The Holdovers (2023) Tangerine (2015)

Nice obfuscation, thank you!

First, we replace with being tangents on the inscribed circle, as in

The inequality thus becomes

which is

or

First, for the upper bound

which is obviously true because is

Second, for the lower bound

Is equivalent to

The sum must be >= 0 because each summand is.

We can choose from any number of factors and as long as they are even (including 0), therefore the total number of combinations is .

I think those are the ones based on

But there are more based on like

If we write down the floor values we see

(note how, conveniently, is just at the border between 44 and 45) or

or

and then using the Sums of powers formula from the suggested page

Empirically .

A second solution is S=M, J=W, B=T.

The shortest path is

If you look at that path in iterations 1-3, you see that it cuts through solid squares in two symmetric ways, in such a way that the next iteration's hole just barely misses it. Due to the fractal nature, this repeats with all subsequent iterations, infinitely.

DST ended last night in Austin

I think the answer is in Why do perfect squares have a odd amount of factors

The squares are exactly those numbers with an odd amount of factors. Because the amount is odd, they are now open.

Lockers 1, 4, 9, 16, 25, 36, 49, 64, 81, 100 are open at the end, all others are closed.

I.e. the square numbers!