Maximum amount of the Letter-Number-game

Question

There is a number-game which you play like this:

1. You think of a number eg. $8$.
2. Then think: How many letters does this number have? In our example: $5$
3. And now do step one with $5$

This goes on recursively till you end up with a number which has the same amount of letters as it's value.
And that number is $4$.

But let us focus on how many times (We will take it as $x$) we had to do step 1 and 2.
When our number is $8$, then: $$ 1. \quad8 \to 5\\ 2. \quad5 \to 4 $$ So, in the case $8$, $x$ is 2.

One more Case with number $2$ $$ 1. \quad2 \to 3\\ 2. \quad3 \to 5\\ 3. \quad5 \to 4 $$

In this case $x$ is $3$

Here comes the question: What is the maximum number $x$ can be, and which case is that?

Answer 1

An easier proof, starting from jafe's idea:

Let $d(n)$ denote the number of letters used in writing down a number $n$. Let $S_0=\{4\}$, and recursively let $S_{n+1}$ be the set of natural numbers $k$ for which $d(k)\in S_n$. (Equivalently, $S_n$ is the set of numbers such that repeating the step $n$ times reaches $4$.)

Claim: $S_n$ is finite for all $n\geq 0$.

Proof (by induction over $n$): Obviously, $S_0$ is finite. Now, let $n>0$. By assumption, $S_{n-1}$ is finite, and thus has a maximal element $M$. Note that there are at most $26^k$ numbers with exactly $k$ letters, so there are at most $1+26+26^2+\cdots+26^k$ numbers using at most $k$ letters. For each $k\in S_n$, we have $d(k)\in S_{n-1}$ and thus $d(k)\leq M$, so in particular $S_n$ can't have more than $1+26+26^2+\cdots+26^M$ elements, and thus is finite.

Now assume there was an $x$ so that the series always terminates after at most $x$ steps: Then clearly $S_x=\mathbb{N}$, a contradiction.

In particular, this generalizes to every language where $d$ has only finitely many fixed points and loops.

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If it can't be assumed from the question, a quick proof that $4$ is always reached*:

Let $n$ be a number with $k$ digits (so $k\leq \log_{10}(n)+1$). Split the string representation of $n$ at each point where a new digit (or "eleven"/"twelve"/"...teen") is "mentioned" (so "one thousand twelve-hundred and thirty-eight" becomes ["one thousand ","twelve-hundred and ","thirty-","eight"]). If $n$ is less than one centillion, then each element of the list will contain at most 32 letters ("three hundred quattuordecillion and"). Otherwise, the upper bound for each component will be $32+\log_{10^{303}}(n)=32+\log_{10}(n)/303$ (ten letters for "centillion" are needed for every factor $10^{303}$). There will at most be $k$ elements in the list.

So in total, $d(n)\leq (\log_{10}(n)+1)(32+\log_{10}(n)/303)$, which is less than $n$ for all $n$ larger than some constant $N$ (it can easily be seen that $N\leq 100$). In particular, repeatedly applying $d$ will always reach a number below $100$. The longest number below $100$ is seventy-seven with $12$ letters. It remains to check that $1,2,\ldots,12$ all reach $4$.

*The above proof doesn't require that, but if it isn't, the result is pretty meaningless - of course, if there was a number that never reaches 4, then there is no $x$ such that applying $d$ for $x$ times always reaches $4$

Answer 2

I claim that $x$ is unbounded.

Let $L$ be the function that counts the number of letters in our number. Consider centillion $=10^{303}$, just as million-million is $10^{6+6}$, centillion-centillion is $10^{303+303}$. For brevity we notate the n-fold concatenation of centillion by (centillion)$^n$. Note that (centillion)$^n = 10^{n\cdot 303}$. More importantly, since each centillion has ten letters we see that $L($(centillion)$^n)=n\cdot 10$.

Note also for small numbers $r$ we have $L($(centillion)$^n+r)=10\cdot n +3+L(r)$. This is because as long as $r$ is less than a hundred we say (centillion)$^n+r$ as centillion-centillion-...- and "r". The centillion-centillion-... part has $10\cdot n$ letters as before, the "and" gives us an extra three letters, and the "r" gives us $L(r)$ letters.

