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

Question

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

I have done some research and several google searches but I have not found the answer to this question:

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!

Thank you in advance!

Edit

Gerry Myerson has (quiet repeatedly) asked me to include the following links for reference: This link and this link.

Answer 1

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

Answer 2

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.

Answer 3

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

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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.

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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.