# What is the largest finite number you can make using no more than 6 characters?

(I am new to the Math Stack Exchange community so tell me if this question is not allowed, however, I checked on Meta first.)

(Also, I don't really know which tags I should assign this question, so please offer tag suggestions if you don't agree with the tags I tagged this question with)

Sorry for the many edits to this question, but I needed to clarify what I was asking.

What is the largest finite number you can make using no more than 6 characters? Characters include numbers and operations, etc. Rules: Parenthesis do not count as characters, and letters are not allowed. You can use the following operations: Basic Arithmetic, Exponents, Factorials, and Tetration (although you may also use Knuth's up-arrow notation).

Clarification Edit: If you are using exponents - keep in mind this rule: 9^9 (9 to the power of 9) would be 3 characters. However, 9^99 (9 to the power of 99) would be 4 characters.

The following are some guesses that I have come up with, but I'm not confident in any of them

((((9!)!)!)!)!

((((9⁹)!)!)!)

$$9^{9^{9{9{{}}}}}$$

If I am not clear enough with the rules regarding what counts as a character please ask.

Any help is appreciated!

Edit

• Yeah, the thing with these kinds of questions is that you need really precise rules on your characters and what they are allowed to represent, in order for this to become an interesting question. I mean, without any stringent rules, I could simply postulate that the symbol $a$ represents $BB(BB(BB(BB(BB(2))))))$ ($BB$ is the busy beaver function ... if you're interested in big numbers, definitely check out that one ...) Dec 10 '20 at 22:19
• Bram28 What rules are you suggesting I imply in my question? Dec 10 '20 at 22:28
• @Ian54 You really need to specify what operations you allow. Many puzzles of this type allow only the basic arithmetic operations, but you allowed factorial. Do you allow also tetration? Trigonometric functions? Dec 10 '20 at 22:46
• You checked on meta, and waited two whole hours for a response. I conclude you did not ask in good faith. Also, you really ought to have linked to the ten-character question I pointed to on meta. Dec 10 '20 at 22:57
• Does use of Knuth's up-arrow notation permit constructions like $9\uparrow^{999}9$? Does exponentiation count as a character (i.e. is $9^9$ two characters or three)? This problem is not well-defined without a very clear description of what notation is permitted. Dec 10 '20 at 23:25

I submit:$$9\uparrow^{ \left({^\left({{^9}9}\right)}9 \right)} 9$$ Using iterated Knuth's up-arrow notation, where we take $$a \uparrow^n b$$ to be $$a\,\underbrace{\uparrow\cdots\uparrow}_n\,b$$

It depends a lot on what notation we're allowed to use. I have proven all the following are optimal for the given notation sets (in addition to digits of course):

1. $$\{!,+,-,\div,\times, a^b\} \hspace{26pt} \left(\left(\left(\left(9^9\right)!\right)!\right)!\right)!$$
2. $$\{!,+,-,\div,\times, a^b, {^b} a\} \hspace{26pt} {^{^{^{^{^9 9}9}9}9}}9$$
3. $$\{!,+,-,\div,\times, a^b, \uparrow\}\hspace{26pt}9\uparrow\uparrow\uparrow\uparrow 9$$
4. $$\{!,+,-,\div,\times, a^b, \uparrow,\uparrow^{k}\} \hspace{26pt} 9\uparrow^{((9^9)!)}9$$
5. $$\{!,+,-,\div,\times, a^b, {^b} a, \uparrow,\uparrow^{k}\} \hspace{26pt}9\uparrow^{ \left({^\left({{^9}9}\right)}9 \right)} 9$$

Proofs: 1: Let $$s_n$$ be the largest number achievable in $$n$$ symbols. Clearly $$s_1=9$$, and it's not hard to check $$s_2 = 9^9$$ (which is quite a bit larger than $$9!$$). Note that $$s_{n+1}$$ is always in the following set:$$\left\{s_n!, 9^{s_n}, s_n^9\right\} \cup \{s_m + s_{n+1-m}\mid 1 \le m\le n\} \cup \{s_m \times s_{n+1-m}\mid 1 \le m< \le n\}$$ (Note: if $$n$$ becomes too large, this might not work anymore, as it might be possible to get an extremely small number by some complicated process and then divide by it, proving that this doesn't ever happen is probably impossible. $$n=6$$ should be small enough to not worry about that).

Observe that the elements of the two sets on the right side there are smaller than $$2s_n$$ and $$s_n^2$$ respectively. However, if $$s_n\ge 1$$, then $$2s_n$$ and $$s_n^9$$ are smaller than all three of $$\{s_n!, 9^{s_n}, s_n^9\}$$. Hence $$s_{n+1}$$ is among $$\{s_n!, 9^{s_n}, s_n^9\}$$. Obviously $$x! > 9^x > x^9$$ for $$x$$ large enough, and one can check $$x=22$$ is large enough (see Wolfram Alpha https://www.wolframalpha.com/input/?i=9%5Ex+%3C+x%21). Since $$9^9 > 22$$, we have that for $$n>2$$, $$s_{n+1} = s_n!$$

2: Same as above, but note that $$^x 9 > x!$$ for $$x$$ larger than about $$2$$.

