Character of $n^{th}$ symmetric square $V_2$ of $SU(2)$ If $V_n$ is $n+1$-dimensional repn of $SU(2)$ how to compute decomposition into irreducibles of $S^nV_2$ efficiently?
Doing some rather tedious computation by picking a basis in $V_2$ and considering action of diagonal representatives of classes of $SU(2)$ then computing coefficients in the character, I arrived at the result:
$$ S^nV_2 = V_{2n} \oplus V_{2n-4} \oplus V_{2n-6} \oplus \dots \oplus V_0$$
Can this be obtained using properties of the symmetric product and some clever observation? The final formula looks rather nice and it'd be nice if it had some intuitive sense...
 A: Unfortunately, there is no nice property of the symmetric product to be used here. Your computation is the $m=2$ case of the plethysm of symmetric powers
$$
S^n(S^m(V_1))\ .
$$
Understanding plethysm in general is a very difficult open problem. That's for the bad news. The good news is for $SL_2$ the decomposition is known since about 1856 and it is called the Cayley-Sylvester formula. It gives the multiplicity of the irreducible module $V_k$ as
$$
\left\{m,n,\frac{mn-k}{2}\right\}-\left\{m,n,\frac{mn-k}{2}-1\right\}
$$
where $\{a,b,w\}$ denotes the number of integer partitions of $w$ which fit inside an $a$ by $b$ rectangle. The easiest proof is what you did, namely a character computation which essentially produces the $q$-binomials or Gaussian polynomials. 
Note that in your particular case, this classically corresponds to counting covariants of degree $n$ of the generic quadratic binary form $Q$. These are linear combinations of $Q^i \Delta^j$ where $\Delta=b^2-4ac$ is the discriminant of the quadratic.
