@orlp: I never did hear what you thought about this.

sec

need to think

@El'endiaStarman that won't be any faster unless you've got some special formula to get a row of the siepinski triangle

That's basically what I'd like to see. If we can figure out a formula, we can possibly work backwards from there and find a closed or iterative/recursive solution to the original problem.

@El'endiaStarman also we can only use this to test upper bound conjectures for periods

actually

if we have a conjecture for a period, and it works we only also need to check its divisors

yeah

oftenpaper.net/sierpinski.htm

@El'endiaStarman there's a formula

pascal's triangle, see if the numbers are odd

I'm not sure that'll work for the CA version.

There are gaps...

@El'endiaStarman actually, there's another problem

to get to generation n you need to generate a triangle with n elements (and then reverse and fold over itself)

so if we need to get to generation 2^100...

Yeah, that's not gonna work.

If we can calculate a row directly, that'd help (I think).

@orlp This is an amazing rabbit hole...

By the way, where he talks about Towers of Hanoi, I think that's the structure we get with the CA.