Now say we want to construct a chain of length $l$, then we simply need to "stack" centillions $l$ times.

For brevity I will denote (centillion)$^n$ by $(c)^n$. Then we see that $L(L((c)^{((c)^n)}))=L((c)^n\cdot 10)=n\cdot 10 + 3 + L(10)= n\cdot 10 +6$. We can stack the centillions high enough so that this continues. However, we may have a problem with the 6 that appeared. Luckily this is not the case. If we continue, we see that the part of our spoken number that is not a string of centillions ends up repeating as apply $L$.

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Proof of the repeating cycle:

Start with $(c)^{n_0}$, where $n_0$ is some stack of centillions, i.e. $n_0=(c)^{n_1}$ and so on.

First iteration: this has $n_0\cdot 10$ letters. It is of the form $(c)^{n_1}+10$

Second iteration: when we say $(c)^{n_1}+10$, we say centillion-centillion-.. and ten. This has $n_1\cdot 10 +3+3$ letters. It is of the form $(c)^{n_2}+6$.

Third iteration: when we say $(c)^{n_2}+6$, we say centillion-centillion-.. and sixteen. This has $n_2\cdot 10 +3 +7$ letters. It is of the form $(c)^{n_3}+20$.

Fourth iteration: when we say $(c)^{n_3}+20$, we say centillion-centillion-.. and twenty. This has $n_3\cdot 10+3+6$ letters. It is of the form $(c)^{n_4}+19$.

Fifth iteration: when we say $(c)^{n_3}+19$, we say centillion-centillion-... and nineteen. This has $n_4\cdot 10 + 3 + 7$ letters. It is of the $(c)^{n_5}+20$.

And now we see our extra letters will repeat switching from 19 to 20. So long as we have enough centillions this process will send centillion-centillion-... and nineteen to centillion-... and twenty.

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Now that we have done the working out we could present a slicker answer. Let $c$ be a centillion. Let $c\uparrow l$ denote an $l$ stack of powers of $c$. From our above working out we see that $L(19+c\uparrow l)=20+c\uparrow (l-1)$, and that $L(20+c\uparrow l)=19+c\uparrow(l-1)$. Therefore the x number of $19+c\uparrow l$ is at least $l$.

Answer 3

There are numbers with arbitrarily large lengths when written out (after all, n letters can only uniquely represent a maximum of 26^n different numbers), so I think we can just start backwards from 4, pick the smallest number with that many letters (or the next smallest etc., if we hit a dead end), and continue indefinitely. Not sure how to prove that we always have valid options, though...

4 -> zero's a dead end, so five
5 -> three
3 -> one is a dead end, so six
6 -> eleven
11 -> twenty-three
23 -> one hundred and twenty-four
124 -> This is already in the millions... Maybe I could find this one, but I'm definitely not looking for the next one by hand!

Answer 4

Here is an easy solution if we assume that every number goes to $4$.

Notation: Let $L(n)$ be the number of letters in our number $n$. Let $S(n)$ be the number of steps it takes to get to $4$, (so $S(n)$ gives us the $x$ number in the question).

Claim: There is no finite upperbound on $S(n)$.

Proof: Suppose for sake of contradiction that there are no chains of length greater than $M$. Let $n$ be a number with $S(n)=M$, i.e. it takes $M$ steps for $n$ to get to 4.

We will now construct a number which we call $N$ that has $n$ digits. This will make a chain of length $M+1$. This contradiction completes the proof.

Let $k$ be the last digit in $n$ and let $p$ the rest of the digits, i.e. if $n=14375$ then $k=5$ and $p=1437$.

If $k$ is one of $\{3,4,5,6,7,8\}$ let $k'$ be $\{$"two","four","three","eleven","fifteen","thirteen"$\}$ respectively. Then $N=$"$k' ($centillion$)^p$" has $n$ digits.

If $k$ is $\{1,2,9\}$ then let $k'$ respectively be $\{$"three","eleven","two",$\}$. Then $N=$ "two-$($centillion$)^{p-1}$ and $k'$" has $n$ digits.

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The assumption that every number goes to $4$ is not valid Consider "four multiplied by five" this is equal to $20$ and has $20$ characters. Therefore this will never go to $4$.