Given $n$ you get a number of partitions of $n$ and let's denote $\lambda_{k,m}$ to be the $k$th part of the $m$th partition. Now I built the following sum, that stimulated the following question: $$ \lim_{n\to\infty}\sum_{m} \prod_k \dfrac{1}{\lambda_{k,m}!}\overset{?}{=}2.5294773\dots $$
(I calculated till $n=50$, tbc...).
Does this sum converge? How to prove that? If so, does the constant have a name?
I asked ISC and a SearchEngine without success.
For any $n > 0$, let
Recall $p(n)$ is the number of solutions for $(x_1, x_2, x_3 \ldots ) \in \mathbb{N}^{\mathbb{Z}_{+}}$ of the equation:
$$x_1 \cdot 1 + x_2 \cdot 2 + x_3 \cdot 3 + \ldots = n$$
and its generating function is given by
$$\begin{align} P(t)\; &\stackrel{def}{=} \sum_{n=0}^\infty p(n) t^n\\ &= (1 + t + t^2 + \cdots ) (1 + t^2 + t^4 + \cdots )(1 + t^3 + t^6 + \cdots )\cdots\\ &= \prod_{k=1}^\infty \left( 1-t^k \right)^{-1} \end{align} $$ To construct the generating function for $q(n)$, we need to associate a factor $\frac{1}{k!}$ to each part of a partition of $n$ with length $k$. This means
$$ \begin{align} Q(t)\; &\stackrel{def}{=} \sum_{n=0}^\infty q(n) t^n\\ &= \left(1 + \frac{t}{1!} + \left(\frac{t}{1!}\right)^2 + \cdots \right) \left(1 + \frac{t^2}{2!} + \left(\frac{t^2}{2!}\right)^2 + \cdots \right) \cdots\\ &= \prod_{k=1}^\infty \left(1 - \frac{t^k}{k!}\right)^{-1} \end{align}$$
Consider following function: $$g_2(t) \stackrel{def}{=} \prod_{k=2}^\infty \left(1 - \frac{t^k}{k!}\right)$$
Since $\displaystyle\;\sum_{k=2}^\infty \frac{t^k}{k!} = e^t - 1 - t\;$ has infinite radius of convergence, the infinite product on RHS defines an entire function over $\mathbb{C}$. The zeros of $g_2(t)$ are located on the circles $|t| = ( k! )^{1/k}, k = 2, 3, \ldots$. The one nearest to $0$ has radius $\sqrt{2}$. This implies $\displaystyle\;\frac{1}{g_2(t)}$ is analytic for $|t| < \sqrt{2}$.
Notice we can express $Q(t)$ in terms of $g_2(t)$.
$$Q(t) = \frac{1}{(1-t)g_2(t)} = \frac{1}{(1-t)g_2(1)} + \frac{1}{1-t}\left(\frac{1}{g_2(t)} - \frac{1}{g_2(1)}\right)$$
and the last term in RHS is analytic for $|t| < \sqrt{2}$. If we pick a $r_2 \in (1,\sqrt{2})$ and Taylor expand the last term at $t = 0$: $$\frac{1}{1-t}\left(\frac{1}{g_2(t)} - \frac{1}{g_2(1)}\right) = \sum_{n=0}^\infty \alpha_n t^n$$ We will have $\alpha_n = o(r_2^{-n})$. As a result, we find
$$q(n) = \frac{1}{g_2(1)} + o(r_2^{-n})\quad\implies\quad \lim_{n\to\infty} q(n) = \frac{1}{g_2(1)} = \prod_{k=2}^\infty \left(1 - \frac{1}{k!}\right)^{-1}$$
As mentioned in comment, the limit $$\frac{1}{g_2(1)} \approx 2.529477472079152648180116154253954242\ldots$$ and I have no idea what other meanings it carries.
Despite this, we can do a little bit better than just deriving the limit. For any $m \in \mathbb{Z}_{+}$, let $$g_m(t) = \prod_{k=m}^\infty \left(1 - \frac{t^k}{k!}\right)$$ It is entire on $\mathbb{C}$ with zeros lying on the circles $|t| = (k!)^{1/k}$ for integer $k \ge m$. In particular,
$$Q(t) = \frac{1}{g_1(t)} = \frac{1}{(1-t)g_2(t)} = \frac{1}{(1-t)\left(1-\frac{t}{\sqrt{2}}\right)\left(1+\frac{t}{\sqrt{t}}\right)g_3(t)}$$ has three simple poles $1, \pm \sqrt{2}$ for $|t| < \sqrt[3]{6}$. If we separate these poles from $Q(t)$, we find:
$$Q(t) = \frac{1}{g_2(1)}\frac{1}{1-t} - \frac{\sqrt{2}+1}{2g_3(\sqrt{2})}\frac{1}{1-\frac{t}{\sqrt{2}}} + \frac{\sqrt{2}-1}{2g_3(-\sqrt{2})}\frac{1}{1+\frac{t}{\sqrt{2}}} + \Delta_Q(t) $$ where $\Delta_Q(t)$ is something analytic for $|t| < \sqrt[3]{6}$.
Pick any $r_3 \in (\sqrt{2},\sqrt[3]{6})$. Above separation leads to following estimation for $q(n)$: $$\begin{align}q(n) &= \frac{1}{g_2(1)} + \left( - \frac{\sqrt{2}+1}{2g_3(\sqrt{2})} + (-1)^n \frac{\sqrt{2}-1}{2g_3(-\sqrt{2})}\right) 2^{-n/2} + o(r_3^{-n})\\ &\approx 2.5294774720791526481 + \left[\begin{array}{c} -2.9161303884150758204 \\ + (-1)^n 0.1628043195679213429\end{array} \right] 2^{-n/2} \end{align} $$ The table below compares the exact values of $q(n)$ with this approximation (upto 20 decimal places) for various $n$:
$$\begin{array}{r:l} n & q_{exact}(n) & q_{approx}(n)\\ \hline 25 & 2.52894\color{grey}{64176574489192} & 2.52894\color{grey}{59448304110955}\\ 50 & 2.529477390023\color{grey}{7933300} & 2.529477390023\color{grey}{6415059}\\ 75 & 2.529477472063311901\color{grey}{8} & 2.529477472063311901\color{grey}{7}\\ 100 & 2.5294774720791502027 & 2.5294774720791502027 \end{array}$$ As one can see, this is a pretty decent approximation. When $n = 50$, it gives a number accurate to $12$ decimal places.