Hearted Youtube comments on Mathologer (@Mathologer) channel.
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SPOILERS SPOILERS SPOILERS: It's uncountable; here's a funny algorithm: given a sequence (n_k) of natural numbers, we start out just as normal, adding positive terms until we overshoot. But rather than turning around immediately, we keep going for n_0 many positive terms, and only then turn around. Next, once we undershoot, we don't immediately turn around, but keep adding n_1 many negative turns before we turn around. Then once we overshoot again, we keep adding n_2 many positive terms before we turn around, etc. Note that the original algorithm is what you get when you start with the constant sequence (0,0,0,...).
Now of course, this isn't guaranteed to converge; you could end up over/undershooting too wildly. But - certainly if all the n_i are 0 or 1, it will converge. And there are already uncountably many sequences of just 0s and 1s.
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@zacharyjoseph5522 It's a bit tricky to explain without diagrams, but I'll do my best. Let us call the whole shape G for glasses. First consider the pair of 2x3 rectangles at the extreme left and right of the glasses. If a domino is placed which crosses the border between the 2x3 and the rest of the glasses, then the 2x3 is left with a region of size 5, so cannot be covered.
We therefore know that the dominoes covering the pair of 2x3 areas are necessarily wholly within the 2x3 areas, so the tilings would be the same if the 2x3 regions were actually disconnected from the glasses. So if we let G' be the glasses without this pair of 2x3 regions and if we let N be the function counting the number of tilings of a shape, then we have N(G) = N(2x3)^2*N(G'), as the tilings of the pair of 2x3 regions and the tiling of G' are independent.
We'll need names for a few other things, so we'll call the boundary of a hole E for eye. So the eye is the 10 squares directly surrounding the hole. Consider the middle 2x2 in G'. Imagine a vertical line L cutting this 2x2 into two pieces. This line divides G' into two equal copies of the same shape, an eye with a 2x1 hanging off one side and a 2x2 hanging off the other. We'll call this shape E+2x2+2x1
Now if we consider tilings with no domino crossing L, then clearly the number is N(E+2x2+2x1)^2, as the tilings of the two shapes on either side of L are independent.
If instead we have a domino crossing L, then we must in fact have 2 dominoes crossing L. This is because otherwise we'd create a region of odd size, which couldn't be tiled. In this arrangement with 2 dominoes crossing L, we again have two identical regions to tile, but this time they are E+2x2, following the naming convention for the previous shape. In this case there are N(E+2x2)^2 tilings.
So we've shown that N(G') = N(E+2x2+2x1)^2+N(E+2x2)^2.
Next we consider E+2x2+2x1. If the 2x1 area is covered by one domino, then we're left with tiling E+2x2. Otherwise, then the tiling is forced all the way to a remaining 2x2 region. Therefore N(E+2x2+2x1) = N(E+2x2)+N(2x2).
Now we consider E+2x2. Imagine a line L' dividing the E from the 2x2. If no domino crosses L', then the E and 2x2 are tiled separately, so we get N(E)*N(2x2) tilings. If instead a domino crosses L', then the rest of tiling is forced, so we get 1 such tiling. Therefore N(E+2x2) = N(E)*N(2x2)+1.
The remainder is not especially hard to check. N(E) = N(2x2) = 2, so N(E+2x2) = 2*2+1 = 5, N(E+2x2+2x1) = 5+2 = 7, N(G') = 7^2+5^2 = 74. Finally N(2x3) = 3, so N(G) = 3^2*74 = 666.
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13:46 I was so annoyed that the pattern was that simple. I had worked out a completely different, more complicated pattern. The sums of the differences were always factors of the double position number they surrounded. (e.g. the position numbers surrounding 2 were 1 and 3, which sum to 4, which is double the original position number). Furthermore, the number needed to multiply the sum to get to double the position number went 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, etc. It was a pattern, just a way more complicated one. Now I'm going to have to try to prove that they are equivalent patterns, which may be quite difficult.
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11:56 challenge question: The Pythagorean triple (153, 104, 185) corresponding to the box
[ 9, 4, 17, 13 ]. If you call the children A,B,C, the 153, 104, 185 is the A'th Child of 'CCC'
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