We have a simple (or single) block of Langton's Ants colony which includes two ants looking in the same direction. Their positions can be interpreted as knight's walk. The distances between each next and previous blocks are the same (and equals $1$ cell), so here you can see $3$ such blocks:

For any $n\in \mathbb{N}$ our colony always oscillates. In other words, it repeats all actions after several steps (you may verify it using program named Golly or any another). Therefore, it has a period. What is nice here is the fact that the period is linear and equals $4(8n+3)$ steps. If we increase distance between blocks from $1$ cell to $2$, so areas where they oscillates do not intersect and period is constant ($44$ steps). It is hard to say what exactly the colony does for small $n$, but when we take large values, we may notice $3$ main steps:

- builds first triangle after ~$6n$ steps
- builds second triangle after ~$12n$ steps
- after $2(8n+3)$ steps the ants repeat everything they did symmetrically in backward order

Also we may notice that:

- Each ant moves in an area bounded by a square with side $n+5$
- The real area where it moves (equivalently number of cells in area bounded by square which ant visit at least once) equals $5(n+3)$
- colony moves in area bounded by rectangle with length $3(n+1)$ and width $2(n+2)$
- real area for colony $3n(n+5)+2$

If we define $T_1(n,k)$ as number of white cells colored to black at the whole period by upper ant in $k$-th block of colony length $n$ blocks (and $T_2(n,k)$ for lower ant), so:

$T_1(n,k)=T_2(n,k)=0, n<k$

$T_1(n,k)=4(5n+1), k=1$ or $k=n$

$T_1(n,k)=2(11n+1), 1<k<n$

$T_2(n,k)=4(3n+2), k=1$ or $k=n$

$T_2(n,k)=10(n+1), 1<k<n$

$T_1(n,k)+T_2(n,k)=4(8n+3), n\geqslant k$

If we define $q(n,k)$ as number of black cells at $k$-th step of colony length $n$ blocks, $q_1(n,k)=q(n+1,k)-q(n,k)$, so:

$q(n,k)=q(n,4(8n+3)±k)$

$q(n,0)=q(n,4m(8n+3))=0$

$q_2(n,k)=q_1(n+1,k)-q_1(n,k)=0, 6n+9\geqslant k$

$q_3(n,k)=q_2(n+1,k+6)-q_2(n,k)=0, 12n+4\geqslant k$

$q_4(n,k)=q_3(n+1,k+12)-q_3(n,k)=0, 20n-4\geqslant k$

$q_5(n,k)=q_4(n+1,k+20)-q_4(n,k)=0, 26n+15\geqslant k$

Finally we have:

$$\sum\limits_{k=1}^{4(8n+3)} q_1(n,k) = 8(77n+31)$$ $$\sum\limits_{k=1}^{4(8n+3)} q(n,k) = 4(77n^2-3(5n-6))$$

How can we prove all (or part) of these facts? Is there another absolute oscillators in LA?