3: Same idea, but we would also have the possibility $$s_{n+1} \in \left\{s_i\uparrow^j s_k \mid i+j+k = n+1\right\}$$ I claim that for $$n\ge 5$$, with this symbol set, $$s_n = 9\uparrow^{n-2} 9$$ is optimal. For $$n=6$$ this gives dbx's solution of $$9\uparrow\uparrow\uparrow\uparrow 9$$ We can prove inductively. Observe that for $$n=5$$ $$9 \uparrow\uparrow\uparrow 9 = \underbrace{9\uparrow\uparrow\left(9\uparrow\uparrow\left(9\cdots9\right)\right)}_{9 \text{ times}}$$ however, the biggest number achievable with 5 non-arrow symbols is $$^{^{^{^9 9}9}9}9 = \underbrace{9\uparrow\uparrow\left(9\uparrow\uparrow\left(9\cdots9\right)\right)}_{5 \text{ times}} = 9\uparrow\uparrow\uparrow 5 < 9\uparrow\uparrow\uparrow 9$$ Now, for each $$n$$ we'll have $$s_{n+1}$$ is either $$s_n!$$ or something involving arrows. If $$s_n = 9 \uparrow^{n-2} 9$$, the $$(n+1)$$symbol arrow-forms other than $$9\uparrow^{n-1} 9$$ are all less than $$s_{n-1} \uparrow^{n-2} s_{n-1}$$ Now for some magic:$$s_{n-1} \uparrow^{n-2} s_{n-1} < s_n \uparrow^{n-2} s_n = \left(9 \uparrow^{n-2} 9\right) \uparrow^{n-2} \left(9 \uparrow^{n-2} 9\right) \le 9 \uparrow^{n-2}\left(9 \uparrow^{n-2} \left(9 \uparrow^{n-2} 9\right)\right) = 9 \uparrow^{n-1} 4 < 9\uparrow^{n-1}9$$ Hence if we allow arrows but not powers of arrows, optimal is $$9\uparrow\uparrow\uparrow\uparrow 9$$

4 & 5: We can use similar reasoning if we're allowed to do powers of arrows instead of writing all arrows consecutively: My claim is, with either symbol set (4 or 5), $$s_{n+3} = 9 \uparrow^{s_n} 9$$ for all $$n\ge 1$$. The proof is almost identical to before using induction. It's easy to check that this is true for $$n=1$$; none of the other 4-symbol expressions are even close to $$9\uparrow^9 9$$. For $$n\ge 2$$, if $$s_{n+3}$$ isn't $$9\uparrow^{s_n} 9$$, then it must be less than $$\begin{eqnarray} s_{n+2} \uparrow^{s_{n-1}} s_{n+2} &=& \left(9\uparrow^{s_{n-1}} 9\right) \uparrow^{s_{n-1}} \left(9\uparrow^{s_{n-1}} 9\right)\\ &<&9 \uparrow^{s_{n-1}} \left(9 \uparrow^{s_{n-1}} \left(9 \uparrow^{s_{n-1}} 9\right)\right) = 9\uparrow^{s_{n-1}+1} 4 < 9\uparrow^{s_n}9 \end{eqnarray}$$

This proves that my solutions at the top $$9\uparrow^{ \left({^\left({{^9}9}\right)}9 \right)} 9 \hspace{20pt}\text{and}\hspace{20pt}9\uparrow^{((9^9)!)}9$$ are optimal if we allow arrow-powers and tetration or arrow powers but not tetration, respectively.

I used a couple of times the inequality for $$x,y,z \ge 2$$, $$k\ge 1$$ $$(x \uparrow^k y) \uparrow^k z \le x \uparrow^k \left(y\uparrow^k z\right)$$ which is easy to show.

• Thank you so much! I finally got the answer I was looking for with a good explanation! Dec 15 '20 at 22:42

I submit $$9\uparrow\uparrow\uparrow\uparrow 9$$. (Wikipedia)

But it's possible something like the busy beaver function (wikipedia again) could be even more effective -- and at any rate once one starts talking about numbers this large it's effectively impossible to determine which is the largest anyway.

I need 50 of reputation to write a comment, but sadly enough I got no reputation because I am new.. So I am putting my comment here.

Using standard notations, the biggest number you can possibly write is $$((((9!)!)!)!)!$$

Then if you are allowed to, you can use notations for very large numbers like these: https://en.wikipedia.org/wiki/Knuth%27s_up-arrow_notation https://en.wikipedia.org/wiki/Conway_chained_arrow_notation https://en.wikipedia.org/wiki/Steinhaus-Moser_notation

• $9^{9^{9^{9^{9^{9}}}}}$ is waaaaaay bigger than $9^{99999}$. Dec 10 '20 at 22:55
• Yes you are of course right, I am going to edit it. I guess that the thing that blocked my brain was my inability to properly formatting these powers. But anyway, my mistake is unforgettable, I am not looking for excuses. Dec 10 '20 at 23:21