Number of binary and constant-weight strings of length m which do not contain k consecutive zeros I am considering the set of all finite and binary strings. In the following I am interested on the number $N(m,k,r)$ of strings with length $m$ and Hamming weight $r$ which do not contain $k$ consecutive zeros. 
The Hamming weight $r$ is defined as the number of non-zero positions in a string.
Example: The binary string $x=1001101$ is given. The Hamming weight of $x$ is $r=4$, since $x$ contains 4 non-zero entries.
Is there a general formulary for computing $N(m,k,r)$?

So you are looking for $N(m,k,r)=$ number of binary strings with:
length $=m$, Hamming weight $=$ number of ones$=r$, and which
"do not contain $k$ consecutive zeros".
Taking the complement of the string (exchanging the zeros and ones), that is the same as:
length $=m$, Hamming weight $=$ number of ones$=m-r$, and which
"do not contain $k$ consecutive ones".

Consider a binary string  with $s$ "$1$"'s and $q$ "$0$"'s in total. Let's put an additional (dummy) fixed $0$ at the start and at the end of the string.
We individuate as a run the consecutive $1$'s between two zeros, thereby including runs of null length. With this scheme we have a fixed number of $q+1$ runs.
If we sequentially enumerate the length of each run so individuated, we construct a bijection with the number of ways of putting
$s$ (undistinguishable) balls into $q+1$ (distinguishable) bins.
Now consider the case in which runs have a max length of $t$ ones, or that the bins have a limited capacity of $t$ balls, or otherwise the
$$ \bbox[lightyellow] {  
N_b (s,r,q + 1) = {\rm No}{\rm .}\,{\rm of}\,{\rm solutions}\,{\rm to}\;\left\{ \matrix{
  {\rm 0} \le {\rm integer}\;x_{\,j}  \le r \hfill \cr 
  x_{\,1}  + x_{\,2}  + \; \cdots \; + x_{\,q + 1}  = s \hfill \cr}  \right.
 }$$
which as explained in this other post is expressed as
$$ \bbox[lightyellow] {  
N_b (s,t,q + 1)\quad \left| {\;0 \le {\rm integers  }s,q,t} \right.\quad  = \sum\limits_{\left( {0\, \le } \right)\,\,j\,\,\left( { \le \,{s \over t}\, \le \,q + 1} \right)} {\left( { - 1} \right)^j \left( \matrix{
  q + 1 \cr 
  j \cr}  \right)\left( \matrix{
  s + q - j\left( {t + 1} \right) \cr 
  s - j\left( {t + 1} \right) \cr}  \right)} 
 }$$
(which essentially is derived from the inclusion-exclusion principle).
Then, using your symbols, the number of strings with
length $=m$, Hamming weight $=$ number of ones$=r$, and which
" contain runs with max $k-1$ consecutive zeros", equivalent to
length $=m$, Hamming weight $=$ number of ones$=m-r$, and which
" contain runs with max $k-1$ consecutive ones", is given by
$$ \bbox[lightyellow] {  
N(m,k,r)=N_b (m - r,k - 1,r + 1)
 }$$
And, as an example $N(7,3,4)=30$, as in the answer above.
Of course the number of strings with runs whatever is $N_b(m-r,m-r,r+1)={m \choose r}$.
So it is easy to compute also the number of strings with "(at least) one run of exactly $k$ consecutive ones"
etc.
See also this other related post.
