$\mathbf{SETUP}$
In this previous question, I show how the sum of all cycles of type defined by non-unity partitions of $n$ relates to the derangement numbers / subfactorial $!n$ (number of permutations with no fixed points), to ask what explains the remarkable arithmetic form of the equality.
Famously, $!n$ tends to $e^{-1}$ as a fraction of the factorial $n!$, meaning we can write:
\begin{align} \lim_{n\rightarrow\infty} \left( \sum_{k=0}^{n} \frac{(-1)^k}{k!} = \sum_{\lambda \in P_{n,1}} \frac{1} {2^{\alpha_2} ... n^{\alpha_n} \times \alpha_2! ... \alpha_n!} \equiv \sum_{\lambda \in P_{n,1}} \frac{1}{\Lambda_\lambda} \right) = e^{-1} \end{align}
In that question I define $P_{n,1} \equiv P_{n, \lambda \nsupseteq \{1\}}$ as the set of cycle types formed by all the non-unity partitions of $n$ (all partitions excluding the use of 1). This excludes all permutations that have any fixed points (1-cycles) in its cyclic structure.
$\mathbf{RESULT}$
If one then excludes transpositions also, one defines the group of cycle types that excludes 1-cycles and 2-cycles $P_{n, \lambda \nsupseteq \{1,2\}} \equiv P_{n,2}$, and performs the same sum, taking the limit $n \rightarrow \infty$:
\begin{align} \lim_{n \rightarrow \infty} \sum_{\lambda \in P_{n,2}} \frac{1} {3^{\alpha_3} ... n^{\alpha_n} \times \alpha_3! ... \alpha_n!} = e^{-\frac{3}{2}} = e^{-(1 + \frac{1}{2})} = e^{-H_2} \end{align}
Doing the same for $P_{n,3}$ gives:
\begin{align} \lim_{n \rightarrow \infty} \sum_{\lambda \in P_{n,3}} \frac{1} {4^{\alpha_4} ... n^{\alpha_n} \times \alpha_4! ... \alpha_n!} = e^{-\frac{11}{6}} = e^{-(1 + \frac{1}{2} + \frac{1}{3})} = e^{-H_3} \end{align}
Again for $P_{n,4}$ gives:
\begin{align} \lim_{n \rightarrow \infty} \sum_{\lambda \in P_{n,4}} \frac{1} {5^{\alpha_5} ... n^{\alpha_n} \times \alpha_5! ... \alpha_n!} = e^{-\frac{25}{12}} = e^{-(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4})} = e^{-H_4} \end{align} and so on, where $H_k$ is the $k$th harmonic number!
I do not have a proof of why this seems to be the case, but computer code I wrote confirms this up to the highest $n$ I tried to do ($n \sim 50$ or so).
$\mathbf{QUESTION}$
What explains this remarkable pattern?
I think I've only found something like this result in this paper by Arratia and Tavaré:
on the cyclic nature of random permutations... but I am not versed enough in its notation to understand it very well, other than simply recognising it must be describing the same underlying pattern. (The 'result' appears as comment at the end of the proof at the top of page 1581.)
$\mathbf{UPDATE / ANSWER?}$
This answer to the "Name Drawing Problem" describes a recurrence relation that gives intuition to why the harmonic numbers appear in this kind of question... I think.