\centerline{\bf Appendix 1}\par
Two formulae of Clebsch-Gordan are:
$$\eqalign{
r_x^ms_y^n&=\sum{{m\choose k}{n\choose k}\over{m+n-k+1\choose k}}
(\overline{rs})^k_{y^{n-k}}(xy)^k\cr
(\overline{rs})^k_{y^{n-k}}&=\sum(-1)^k
{{m\choose k}{n\choose k}\over{m+n\choose k}}
(rs)^\nu r_x^{m-\nu}s_y^{n-\nu}(xy)^\nu\cr}$$

I prefer to formulate it in terms of matrices. Assume $m\leq n$.
Let $A=(a_{ij})$, with $0\leq i,j\leq m$, where
$$a_{ij}=\cases{
0&if $i>j$,\cr
1& if $i=j$,\cr
{{m-i\choose j-i}{n-i\choose j-i}\over{m+n-i-j+1\choose j-i}}&if
$i<j$.\cr}$$
Similarly, let $B=(b_{ij})$, with the same range of indices, where
$$b_{ij}=\cases{
0&if $i>j$,\cr
1& if $i=j$,\cr
(-1)^{i+j}{{m-i\choose j-i}{n-i\choose j-i}\over{m+n-2i\choose j-i}}&if
$i<j$.\cr}$$

Then Gordan's inversion formula amounts to the non-obvious assertion that
$B$ is the inverse of $A$.

I want to record one observation I have made about Gordan's
coefficients. The matrix $A$ is an upper triangular unipotent matrix.
Let $C=(c_{ij})$ be the logarithm of $A$. Then for $j>i$ we have
$$C_{ij}=\kappa_{j-i}A_{ij}\prod_{k=0}^{j-i-2}{1\over n+m-2i-k},$$
where $\kappa=\{\kappa_r\}_{r=1,2,3,\dots}$ is a sequence of integers
independent of $m,n$. We present the first 20 terms of the sequence
$\kappa$
below.
\halign{\quad$#$\hfil&\quad$#$\hfil\cr
r&\kappa_r\cr
1&1\cr
2&1\cr
3&3\cr
4&14\cr
5&80\cr
6&468\cr
7&2268\cr
8&10224\cr
9&313632\cr
10&9849600\cr
11&21954240\cr
12&-8894136960\cr
13&-105857556480\cr
14&20609598562560\cr
15&650835095904000\cr
16&-80028503341516800\cr
17&-5018759207362252800\cr
18&503681435808239001600\cr
19&56090762228110443724800\cr
20&-4811020757268709294080000\cr}

