Expected disjoint blocks of five black cells

Twenty squares sit in a row, and each is independently painted black or white with probability $\tfrac12$. A maximal run of $m$ consecutive black squares is counted as containing $\lfloor m/5\rfloor$ non-overlapping five-blocks (so a run of $6$ counts once, a run of $10$ counts twice). Let $X$ be the total number of five-blocks in the row. Writing $E[X]=\tfrac pq$ in lowest terms, find $p+q$.

$1333361$. Let $C_k=\tfrac{21-k}{2^k}$ be the expected number of (overlapping) windows of $k$ black squares. A run of length $m$ should count $\lfloor m/5\rfloor$, which is achieved by adding windows of length $5,10,15,20$ and subtracting windows one longer ($6,11,16$); this yields $E[X]=\tfrac{284785}{2^{20}}$, confirmed by enumerating all $2^{20}$ rows.