drc wrote:This rule doesn't seem to have gliders, but it has oscillators:
I looked at this rule back in 1987, and found most of the oscillators that you mentioned. I also found these p5 oscillators:
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#C p5 oscillators
x = 23, y = 9, rule = B2/S34H
bo$2bo2bo$2b4o8bobo$3o2bo8b3o$2bo3bo9b3obo$3bo2b3o7bob3o$3b4o13b3o$3bo
2bo13bobo$7bo!
Here are two of the p4s that you mentioned:
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#C p4 oscillators
x = 14, y = 4, rule = B2/S34H
o8bo$3o6b3o$2bo8bobo$13bo!
By stacking more of those components we can double the period to 8, in infinitely many ways. For example:
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#C p8 oscillators
x = 54, y = 8, rule = B2/S34H
o12bo12bo14bo$3o10b3o10b3o12b3o$2bobo10bobo10bobo12bobo$4bobo10b3o10bo
bo12bobo$6b3o10bobo10b3o12b3o$8bo12bo12bobo12bobo$36bo14b3o$53bo!
It's easy to show that there are no finite, nonempty stable patterns. However, there's a structure that serves as a wall; it's stable except for its ends, which quickly become period 5. (Your p5 and the second p5 shown above are short forms of this.)
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#C wall
x = 43, y = 23, rule = B2/S34H
o$3o$2b3o$4b3o$6b3o$8b3o$10b3o$12b3o$14b3o$16b3o$18b3o$20b3o$22b3o$24b
3o$26b3o$28b3o$30b3o$32b3o$34b3o$36b3o$38b3o$40b3o$42bo!
The wall can repair itself if one of its central cells is deleted; the wall recovers in 14 gens, and some debris dies out in gen 44:
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#C self-repair
x = 43, y = 23, rule = B2/S34H
o$3o$2b3o$4b3o$6b3o$8b3o$10b3o$12b3o$14b3o$16b3o$18b3o$20bobo$22b3o$
24b3o$26b3o$28b3o$30b3o$32b3o$34b3o$36b3o$38b3o$40b3o$42bo!
Also, the wall forms an impenetrable barrier between its two sides. As long as all of the cells that touch 2 wall cells are empty, they'll remain that way and the wall won't be damaged. We can combine 6 walls to make a hexagon and create many billiard tables. The corners where the walls meet can have 2 shapes, one with period 2 outside and the other with period 3. (The corners can be destroyed from the outside, but not from the inside.)
Here are a few examples of billiard tables:
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#C p6: p6 inside, p2 outside
x = 15, y = 19, rule = B2/S34H
3bo$3bo$ob4o$b3ob4o$2bo4bo$2b2o3b2o$3bo4bo$3b2o3b2o$4bob2obo$4b2o3b2o$
5bob2obo$5b2o3b2o$6bo4bo$6b2o3b2o$7bo4bo$6b4ob3o$9b4obo$11bo$11bo!
That's emulating a 1-dimensional XOR rule, and we can get infinitely many periods from longer versions of it. For example:
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#C p2044: p2044 inside, p2 outside
x = 29, y = 47, rule = B2/S34H
3bo$3bo$ob4o$b3ob4o$2bo4bo$2b2o3b2o$3bo4bo$3b2o3b2o$4bob2obo$4b2o3b2o$
5bo4bo$5b2o3b2o$6bo4bo$6b2o3b2o$7bo4bo$7b2o3b2o$8bo4bo$8b2o3b2o$9bo4bo
$9b2o3b2o$10bo4bo$10b2o3b2o$11bo4bo$11b2o3b2o$12bo4bo$12b2o3b2o$13bo4b
o$13b2o3b2o$14bo4bo$14b2o3b2o$15bo4bo$15b2o3b2o$16bo4bo$16b2o3b2o$17bo
4bo$17b2o3b2o$18bo4bo$18b2o3b2o$19bo4bo$19b2o3b2o$20bo4bo$20b2o3b2o$
21bo4bo$20b4ob3o$23b4obo$25bo$25bo!
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#C p12: p4 inside, p3 outside
x = 11, y = 11, rule = B2/S34H
2b2o$b5o$3o2b3o$2o5b2o$bo2b2o2bo$b2obobob2o$2bo2b2o2bo$2b2o5b2o$3b3o2b
3o$5b5o$7b2o!
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#C p14: p14 inside, p2 outside
x = 15, y = 15, rule = B2/S34H
4bo$4bo$3b4o$ob3ob3o$b3o4b4o$2bo4bo2bo$2b2o6b2o$3bo7bo$3b2o6b2o$4bo4bo
2bo$3b4o4b3o$6b3ob3obo$8b4o$10bo$10bo!
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#C p16: p16 inside, p2 outside
x = 15, y = 19, rule = B2/S34H
3bo$3bo$ob4o$b3ob4o$2bo4bo$2b2o2b3o$3bo2bobo$3b2o3b2o$4bo4bo$4b2o3b2o$
5bo4bo$5b2o3b2o$6bo4bo$6b2o3b2o$7bo4bo$6b4ob3o$9b4obo$11bo$11bo!
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#C p18: p9 inside, p2 outside
x = 15, y = 15, rule = B2/S34H
3bo$4bo$3b4o$2b3ob3obo$4o4b3o$2bo7bo$2b10o$3bo7bo$3b10o$4bo7bo$4b3o4b
4o$4bob3ob3o$8b4o$10bo$11bo!
