TITLE: Counting cyclic binary sequences of length $n$ where ones appear in blocks of length at least $k$
QUESTION [6 upvotes]: How many binary cyclic sequences of length $n$ exist, where ones only appear in blocks of length at least $k$? We do not consider sequences that result from each other by a cyclic shift equivalent.
Example: Let $n=6$ and $k=2$, i.e. we have no isolated one. Then
[0,0,0,0,0,0]
[1,1,0,0,0,0] and 5 cyclic shifts of it
[1,1,1,0,0,0] and 5 cyclic shifts of it
[1,1,1,1,0,0] and 5 cyclic shifts of it
[1,1,0,1,1,0] and 2 cyclic shifts of it
[1,1,1,1,1,0] and 5 cyclic shifts of it
[1,1,1,1,1,1]
are all of the 29 possible sequences.
The case $k=2$ is covered by https://oeis.org/A109377

REPLY [6 votes]: Consider the monoid $M_k$ in the generators $A_k=\lbrace 0, 01^k,
01^{k+1}, 01^{k+2},...\rbrace$, where $1^r$ denotes a string of $r$
ones. This contains all binary sequences with no strings of 1's of
length less than $k$, except for the words $1^r$ for $r\geq k$. The
monoid $M_k$ is freely generated by $A_k$. Hence if
  $$ F_k(x) = x+x^{k+1}+x^{k+2}+\cdots = x+\frac{x^{k+1}}{1-x}, $$
then the generating function for the
number of words of length $n$ in $M_k$ is
   $$ G_k(x) = \frac{1}{1-F_k(x)}. $$
It is easy to check that $M_k$ is in fact very pure in the
sense of Section 4.7.4 of Enumerative Combinatorics, vol. 1,
second ed. Thus by Proposition 4.7.13, the generating function for
cyclic words is
  \begin{eqnarray*} H_k(x) & = &\frac{xF'_k(x)}{1-F_k(x)}\\
      & = &
      \frac{x(1-2x+x^2+(k+1)x^k-kx^{k+1})}{(1-x)(1-2x+x^2-x^{k+1})}.
   \end{eqnarray*}
To account for the words $1^r$, we merely need to add $x^k/(1-x)$ to
$H_k(x)$.