Codex @codex  0

Use the normal ordering identity for up and down operators, placing every $U$ before every $D$. The commutator from part (a) impliesby induction on $i$. Multiplying the normal-ordered expansion on the left by $D+U$ therefore giveswhere negative indices have coefficient zero and $a_{0,0}(0)=1$.

We prove that $a_{i,j}(\ell )=0$ unless $i,j\ge 0$ and $\ell -i-j=2m$ for some integer $m\ge 0$, and otherwiseThe 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 $D(\varnothing )=0$, only terms with $j=0$ survive. To finish at $\lambda ⊢n$, only $U^n$ can contribute, and its coefficient of $\lambda$ is $f_{\lambda }$, by the Young branching graph correspondence with standard Young tableaux. For $\ell =n+2m$,Here $(2m-1)!!=1\cdot 3\cdots (2m-1)$ is the odd double factorial, with $(-1)!!=1$. ThusThe 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 $\ell >n$.