Is there any mathematical formula or method to compute the values of Landau function (by hand)? I know that the value of Landau function $g(n)$ for any natural number $n$ gives us the largest order of an element of the symmetric group $S_{n}$ Namely, $g(n)=\max(\text{ord}(k) \mid \forall k \in S_{n})$. But, this cannot help us compute $g(n)$ for large $n\in \mathbb{N}$. Is there any possible mathematical method to compute $g(n), \forall n\in \mathbb{N}$ ?
Thank you for your time in advance!
 A: This paper describes a method for computing it that the authors claim works up to $n=10^{15}$.
Edit: I see now that you want to do this by hand, and of course we are talking here about a method intended for a computer. I don't know whether it can easily be done by hand for smaller $n$. I think it differs from earlier methods in that it does not require prior computation of $g(m)$ for all $m<n$.
A: $g(n)$ is the maximum value of $\text{lcm}(n_1, \dots n_k)$ where $\sum n_i = n$. Suppose some $n_i$ can be written as $n_i = ab$ where $a, b \ge 2$ and $\gcd(a, b) = 1$ (this is possible iff $n_i$ is not a prime power). Then we can contemplate the effect of removing $n_i$ and replacing it by $a$ and $b$. This doesn't affect the lcm but $n_i > a + b$ since
$$(a - 1)(b - 1) \ge 1 \Leftrightarrow ab \ge a + b + 1$$
so the sum has gotten smaller, which means we can now increase the lcm by adding more terms or making other terms larger. If $\{ n_i \}$ is a maximal partition then either this move is impossible or making it does not increase the lcm; either way we can make it WLOG. In other words:

Lemma 1: The maximum value of $\text{lcm}(n_1, \dots n_k)$ is achieved by a partition in which the $n_i$ are all prime powers.

Furthermore suppose $n_i$ and $n_j$ are powers of the same prime, say $n_i = p^a$ and $n_j = p^b$ where $a \ge b$ WLOG. Then we can remove $n_j$ without affecting the lcm, so we can again now increase the lcm by either adding more terms or making other terms larger. So:

Lemma 2: The maximum value of $\text{lcm}(n_1, \dots n_k)$ is achieved by a partition in which the $n_i$ are powers of distinct primes.

This gives us the following "formula" for $g(n)$, which you can find on the OEIS entry for Landau's function:
$$\boxed{ g(n) = \text{max} \left\{ \prod p_i^{e_i} : \sum p_i^{e_i} \le n \right\} }$$
(the $\le$ is because we can always pad with $1$s). This is still an optimization problem but over a smaller set than initially (over what we might call the prime-power-partitions of $n$), and involving a product instead of an lcm which is easier to understand the behavior of.
As an application I believe that $g(52) = 4 \cdot 9 \cdot 5 \cdot 7 \cdot 11 \cdot 13 = 180180$ but the strategy I had in mind for writing down a proof by hand got very complicated and involved bounding and computing several smaller values of $g$.
