A function from partitions to natural numbers - is it familiar? I've come across a function from the set of integer partitions to the natural numbers which I don't recognise but which probably ought to be familiar; it arises in the homogeneous Garnir relations for graded Specht modules (see Kleshchev, Mathas & Ram: "Universal graded Specht modules", Proc. LMS 105). I hope someone will recognise it and point me to a useful reference.
The function (let's call it $f$) is defined recursively; to begin with, $f(\varnothing)=1$.  Now suppose we have a non-empty partition $\lambda$, and take a box $(i,j)$ in the Young diagram for which $i+j$ is maximised.  Let $\mu$ be the partition you get by removing the first $i$ rows from the Young diagram and let $\nu$ be the partition you get by removing the first $j$ columns from the Young diagram.  Now recursively define
$f(\lambda) = \binom{i+j}if(\mu)f(\nu)$.
For example, take $\lambda=(5,3,3,1)$. Let $(i,j)=(3,3)$, so that $\mu=(1)$ and $\nu=(2)$. One can calculate $f(\mu)=2$ and $f(\nu)=3$, so that $f(\lambda)=120$.
It's not hard to show that $f$ is well-defined (i.e. doesn't depend on the choice of $(i,j)$).  I have two other definitions of $f$ (also recursive) which I'll omit for now (it's non-trivial to show that they are equivalent).
Please let me know if you've seen this function (or a similar-looking one) before.  Since it's a function from partitions to positive integers, it ought to count tableaux of some kind.
 A: Okay, I agree that this answer comes actually a little late...
If you type some values of your function into FindStat, you will see David Speyer's answer automatically generated, since it is obtained as a natural statistic obtained after applying a natural combinatorial map.
To have some values, for size up to $4$ it is given by
$$
[1] => 2,
[2] => 3,
[1,1] => 3,
[3] => 4,
[2,1] => 6,
[1,1,1] => 4,
$$
$$
[4] => 5,
[3,1] => 8,
[2,2] => 6,
[2,1,1] => 8,
[1,1,1,1] => 5.
$$
A: Find a large enough $N$ that $i+j \leq N$ for all boxes $(i,j)$ of the partition. So your partition is contained in the staircase partition $(N,N-1,N-2, \ldots, 2,1)$. Turn your partition into a planar rooted forest on $N+1$ vertices using the bijection I'll list below. Your statistic is the number of ways to order the vertices of the tree so that every vertex is labeled before its children.
I'll use your example $(5,3,3,1)$ to illustrate. It embeds in $(5, 4,3,2,1)$. Here I show the original partition with $\ast$'s and the extra elements of the staircase with $\cdot$'s. 
$$\begin{matrix} 
\ast &\ast & \ast & \ast & \ast \\
\ast & \ast & \ast & \cdot & & \\
\ast & \ast & \ast & & \\
\ast & \cdot & & & \\
\cdot & & & & \\
\end{matrix}$$
Read along the southwest edge of the partition to make a matching parentheses string: Up is $($ and right is $)$; pad with an extra $($ at the start and $)$ at the end.
So we get $(()())\ (())\ ()$. The vertices of my forest will be matching pairs, with edges going up for containment. So
$$\begin{matrix}
& & a & b & c \\
& \nearrow &\uparrow & \uparrow & \\
d & & e & f & \\
\end{matrix}$$
We are counting was to order $(a,b,c,d,e,f)$ so that $a$ comes before $d$ and $e$ and $b$ comes before $f$. Sure enough, the count is $6!/(3 \cdot 2) = 120$.
Proof Sketch: If I choose $N$ so large that the Dyck path never touches the boundary of the staircase, then the tree has a single connected component. The root of that component must be ordered first. Deleting that root has the same effect as reducing $N$ by $1$. So we may assume that we choose $N$ so that the Dyck path does touch the boundary, say at $(i,j)$ with $i+j = N$.
Then the forest may be divided into two halves: Components coming from the left of $(i,j)$ and components coming from the right. The number of ways to allocate the labels between those halves is $\binom{i+j}{i}$, and we then must recursively choose how to order each of them. This replicates your recursion.
