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An oscillating tableau is a sequence of Young diagrams, each obtained from its predecessor by adding or removing one box. These are paths in the Young branching graph with both directions allowed. A path of length $\ell$ from the empty diagram to $\lambda ⊢n$, with $\ell -n=2m\ge 0$, has count $\ell !f_{\lambda }/(2^{m}n!m!)$. This differs from a saturated upward path, whose length must equal $n$.

The Young lattice, or Young poset, has one vertex for each partition of an integer, including the empty partition, ordered by inclusion of their Young diagrams. A diagram covers another exactly when it adds one box. The Young branching graph is its graded graph: vertices at level $i$ are the partitions of $i$, and edges join diagrams differing by one box.

A descending path from $\lambda ⊢n$ to $(1)$ determines a standard Young tableau: label the successive removed boxes $n,n-1,\dots ,2$, then label the remaining box $1$. Every removed box is a Removable node of a Young diagram, so the resulting labels increase along rows and columns. Conversely, deleting boxes in decreasing label order from a standard Young tableau gives the path. Extending by the unique edge from $(1)$ to the empty diagram yields the usual path-to-tableau correspondence.

For the operator identity, compare the coefficients of each $\mu ⊢i$ in $DU(\lambda )$ and $UD(\lambda )$. If $\mu \ne \lambda$, a common upper cover exists exactly when a common lower cover exists, and each is then unique: the two diagrams differ by exchanging one box, their union is the upper cover and their intersection the lower cover. Since unions and intersections of partition diagrams are again partition diagrams, these off-diagonal coefficients agree.

The coefficient of $\lambda$ in $DU(\lambda )$ is the number of addable nodes of a Young diagram; its coefficient in $UD(\lambda )$ counts the choices of a Removable node of a Young diagram. Along the diagram's boundary the two types of corners alternate, beginning and ending with addable corners. Hence there is exactly one more addable than removable corner. For the empty partition the counts are $1$ and $0$.

Therefore the up and down operators satisfyor $D_{i+1}{U}_i-U_{i-1}{D}_i=I_i$ at level $i$. At level zero, interpret $D_0$ and the lowering contribution as zero. This makes the Young lattice a differential poset.

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 $\ell >n$.