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Langton's ant runs on an infinite white grid. At every white square, it turns right, flips the color of the square, and moves forward one square. At every black square, it turns left, flips the color of the square, and moves forward one square. After many interations, you get complex emergent behavior, such as "recurrent highways" starting at step ~10,000.

Let's suppose that Langton's ant awakens instead on a torus of size $n times n$. It follows its two rules, thus changing the coloring of the torus. At some point, however, it encounters a colouration it has seen before, while being in the same spot as before, and finds itself in a cycle. How do we find the length of a cycle for a size $n$ torus, and is this result known? We can get a stupid upper bound by observing that there are at most $2^n^2$ colorings, $n^2$ positions, and $4$ orientations, so a cycle cannot be longer than $2^n^2+2n^2$. Is there a closed form for this, or at least some tighter bounds?

EDIT 1

I ran some quick calculations just to get a feel for the magnitudes of the numbers¹. Here's an animation of Langton's ant on a $3 times 3$ torus, where the cycle takes 22 steps:

Some more results I got are:
$$beginmatrix
textSize & textSteps & textFactorization\
hline
1 & 2 & 2\
2 & 8 & 2^3\
3 & 66 & 2 cdot 3 cdot 11\
4 & 96 & 2^5 cdot 3\
5 & 11,710 & 2 cdot 5 cdot 1171\
6 & 4,592 & 2^4 cdot 7 cdot 41\
7 & 64,165,598 & 2 cdot 7^2 cdot 31 cdot 21121\
8 & 11,502,464 & 2^7 cdot 73 cdot 1231\
9 & 8,271,515,006,982 & 2 cdot 3^4 cdot 51058734611
endmatrix$$

The value for $n=9$ is due to Connor Harris.

This does not, to my knowledge, match any sequence in OEIS.

EDIT 2

The only real pattern I've found thus far is in the prime factorizations—for odd-sized tori, there is (so far) exactly one factor of 2 in the factorization. However, in even-sized tori, the factors of 2 have multiplicities 3, 5, 4, and 7, which seems interesting. Is there any reason to believe this is true for all even/odd-sized periods?

Footnotes

1: as per Connor Harris's comment, I only checked until the initial state reappeared (i.e., the torus became blank).

Not an answer, but a quick table with the number of turns until the recurrence of the initial position on rectangular grids small enough to be testable by computer.

The ant is initially facing up (that is, along a column); this matters for non-square boards that may have different periods depending on whether the ant starts by facing in the "long" or the "short" direction. Numbers show the length of time until the first full recurrence: an all-white board with the ant in the original position facing up. A bracketed number before the entry indicates that a "quasi-recurrence," or an all-white or all-black grid with the ant in any cell facing either up or down (or left and right, in the case of a square grid), occurs before the first full recurrence. The development of the board after a quasi-recurrence is isomorphic to that of the actual initial position, but translated, rotated, and (if the board is all black at the beginning of the quasi-recurrence) mirrored across the ant's initial line of sight. The bracketed number shows the number of quasi-recurrences up to and including the first full recurrence; thus, an entry of $[3]~66$ means quasi-recurrences on turns 22 and 44 before a full recurrence on turn 66.

beginarrayr
downarrowtextrows/colsrightarrow&2&3&4&5&6&7&8&9&10 \
hline
2 & [2]~8 & 16 & 16 & 16 & 16 & 16 & 16 & 16 & 16\
3 & 8 & [3]~66 & 72 & [3]~954 & 196 & 208 & 3,008 & 6,064 & 304 \
4 & 8 & [2]~56 & [2]~96 & 624 & 696 & 3,448 & 2,336 & 13,360 & 2,608\
5 & 8 & [5]~170 & 96 & [5]~11,710 & 16,804 & [5]~344,300 & 606,688 & [5]~14,988,170 & 1,544,720\
6 & 8 & 120 & 96 & 4,184 & 4,592 & [3]~296,736 & 507,056 & 1,824,688 & 2,045,304 \
7 & 8 & [7]~322 & 96 & 17,432 & 714,592 & [7]~64,165,598 & 34,882,576 & 299,407,462 & 58,495,320 \
8 & 8 & 208 & 96 & [2]~31,600 & 147,424 & 10,003,800 & [2]~11,502,464 & [2]~1,634,057,664 & [2]~4,622,916,480 \
9 & 8 & [9]~522 & 96 & [9] 1,568,880 & 8,066,144 & [9]~2,508,401,214 & 3,586,271,200& [9] 919,057,222,998 & >10^10\
10 & 8 & [5]~320 & 96 & [5]~5,445,220 & 10,426,496 & [5]~2,107,770,700 & 2,084,996,112 & 133,648,022,836 & > 1.3 cdot 10^10
endarray

These are usually much smaller than the trivial upper bound in your comment. Periods seem quite erratic in general, especially for boards with at least one small dimension: for example, the periods of boards with 3 rows and 18, 19, and 20 columns are respectively 10,930,388, 592, and 30,519,840. The relative infrequency of quasi-recurrences is striking, as a randomly chosen state of the board is much more likely to be a quasi-recurrence than a the initial state.

Edit: Some other observations. The square toruses of odd side-length $n in 3, 5, 7, 9$ all have a period divided into $n$ quasi-recurrences. Furthermore, none of their period lengths is divisible by $4$. Furthermore, with the exception of the $2 times 2$ torus (for which the quasi-recurrence occurs after turn $4$ with a black grid and the ant rotated 180 degrees in its starting square), every quasi-recurrence I have found has left a white grid with the ant moved forward or backward within its initial column, but not moved side to side or rotated. For anyone who wants to do their own experimenting, I have a C program (not especially well designed, but workable) here.

We can tighten the upper bound by looking at the symmetries on the torus. The ant's behavior will be the same no matter which of the $n²$ squares it starts on or which of the $4$ directions it is facing. In addition, a reflection coupled with a color inversion is another symmetry. These correspond to the symmetry groups $mathbbZ_n times mathbbZ_n$, $mathbbZ_4$, and $mathbbZ_2 times mathbbZ_2$, respectively. The maximum order of any element of $mathbbZ_n times mathbbZ_n$ is $n$. This means that the ant will find itself on a blank grid but possibly in the wrong position after $2^n²+2$ steps, but after doing this $n$ times it must be in the right position. Similarly, the maximum order of any element of $mathbbZ_2 times mathbbZ_2$ is $2$, so after $2^n²n²$ steps we may end up reflected and inverted, but doing this twice we'll end up where we started. The maximum order of $mathbbZ_4$ is $4$, so the symmetry of rotation doesn't help us.

Combining these two symmetries, a better upper bound is $mathbf2^n²+1n$.