By attaching a hexagon with p3 corners to each of the p2 corners of that, we can turn this into a true p9 oscillator. There's probably a simpler way.
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#C p9: p9 inside, p3 outside
x = 37, y = 37, rule = B2/S34H
10b2o$9b5o$8b3o2b3o$8b2o5b2o8b2o$9bo6bo7b5o$2b2o5b2o5b2o5b3o2b3o$b5o4b
o6bo5b2o5b2o$3o2b3o2b2o5b2o5bo6bo$2o5b2o2b3o2b3o5b2o5b2o$bo6bo4b5o7bo
6bo$b2o5b2o5b2o8b2o5b2o$2bo6bo6bo7b5o2b3o$2b2o5b2o4b4o4b3o2b5o$3b3o2b
5ob3ob3ob3o5b2o$5b5o2b4o4b4o$7b2o5bo7bo$14b10o$15bo7bo$15b10o3b2o$16bo
7bo2b5o$8b2o5b4o4b6o2b3o$7b5o2b3ob3ob3ob2o5b2o$6b3o2b5o4b4o3bo6bo$6b2o
5b2o7bo4b2o5b2o$7bo6bo7b2o4bo6bo$7b2o5b2o7bo4b2o5b2o$8bo6bo7b2o4b3o2b
3o$8b2o5b2o5b5o4b5o$9b3o2b3o4b3o2b3o4b2o$11b5o5b2o5b2o$13b2o7bo6bo$22b
2o5b2o$23bo6bo$23b2o5b2o$24b3o2b3o$26b5o$28b2o!
Here's an unrelated p9 oscillator, obtained by changing one corner of a p3 hexagon:
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#C p9: stable inside, p9 outside
x = 13, y = 13, rule = B2/S34H
3bo$3b2o5bo$2b5o4bo$4o2b4o$b2o5b2o$2bo6bo$2b2o5b2o$3bo6bo$3b2o5b2o$3b
4o2b4o$6b5o$8b2o$9bo!
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#C p42: p21 inside, p2 outside
x = 17, y = 23, rule = B2/S34H
2bo$3bo$2b4obo$4ob3o$2bo4bo$2b2ob4o$3bo4bo$3b2o3b2o$4bo4bo$4b2o3b2o$5b
o4bo$5b2o3b2o$6bo4bo$6b2o3b2o$7bo4bo$7b2o3b2o$8bo4bo$8b4ob2o$9bo4bo$9b
3ob4o$9bob4o$13bo$14bo!
We could make that a true p21 in the same way that we got a true p9.
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#C p64: p64 inside, p2 outside
x = 17, y = 23, rule = B2/S34H
2bo$3bo$2b4obo$4ob3o$2bo4bo$2b7o$3bob2obo$3b2o2b3o$4bobo2bo$4b2o3b2o$
5bo4bo$5b2o3b2o$6bo4bo$6b2o3b2o$7bo4bo$7b2o3b2o$8bobo2bo$8b3o2b2o$9bo
4bo$9b3ob4o$9bob4o$13bo$14bo!
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#C p87: p29 inside, p3 outside
x = 12, y = 13, rule = B2/S34H
2b2o$b5o$3o2b3o$2o2bo2b2o$bo6bo$b2o2bo2b2o$2bo6bo$2b2o2bo2b2o$3bo6bo$
3b2o2bo2b2o$4b3o2b3o$6b5o$8b2o!
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#C p126: p126 inside, p2 outside
x = 18, y = 25, rule = B2/S34H
2bo$3bo$2b4obo$4ob3o$2bo4bo$2b2ob4o$3bo4bo$3b2o3b2o$4bo4bo$4b2o3b2o$5b
o4bo$5b2o3b2o$6bo4bo$6b2o3b2o$7bo4bo$7b2o3b2o$8bo4bo$8b2o3b2o$9bo4bo$
9b4ob2o$10bo4bo$10b3ob4o$10bob4o$14bo$15bo!
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#C p132: p44 inside, p3 outside
x = 14, y = 15, rule = B2/S34H
3bo$3b2o$2b5o$4o2b4o$b2o5b2o$2bo6bo$2b2o5b2o$3bo6bo$3b2o5b2o$4bob2o3bo
$4b2obobob2o$4b4o2b4o$7b5o$9b2o$10bo!
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#C p236: p236 inside, p2 outside
x = 19, y = 27, rule = B2/S34H
2bo$3bo$2b4obo$4ob3o$2bo4bo$2b2o3b2o$3bo4bo$3b2o3b2o$4bo2bobo$4b2o3b2o
$5bo4bo$5b2o3b2o$6bo4bo$6b2o3b2o$7bo4bo$7b2o3b2o$8bo4bo$8b2o3b2o$9bo4b
o$9b2o3b2o$10bo2bobo$10b3o2b2o$11bo4bo$11b3ob4o$11bob4o$15bo$16bo!
We could also make hexagons with some p2 corners and some p3; I haven't explored that.
I never found any spaceships or puffers, but I'd bet that some exist. The front end of this pattern reappears every 8 gens until gen 104; after that the exhaust destroys it:
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x = 15, y = 8, rule = B2/S34H
2o5bo$b2o3bo3b2o$bo3bobo4bo$7bo3b3o$2b2o2b2o4b2o$2bobo3b3o3bo$8bo4bo$
5bo4bo!
