If $V_n$ is the $n+1$ dimensional irreducible complex rep of $SU(2)$ I'd like to find closed form formula for:
$$ V_1^{\otimes n}$$
Using tensor distributivity over direct sum and Clebsh formula I've computed first couple powers, but I'm clueless how to go from there.
$$ V_1^{\otimes 2} = V_1 \otimes V_1 = V_2 \oplus V_0$$ $$ V_1^{\otimes 3} = V_1 \otimes (V_2 \oplus V_0) = V_3 \oplus V_1 \oplus V_1$$ $$ V_1^{\otimes 4} = V_1 \otimes (V_3 \oplus V_1 \oplus V_1) = V_4 \oplus V_2 \oplus V_2 \oplus V_2 \oplus V_0 \oplus V_0$$
If one presents number of copies of $n^{th}$ irrep in a table this can be summarized in following table:
\begin{matrix} V_0 & V_1 & V_2 & V_3 & V_4 & V_5\\ & 1\\ 1&&1\\ &2&&1\\ 2&&3&&1\\ &5&&4&&1 \end{matrix}
Is there some simple pattern here I fail to see? Is this a well known sequence of integers?
The multiplicity of $V_k$ in $V_1^{\otimes n}$ can be given a pretty nice combinatorial description; I wrote a blog post about this a long time ago. It's called Catalan's triangle (the table here is oriented differently from yours, its diagonals going down and to the left are your rows) and has a fairly nice closed form. It is A053121 on OEIS.