For a gaussian vector variable $w\sim N(0,I\_{n\times n})$, the moments of square norm are $\mathbb{E} \|w\|^{2 r} = \prod\_{t=0}^{r-1} (n + 2 t)$.


Based on [Isserlis' theorem](https://www.wikiwand.com/en/Isserlis%27_theorem), $\mathbb{E} \|w\|^{2 r}$ can also be evaluated as
$$\sum\_{\pi\in \mathcal{P}([r]), |\pi|\leq n}\frac{n!}{(n-|\pi|)!}\prod\_{p\in\pi}(2 |p|-1)!!$$
where $\mathcal{P}([r])$ means all partitions on set $[r]=\{1,\dots,r\}$, $\pi$ is a partition, $p$ is one block in a partition, $|\pi|$ and $|p|$ are number of blocks and number of elements in a block.


Now consider a variant of the above problem.
$$\sum\_{\pi\in \mathcal{P}([r]), |\pi|\leq n}\frac{n!}{(n-|\pi|)!}\prod\_{p\in\pi}\frac{1}{2}~(2 |p|-1)!!$$
The above formula only differs from moments of square norm of gaussian vector variable with a factor $\frac{1}{2}$.
Is there a similar finite product solution and probability interpretation for the above formula?



A combinatorics problem and the probability interpretation
Fix $n$. Let
$$ G(x) = \sum\_{i=0}^n \frac{n!}{(n-i)!}\frac{x^i}{i!} = (1+x)^n. $$
Let
$$ F(x) = \sum\_{j\geq 1}\frac 12 (2j-1)!!\frac{x^j}{j!} =
 \frac{1}{2\sqrt{1-2x}}-\frac 12. $$
By the Compositional Formula (Theorem 5.1.4 of *Enumerative
Combinatorics*, vol. 2), the number you want is $r!$ times the
coefficient of $x^r$ in
$$ G(F(x)) = \frac{1}{2^n}\left(
 1+\frac{1}{\sqrt{1-2x}}\right)^n. $$
You can expand this by the binomial theorem and then expand each term
into a power series to get a formula for your number as a sum with $n$
terms.


