow n$.

For Bowers arrays the situation is more extreme. Any expression of the form {2,2,...} will evaluate to the number 4, as applying the evaluation rules will either keep the first two entries the same, or will replace the array with {2, {2,1,...},...} = {2,2,...}. So the array will eventually evaluate to {2,2} = 4.

Further, any array {2,b,c,d,...} with four or more entries (not counting trailing 1's) will evaluate to 4 as well, as {2,b,c,d,...} will evaluate to {2,b',c-1,d,...} and then {2,b'',c-2,d,...} eventually reaching {2,n,1,d,...}. This evaluates to {2,2,{2,n-1,1,d,...},d-1,...}, which we have already determined equals 4.

In particular, {2,3,4,...,n} will evaluate to 4 for n > 4, whereas {3,2,4,...,n} will grow extremely fast.

• Yes, I have forgotten these easy facts. There is one exception : {2,1,...} solves to $2$, but this does not occur in my question. So, can we be sure that $3,2,4,5,...$ is maximal in both cases ? – Peter Jan 21 '15 at 15:01
• For Conway Chained Arrows, I'm 100% sure that $3,2,4,5...$ is maximal. It's a pain to prove formally, but it is easy to see once you play around with some chains of length $4-6$ – Guri Harari Jan 21 '15 at 15:30

For Conway Chained Arrow Notation, the largest possible number is always: $$3→2→4→...→n$$

For $n>4$ this follows directly from the recursion process given by Conway's notation plus the fact that $3^2>2^3$. For $n=4$, one must calculate it directly ($3↑↑↑↑2$ vs $2↑↑↑↑3$) to show that $3↑↑↑↑2$ is indeed larger.

As for Bowers Arrays, I'm not too familiar with them, but since $\lbrace{a,b,c}\rbrace=a→b→c$, then the largest 3-element array made with $2,3,4$ would be $\lbrace{3,2,4}\rbrace$}.

• Now, I remember that the conway chain $3,2,X$ is always greater than $2,3,X$ for any chain $X$. Thanks for correcting this. – Peter Jan 21 '15 at 15:03