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The content vector of a standard Young tableau $T$ is $(c_T(1),\dots ,c_T(n))$, where $c_T(r)$ is the Content of a Young-diagram cell containing $r$. Such vectors are exactly the integer vectors satisfying: the first coordinate is zero; each subsequent coordinate has an earlier neighbor differing by one; between two occurrences of $a$ both $a-1$ and $a+1$ occur. To reconstruct the tableau, insert each entry in the next available cell on its prescribed diagonal. The neighbor conditions supply its required predecessors, so each insertion is an addable node of a Young diagram.

Necessity is a property of standard Young tableaux alone. The first entry occupies $(1,1)$, with Content of a Young-diagram cell zero. Every subsequent cell has a cell immediately above or to its left, of content one larger or one smaller, already present. If two entries occupy the same diagonal, the later cell lies strictly southeast of the earlier. The cells immediately right of and immediately below the earlier one exist and have contents $a+1,a-1$; their entries lie strictly between the two given entries. This proves all three conditions for content vectors of standard Young tableaux.

For sufficiency, build the Young diagram one cell at a time. All coordinates are integers, because each new coordinate differs by one from an earlier coordinate, starting at zero. Cells of any fixed content $a$ are ordered northwest to southeast. For $a\ge 0$ their positions are $(r,r+a)$, $r=1,2,\dots$; for $a<0$ they are $(r-a,r)$.

At the first occurrence of a positive $a$, the candidate is $(1,a+1)$. The occurrence of $a+1$ earlier would already force a cell of content $a$ in its row, so the neighbor condition must instead supply the content $a-1$. This supplies the left predecessor. The negative case is symmetric, supplying the upper predecessor, and the first zero gives $(1,1)$.

At any later occurrence of $a$, take the next position on its diagonal. The preceding occurrence already supplied all but the next necessary predecessor on each neighboring diagonal. The repeated-entry condition supplies both $a-1$ and $a+1$ after that preceding occurrence, so those next predecessors are present. More explicitly, for $a>0$ and the $r$th occurrence, the left predecessor is the $r$th cell of content $a-1$, and the upper predecessor is the $(r-1)$st cell of content $a+1$. For $a=0$ both are the $(r-1)$st cells on their respective diagonals; for $a<0$ the roles are reversed. These are exactly the predecessors supplied by the two new neighboring occurrences.

The new cell is therefore an addable node of a Young diagram. Induction produces a partition of an integer at every stage, and filling the added cell by its insertion time gives a standard Young tableau. No choice was possible because addable nodes of a Young diagram have distinct contents. Thusand the construction also proves uniqueness of the tableau with a specified vector.

Fix $\lambda ⊢n$ and $\rho ⊢n-1$. Write $\rho \nearrow \mu$ when a single addable node of a Young diagram turns $\rho$ into $\mu$. For each positive part ${\lambda }_i$, let ${\lambda }^{\downarrow i}$ be the partition of an integer obtained by decreasing that part by one, sorting, and omitting zeros. The Vershik linear relations areEqual row lengths must be counted separately on the right. Equivalently, if $c(\lambda ,\gamma )$ is the number of rows whose decrement produces $\gamma$, the right side is ${\sum }_{\gamma \nearrow \lambda }c(\lambda ,\gamma )M(\rho ,\gamma )$. There is no multiplicity coefficient on the left because irreducible restriction branching rule for a symmetric group is simple.

To prove the relations, restrict the Young permutation module $M^{\lambda }$ to the subgroup $S_{n-1}$ fixing $n$. The tabloids split into orbits of a group action according to which labeled row contains $n$. Removing that entry gives, for row $i$, the Young permutation module with composition $\lambda -e_i$, which is isomorphic to the one for its sorted partition. HenceIts $V^{\rho }$ multiplicity is the right side. On the other hand decompose $M^{\lambda }$ into irreducible representations first and apply the restriction branching rule for a symmetric group to each summand. Its $V^{\rho }$ multiplicity is then the left side. Equality proves the relation.

For example $\lambda =(2,2)$ restricts to $2M^{(2,1)}$, so the coefficient is two rather than one. For $\lambda =(1^n)$, the regular representation, the relation reduces to ${\sum }_{\mu :\rho \nearrow \mu }\dim V^{\mu }=n\dim V^{\rho }$. These checks emphasize why counting distinct resulting partitions without their row multiplicities would give a false recurrence.

A standard Young tableau fills a Young diagram bijectively with $1,\dots ,n$, increasing along rows and down columns. The cells occupied by the first $r$ entries form a diagram, so a tableau is equivalently a path formed by adjoining one addable node of a Young diagram at each step. The number of standard tableaux of shape $\lambda$ is the dimension of the complex Specht module $S^{\lambda }$.