The product s(2,1)s(1) by Pieri

Given: AC, an integer $r\ge 2$ and the rank-$r$ Schur polynomials.

Horizontal Pieri: for $\mathrm{\ell }(\lambda )\le r$ and $d\ge 0$, the nonzero Schur summands in $S_{\lambda }(V)\otimes {\mathrm{Sym}}^d(V)$ are exactly those indexed by horizontal strips $\nu \mathrm{/}\lambda$ with $\mathrm{\ell }(\nu )\le r$, each with multiplicity one; the character identity is $s_{\lambda }{h}_d={\sum }_{\nu \mathrm{/}\lambda \text{ horizontal}}{s}_{\nu }$, where terms with $\mathrm{\ell }(\nu )>r$ are zero. Here $S_{(1)}(V)=V$ and $h_1=s_{(1)}=x_1+\cdots +x_r$ (The horizontal Pieri rule, Schur modules and their characters).

A skew diagram $\nu \mathrm{/}(2,1)$ for a partition $\nu$ of $4$ is a horizontal strip of size one exactly when $\nu$ is obtained by adding one box to $[(2,1)]$, and additions are legal exactly at the ends of rows, giving the three partitions $(3,1)$, $(2,2)$, $(2,1,1)$ (Skew diagrams and semistandard skew tableaux, Partitions, English diagrams, and conjugation).

For a three-row partition $(a,b,c)$, padded by zeros, deleting all entries $3$ from a semistandard tableau leaves a two-row shape $(p,q)$ with $a\ge p\ge b\ge q\ge c$: equal entries $3$ cannot share a column. Conversely these inequalities make the removed boxes a horizontal strip, so any tableau of shape $(p,q)$ on $\{1,2\}$ extends uniquely by filling the removed boxes with $3$. Its $q$ columns of height two are forced to be $1$ above $2$, and the remaining $p-q$ first-row boxes contain a weakly increasing string of $1$'s followed by $2$'s, with $p-q+1$ choices. Thus $s_{(a,b,c)}(1,1,1)={\sum }_{p=b}^a{\sum }_{q=c}^b(p-q+1)=(a-b+1)(b-c+1)(a-c+2)\mathrm{/}2$. At rank two the same column argument gives $s_{(a,b)}(1,1)=a-b+1$. Hence the rank-two values for $(3,1),(2,2),(2,1)$ are $3,1,2$, and the rank-three values for $(3,1),(2,2),(2,1,1),(2,1),(1)$ are $15,6,3,8,3$. The shape $(2,1,1)$ vanishes at rank two (Semistandard tableaux expand Schur characters, Stable Schur functions from bialternants).