Denote the number of $\binom{n}{k}$ such that the longest consecutive block of things chosen is exactly $j$ things long as ${\binom{n}{k}}_{j}$, where $j\leq k \leq n$; the goal is to derive a general or (more likely) recursive formula.
For $j=0$, notice ${\binom{n}{k}}_{0}=1$ when $k=0$ and ${\binom{n}{k}}_{0}=0$ when $k>0$.
For $j=1$, ${\binom{n}{k}}_{1} \quad (k>0) \quad $ is equivalent to the number of ways to choose where none are consecutive, which is given by the well known formula $\binom{n+1-k}{k}$. Rewrite $$\binom{n+1-k}{k} = \binom{n-k}{k-1} + \binom{n-k}{k} = \binom{(n-2)+1-(k-1)}{(k-1)} + \binom{(n-1)+1-(k)}{(k)}$$ using Vandermonde's Identity, and recursion is established for $j=1$ $(k>1)$: $$\binom{n}{k}_1 = \binom{n-2}{k-1}_1 + \binom{n-1}{k}_1$$ However, this does not hold for $j\geq2$ and I'm stuck. It feels intuitive that a recursive method should exist.
You can obtain a recursive formula by conditioning on the first block (the element 1 does not appear, or 1 appears in a block of length $r<j$, or 1 appears in a block of length $j$): $$\binom{n}{k}_j=\binom{n-1}{k}_j+\sum_{r=1}^{j-1} \binom{n-r-1}{k-r}_j+\sum_{s=0}^j\binom{n-j-[j<n]}{k-j}_s$$