I am interested in whether the following problem is known.
For a given binary vector $V$ of length $n\geq m$, let $S$ be a subset of the possible subvectors of $V$ of length $m$ and say that the size of $S$ is simply the number of subvectors it contains.
For fixed $n$ and $m$ and maximising over all possible binary vectors of length $n$, what is the maximum size $S$ that has the property that the Hamming distance between all pairs $v_1,v_2 \in S$ is at least $d$?
If subvectors were not allowed to overlap then this would be a basic coding theory question.
(Clarification: A subvector has consecutive coordinates.)
Let us simplify by setting n = 2m-1 so there are $m$ subvectors of a fixed $V$ each with length $m$.
Set $n = 7$, $m = 4$ and $d=2$. The vector $(1, 0, 1, 0, 0, 1, 1)$ has subvectors $(1,0,1,0), (0,1,0,0), (1,0,0,1)$ and $(0,0,1,1)$ which all have pairwise Hamming distance at least $2$. So for these values the answer is in fact $4$ which is as high as it can be.
Set $n = 7$, $m = 4$ and $d=3$. Over all vectors $V$ of length $7$, the largest set $S$ of subvectors of length $4$ all of which have pairwise distance $3$ from each other is $2$.
Set $n=9$, $m=5$ and $d=3$. Vector $V=(0, 0, 0, 1, 1, 0, 1, 0, 0)$ gives you the answer $4$ and is the maximum possible for these values of $n,m,d$.
Set $n=11$, $m=6$ and $d=3$. Vector $V=(0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 1)$ gives you the answer $6$ which is as high as it can be.
Set $n=13$, $m=7$ and $d=4$. Vector $V=(0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 1, 0, 1)$gives you the answer $7$ which is as high as it can be.
Clarification II. I would be happy with bounds rather than exact answers. Is there, for example, an equivalent of the Hamming bound for this setup?