= Solution
Use the <normal ordering identity for up and down operators>, placing every $U$ before every $D$. The <commutator> from part (a) implies
$$
DU^i=U^iD+iU^{i-1},
$$
by induction on $i$. Multiplying the normal-ordered expansion on the left by $D+U$ therefore gives
$$
a_{i,j}(\ell+1)
=a_{i-1,j}(\ell)+a_{i,j-1}(\ell)+(i+1)a_{i+1,j}(\ell),
$$
where negative indices have coefficient zero and $a_{0,0}(0)=1$.
We prove that $a_{i,j}(\ell)=0$ unless $i,j\geq0$ and $\ell-i-j=2m$ for some integer $m\geq0$, and otherwise
$$
\boxed{a_{i,j}(\ell)=\frac{\ell!}{2^m i!j!m!}}.
$$
The formula holds at $\ell=0$. For the next step with $\ell+1-i-j=2m$, the three contributions in the recurrence, after factoring out $\ell!/(2^m i!j!m!)$, are respectively $i$, $j$, and $2m$. Terms with a negative index or $m-1<0$ are simply absent. Their sum is $i+j+2m=\ell+1$, giving the required $(\ell+1)!$ numerator. Parity and nonnegativity exclude every remaining case. This proves the coefficient formula.
Apply the identity to the empty partition. Since $D(\emptyset)=0$, only terms with $j=0$ survive. To finish at $\lambda\vdash n$, only $U^n$ can contribute, and its coefficient of $\lambda$ is $f_\lambda$, by the <Young branching graph> correspondence with <standard Young tableaux>. For $\ell=n+2m$,
$$
N(\ell,\lambda)
=\frac{\ell!}{2^m n!m!}f_\lambda
=\boxed{\binom{\ell}{n}(2m-1)!!\,f_\lambda}.
$$
Here $(2m-1)!!=1\cdot3\cdots(2m-1)$ is the odd <double factorial>, with $(-1)!!=1$. Thus
$$
\boxed{N(2m,\emptyset)=\frac{(2m)!}{2^m m!}=(2m-1)!!}.
$$
The operator $(D+U)^\ell$ counts <oscillating tableaux>, allowing both upward and downward steps. A strictly upward path to level $n$ would necessarily have length $n$; the printed operator specification is what determines $N$ when $\ell>n$.
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