0$ we let $M=8(k+1)$ and get that
$$\dist=\{ 0,\pm 1,\pm 2,\pm (4k+3),\pm (4k+4), \pm (4k+5)\},$$
and therefore
$$\dist\,(\mbox{mod}\,M)=\{ 0,1,2, 4k+3,4k+4,4k+5, 8k+6,8k+7\}. $$
Write
$B=[3,4k+2]\cap\nn$ and $C=[4k+6,8k+5]\cap\nn$. Then $B\cup C$ is the
complement of $\dist\,(\mbox{mod}\,M)$ in $\{0,\dots,M-1\}$. Assume that a
Steinhaus set of period $M$ for $(1,1,4k+3)$ existed. Then by Proposition
\ref{pSM} there are $x_1,\dots,x_{2k+1}\in B\cup C$ with $x_i-x_j\in B\cup
C\,(\mbox{mod}\,M)$. Let $\{x_1,\dots,x_{2k+1}\}\cap B=B_0$, $h=|B_0|$,
$C_0=\{x_1,\dots,x_{2k+1}\}\cap C$ and $l=|C_0|$. Then $h+l=2k+1$. We note the
following facts. For each $x\in B_0$ such that $x\leq 4k$, the numbers $4k+3+x,
4k+4+x$ and $4k+5+x$ are in $C\cap (x+\dist)$, hence in particular are not in
$C_0$. Moreover, since $1,2\in \dist$, any two elements of $B_0$ differ by at
least 2, the two sets of three numbers associated with them have an empty
intersection. Using this we obtain the following inequality $$ l+(h-1)\cdot
3+1\leq 4k, $$ where the left hand side counts the number of elements in $C_0$
and the number of elements excluded from $C_0$ for being associated with one of
the numbers in $B_0$ (in a worst case scenario), and the right hand side the
size of $C$. By a symmetric and similar argument we also obtain $$ h+(l-1)\cdot
3+1\leq 4k,$$ by reversing the roles of $B$ and $C$. Thus we get $$
4(h+l)-4\leq 8k \mbox{ or } h+l\leq 2k+1. $$ This is not a contradiction yet
but it follows from $h+l=2k+1$ that all elements of $B$ and $C$ are accounted
for in the above counting in order to establish the inequalities. This is to
say that every element of $B$ is either an element of $B_0$ or else is
associated with an element of $C_0$, and vice versa for elements of $C$. Now we
claim that either $3\in B_0$ or $4k+6\in C_0$. It is easy to see that if
$4k+6\not\in C_0$, then the only element of $B$ with which it is associated is
$3$. In the case $3\in B_0$ it is straightforward to check that $3+4i\in B_0$
for all $i=1,\dots,k-1$ and $4k+5+4j\in C_0$ for all $j=1,\dots,k$, but this is
a contradiction since there are only $2k$ of them. Similarly in the case
$4k+6\in C_0$ it is straightforward to check that $4i\in B_0$ for $i=1,\dots,k$
and $4k+6+4j\in C_0$ for $j=1,\dots,k-1$, which is a total of $2k$ of them,
contradiction again. \end{proof}
Note that the Proposition \ref{1time} and Corollary \ref{2times} give
polynomial time computable criteria for the existence of Steinhaus sets of
period $n+1$ and $2(n+1)$, respectively. However, the criterion in Proposition
\ref{pSM} is NP in $(a_1,\dots,a_n)$ and $M$. We do not know whether the
existence of Steinhaus sets for a given period is computable in polynomial
time. The following corollary provides an easy sufficient condition for this
existence.
\begin{question}
Determine all pairs of positive integers $a,b$ such that sets of
type $(a,a,b)$ are Jackson.
\end{question}
\begin{corollary} Let $(a_1,\dots,a_n)$ be a sequence of positive integers and
let $s=a_1+\dots+a_n$. If $M=(k+1)(n+1)>s$ and there is an integer $0