Let $\beta_n$ denote the flag $h$-vector (as defined in EC1, Section
3.13) of the partition lattice $\Pi_n$ (EC1, Example 3.10.4). Then
$$ \mathrm{mag}_{n,{n-1\choose 2}-j} = \sum_S \beta_n(S), $$
where $S$ ranges over all subsets of $\lbrace 1,2,\dots,n-2\rbrace$ whose
elements sum to $j$. An explicit formula for $\beta_n(S)$ is given by
$$ \beta_n(S) = \sum_{T\subseteq S} (-1)^{|S-T|}
\alpha_n(T), $$
where if the elements of $T$ are $t_1<\cdots < t_k$, then
$$ \alpha_n(T) = S(n,n-t_1)S(n-t_1,n-t_2)
S(n-t_2,n-t_3)\cdots S(n-t_{k-1},n-t_k). $$
Here $S(m,j)$ denotes a Stirling number of the second kind.
Addendum. A combinatorial description of the mag numbers is
somewhat complicated. Consider all ways to start with the $n$ sets
$\lbrace 1 \rbrace,\dots, \lbrace n \rbrace$. At each step we take two
of our sets and replace them by their union. After $n-1$ steps we will
have the single set $\lbrace 1,2,\dots,n \rbrace$. An example for
$n=6$ is (writing a set like $\lbrace 2,3,5\rbrace$ as 235)
1-2-3-4-5-6, 1-2-36-4-5, 14-36-2-5, 14-356-2, 14356-2, 123456. At the
$i$th step suppose we take the union of two sets $S$ and $T$. Let
$a_i$ be the least integer $j$ such that $j$ belongs to one of the
sets $S$ or $T$, and some number smaller than $j$ belongs to the other
set. For the example above we get $(a_1,\dots,a_5)=(6,4,5,3,2)$. If
$\nu$ denotes this merging process, then let $f(\nu) = \sum i$,
summed over all $i$ for which $a_i>a_{i+1}$. For the above example,
$f(\nu) = 1+3+4=8$. (The number $f(\nu)$ is called the major index
of the sequence $(a_1,\dots,a_{n-1})$.) Then
$\mathrm{mag}_{n,{n-1\choose 2}-j}$ is the number
of merging processes $\nu$ for which $f(\nu)=j$. This might look
completely contrived to the uninitiated, but it is very natural within
the theory of flag $h$-vectors.
Following up on David's nice answer, there is a different parametrization that makes the pattern much more obvious. Namely, let $s_1,s_2,\dots$ be arbitrary, then you can write
$$A=\Big(h_{i-j}(s_1,s_2,\dots, s_{j+1})\Big)_{i,j=0}^{\infty}$$
$$B=\Big((-1)^{i-j}e_{i-j}(s_1,s_2,\dots, s_{i})\Big)_{i,j=0}^{\infty}$$
And checking that the matrices and the corresponding generating functions are inverses becomes a little more clear. The $h_i, e_i$ denote the complete homogeneous and elementary symmetric functions respectively, with the convention $h_0, e_0 =1$.
Best Answer
The number of pairs $(P,Q)$ of standard Young tableaux of the same shape and with $n$ squares is $n!$.
The number of oscillating tableaux of length $2n$ and empty shape is $1\cdot 3\cdot 5\cdots (2n-1)$.
The number of leaf-labeled complete (unordered) binary trees with $n$ leaves is $1\cdot 3\cdot 5\cdots (2n-3)$ (Schröder's third problem).
The number of compact-rooted directed animals of size $n$ is $3^n$. See MathSciNet MR0956559 (90c:05009).
Let $f(n)$ be the number of $n\times n$ matrices $M=(m_{ij})$ of nonnegative integers with row and column sum vector $(1,3,6,\dots,{n+1\choose 2})$ such that $m_{ij}=0$ if $j>i+1$. Then $f(n)=C_1C_2\cdots C_n$, where $C_i$ is a Catalan number. No combinatorial proof of this result is known. See Exercise 6.C12 on page 38 (solution on page 84) of http://math.mit.edu/~rstan/ec/catadd.pdf