it is straightforward to verify (39). First, one easily checks that (39) is correct for $k=1$ with $\alpha_1$ equal to 1/2. Let us then assume that (39) is true for $k = 0, 1 ...$ up to $p$. One then obtains $${\renewcommand{\arraystretch}{.7}     \begin{array}[b]{@{}c@{}} {\scriptscriptstyle (p)} \\ D \end{array}} = \beta_p  {\renewcommand{\arraystretch}{.7}     \begin{array}[b]{@{}c@{}} {\scriptscriptstyle (p)} \\ S \end{array}}$$ with $\beta_p$ given by $$\beta_p = (-)^{p+1} \alpha_p - \sum_{k=0}^{p-1} (-)^{p(k+1)} \alpha_{k+1}
\alpha_{p-k} = - \sum_{k=1}^{p-2} (-)^{p(k+1)} \alpha_{k+1} \alpha_{p-k}$$ from which one gets, using (\[eq:sigmaS\]), that ${{\renewcommand{\arraystretch}{.7}     \begin{array}[b]{@{}c@{}} {\scriptscriptstyle (p+1)} \\ \Omega \end{array}}}$ is indeed given by (39) with $\alpha_{p+1}$ equal to $$\alpha_{p+1} = \frac{\beta_p}{p+2}$$ Observe now that $\alpha_k = 0$ for $k$ odd, $k \neq 1$. This can again be shown by recurrence. First note that $\alpha_3$ = 0. Now let $p$ be even, $p > 3$. Suppose $\alpha_k = 0$ for $k$ odd, $1 < k < p$. All terms in the relation defining $\beta_{p}$ are proportional to an $\alpha_m$ with $m$ odd, $1 < m < p$, since $k+1$ and $p-k$ have opposite parities. Therefore, $\beta_p = 0 = \alpha_{p+1}$ and thus $\alpha_k
= 0$ for $k$ odd, $k > 1$. Accordingly only $\alpha_k $ with $k$ even can be 