The period of that oscillator is 2,147,483,646 (2^31-2), so you would've had to run oscar.lua for quite a while to find it. Fortunately, this is a well-studied type of oscillator and so the solution can be found in one of 2 ways:
Big lookup table (lazy way):
A while ago, someone at Wolfram evaluated the periods of this type of oscillator on a
toroidal grid (people suspect that some numbers may be small by various prime factors) (see
here for the document and some commentary). Take the length of the replicator region (in this case, 76 cells), then add one and double it. The number under that index in the table will be the period due to math reasons.
(basically, if you look at the replicator strip as a circle, the initial cell and the one opposite it will never be occupied because symmetry means that if live cells are next to them, then both of those live cells will be on, so it effectively splits into 2 bounded replicator strips)
Using modulo arithmetic (less lazy way):
Over a decade ago, conwaylife.com's own Nathaniel wrote a
paper in which he proved (section 5) that for these types of oscillators, the period is equal to
2*((2^k)-1) where k is the smallest number such that
2^k mod (length+1) is ±1 (he says 2n+1 but the length of each replicator unit is 2 so length 2n means n replicator units).
Note: those oscillators may look very different, but their outer edges simulate the same cellular automaton and both families have the same periods. I hope this was helpful.
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