  Given a sequence ${\bf g}: g_0,\ldots, g_{m}$, in a finite group $G$ with
$g_0=1_G$, let ${\bf \bar g}: \bar g_0,\ldots, \bar g_{m}$, be the sequence
defined by $\bar g_0=1_G$ and $\bar g_i=g_{i-1}^{-1}g_i$ for $1\leq i \leq m$.
We say that $G$ is doubly sequenceable if there exists a sequence ${\bf g}$ in
$G$ such that every element of $G$ appears exactly twice in each of ${\bf g}$
and ${\bf \bar g}$. If a group $G$ is abelian, odd, sequenceable,
R-sequenceable, or terraceable, then $G$ is doubly sequenceable. In this paper,
we show that if $N$ is an odd or sequenceable group and $H$ is an abelian
group, then $N \times H$ is doubly sequenceable.
