According to Wikipedia, the $n$th cumulant $\kappa_n$ is related to the central-moments $\theta_n$ by the following recurrence:
$$\kappa_n = \theta_n - \sum_{m=1}^{n-1} \binom{n-1}{m-1} \kappa_m \theta_{n-m}$$
which allows one to compute the cumulants sequentially, given the central-moments. The Wikipedia page offers no reference or proof of this formula. What's the proof of this formula?
The equation given by Wikipedia connects cumulants to moments (generally).
A proof of a formula connecting cumulants to central moments is found in A Recursive Formulation of the Old Problem of Obtaining Moments from Cumulants and Vice Versa
Letting $K(t)$ be the cumulant-generating function, and $M(t)$ the moment-generating function. The relationship between the two is \begin{equation} M(t)=\exp{\left[K(t)\right]} \end{equation} The proof follows by differentiation of this expression and noting that the $n$th derivative can be written as \begin{equation} D^n[M(t)]=\sum_{i=0}^{n-1}\binom{n-1}{i}D^{n-i}[K(t)]D^i[M(t)] \end{equation} Where $D^k$ denotes the $k$th derivative. Now setting $t=0$: \begin{equation} \theta_n=\sum_{i=0}^{n-1}\binom{n-1}{i}\kappa_{n-i}\theta_i\\ \theta_n=\kappa_n+\sum_{i=1}^{n-1}\binom{n-1}{i}\kappa_{n-i}\theta_i\\ \end{equation} Rewriting yields: \begin{equation} \kappa_n = \theta_n-\sum_{i=1}^{n-1}\binom{n-1}{i}\kappa_{n-i}\theta_i \end{equation} In terms of the central moments and cumulants.
The paper mentioned and the formula cited by Itronneberg still refer to "raw" (non-central, "at the origin") moments. To verify this, take n=3: you get $\kappa_3 = \theta_3 - \kappa_1\theta_2$. (EDIT: fixed according to comment by Gâteau-Gallois) Hence, $\theta_n$ clearly denotes raw moments: the cumulant and the central moment should coincide for $n=3$.
Indeed, the paper quoted doesn't have any reference to central moments.
Another paper, however (which also mentions the former paper), does: Relationships Between Central Moments and Cumulants, with Formulae for the Central Moments of Gamma Distributions. And it provides the following formula (eq. (2.2)):
$$ \kappa_r = \mu_r - \sum_{j=1}^{r-2} {r-1 \choose j} \mu_j \kappa_{r-j} \qquad r \geq 2. $$
Indeed, if you set $r=2$, you get that the summation is empty, and $\kappa_2 = \mu_2$.
