\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}