Extrapolated tiling counts for $2s\times(2s+2)$ rectangles

In fact, both families of conjectures follow from the following uniform enumeration. Let $0\le t<s$ and put

$$B=\binom{t+2}{2}.$$

Then

$$\begin{aligned} T_{2s\times(2s+t)}(s,2)&=(s+t+1)^2+B((s+1)^2-2),\\ T_{2s\times(2s+t)}(s,3)&=2B(s+t+1)+(s-1)\frac{(t+1)(t+2)(2t+3)}3,\\ T_{2s\times(2s+t)}(s,4)&=B^2. \end{aligned}$$

A large square is determined by its upper-left corner $(r,c)$, where $0\le r\le s$ and $0\le c\le s+t$. Two such squares are disjoint precisely when

$$|r-r'|\ge s\quad\text{or}\quad |c-c'|\ge s.$$

Vertical separation is therefore possible only between boundary rows $0$ and $s$. The number of unordered column pairs with separation at least $s$ is

$$\sum_{d=s}^{s+t}(s+t+1-d)=\binom{t+2}{2}=B.$$

Moreover, since $s+t<2s$, three pairwise horizontally separated squares cannot occur.

For two squares, vertically separated pairs contribute $(s+t+1)^2$. For each of the $B$ horizontally separated column pairs there are $(s+1)^2$ ordered row choices, exactly two of which were already counted vertically. This gives the first formula.

For three squares supported entirely on boundary rows, choose which row carries two squares, its separated column pair, and the opposite-row column. This gives $2B(s+t+1)$. Otherwise exactly one square occupies an interior row. For its column $c$, write

$$q(c)=\#\{d\in\{0,\ldots,s+t\}:|d-c|\ge s\}.$$

All such columns lie on one side of $c$ and have mutual separation less than $s$, so the remaining two squares must occupy the two different boundary rows. Their columns can be chosen independently in $q(c)^2$ ways. The nonzero values of $q(c)$ are $t+1,t,\ldots,1$ at one end and $1,\ldots,t,t+1$ at the other, whence

$$\sum_cq(c)^2=2\sum_{j=1}^{t+1}j^2=\frac{(t+1)(t+2)(2t+3)}3.$$

There are $s-1$ interior rows, proving the second formula.

For four squares, an interior-row square can coexist with at most two others. Thus all four lie on the boundary rows, with exactly two on each row. The separated column pair on each boundary row can be chosen independently, yielding $B^2$.

Taking $t=2$, so $B=6$, gives exactly

$$T_{2s\times(2s+2)}(s,2)=7s^2+18s+3,\quad T_{2s\times(2s+2)}(s,3)=40s+8,\quad T_{2s\times(2s+2)}(s,4)=36$$

for every $s>2$. The same theorem with $t=1$ also gives $4s^2+10s+1$, $16s+2$, and $9$, settling the companion formulas for $2s\times(2s+1)$ rectangles.