1$, and $c\in(1/2, c_{\operatorname{vol}}]$. So, we can perform a $\mathfrak{R} \mathfrak{C}$ move to get $$v_{p-1} =
  \left( \lambda_{p-1}\,;\lambda_{p-1}-1, c^{\times 2}, (1-c)^{\times (2p-1)} \right)$$ We now consider two cases depending on whether $v_{p-1}$ is ordered or not. : Suppose $v_{p-1}$ is not ordered, that is, $\lambda_{p-1}-1<c$. Since $\lambda_{p-2}-1\geq c$, we must have $$(\lambda_{p-1}-1) - (1-c) =
  \lambda_{p-2}-(2c-1)-2+c =
  (\lambda_{p-2}-1)-c\geq 0,$$ so that $$\hat{v}_{p-1} =
  \left(\lambda_{p-1}\,; c^{\times 2}, \lambda_{p-1}-1, (1-c)^{\times(2p-1)} \right)$$ is ordered with defect $d_{p-1}=2c-1>0$. Performing another $\mathfrak{R} \mathfrak{C}$ move gives $$\hat{v}_{p} =
  \left(\lambda_{p}\,; (1-c)^{\times (2p+1)}, \lambda_{p}-1\right).$$ Since $d_{p}=3(1-c)-\lambda_{p}=3(1-c)-\lambda_{p-1}+2c-1=(1-c)-(\lambda_{p-1}-1)
\leq 0$, the vector $\hat{v}_{p}$ is reduced. It is positive if and only if $\lambda_{p}-1\geq 0$. Solving for $c$ in the equation $\lambda_{p}-1\geq0$, we obtain a new upper bound for the generalized width ${\mathsf{w}}_{2p+1}$, namely $${\mathsf{w}}_{2p+1}\leq\frac{\mu+p}{2p+1}.$$ We note that this bound is stronger than the previous bound ${\mathsf{w}}_{2p+1}\leq 1$ only when $\mu\leq p+1$, while it is stronger than the volume condition whenever $\mu\in [\alpha_{-},\alpha_{+}]$, where $\alpha_{\pm} = p+1\pm\sqrt{2p+1}$.

: Suppose that $v_{p-1}$ is ordered. Performing a $\mathfrak{R} \mathfrak{C}$ move yields $$\hat{v}_{p} = \left(\lambda_{p}\,;\lambda_{p}-1, (1-c)^{\times(2p+1)}\right)$$ with defect $$d_{p}=2(1-c)-1=1-2c<0.$$ Therefore, $\hat{v}_{p}$ is reduced and positive, so that there is an embedding of $2p+1$ balls of size $c$ into $M_{\mu}^{0}$. That occurs unless $\lambda_{p-1}-1<c$. Solving 