The goal of the four fours puzzle is to represent each natural number using four copies of the digit $4$ and common mathematical symbols.

For example, $165=\left(\sqrt{4} + \sqrt{\sqrt{{\sqrt{4^{4!}}}}}\right) \div .4$.

If we remove the restriction on the number of fours, let $f(N)$ be the number of fours required to be able to represent all positive integers no greater than $N$. What is the asymptotic behaviour of $f(N)$? Can it be shown that $f(N) \sim r \log N$ for some $r$?

To be specific, let’s restrict the operations to the following:

- addition: $x+y$
- subtraction: $x-y$
- multiplication: $x\times y$
- division: $x\div y$
- exponentiation: $y^x$
- roots: $\sqrt[x]{y}$
- square root: $\sqrt{x}$
- factorial $n!$
- decimal point: $.4$
- recurring decimal: $. \overline 4$

It is easy to see that $f(N)$ is $O(\log N)$. For example, with four fours, numbers up to $102$ can be represented (see here for a tool for generating solutions), so, since $96 = 4\times4!$, we can use $6k-2$ fours in the form $(\dots((a_1\times 96+a_2)\times 96+a_3)\dots)\times96+a_k$ to represent every number up to $96^k$.

On the other hand, we can try to count the number of distinct expressions that can be made with $k$ fours. For example, if we (arbitrarily) permit factorial only to be applied to the digit $4$, and allow no more than two successive applications of the square root operation, we get $\frac{216^k}{18}C_{k-1}$ distinct expressions where $C_k$ is the $k$th Catalan number. (Of course, many of these expressions won’t represent a positive integer, many different expressions will represent the same number, and the positive integers generated won’t consist of a contiguous range from $1$ to some $N$.)

Using Stirling’s formula, for large $k$, this is approximately $\frac{864^k}{72k\sqrt{\pi k}}$. So for $f(N)$ to grow slower than $r\log N$, we’d need to remove the restrictions on the use of unary operations. (It is well-known that the use of logs enables *any* number to be represented with only *four* fours.)

Can this approach be extended to show that $f(N)$ is $\Omega(\log N)$? Or does unrestricted use of factorial and square roots mean that $f(N)$ is actually $o(\log N)$? Is the answer different if the use of $x\%$ (percentages) is also permitted?