I haven't seen this before, and it's new to me. The specific case I show in the image below was important to my solving path today, but I took a while to find it.

In a hexagonal pattern, imagine two numbers horizontally next to each other, and think about the number of edges on the right half of the left number, and the left half of the right number. Each of these sections has 3 potential edges and the middle edge is shared. It turns out the only possible combinations for the number of edges in the two sections are 0 and 0; 1 and 1; 1 and 3; 2 and 2; or 3 and 1.

Here's the specific example - I think this logic can be applied to many others. The right half of the right 4 has two edges; therefore the left half has two edges; therefore the right half of the left 4 has two edges; therefore the left half has two edges; therefore the right half of the 2 has two edges; therefore the left half of the 2 has zero edges.

Oooh, fascinating. I wonder whether there's anything this corresponds to in the Standard square grids? Probably the various patterns of propagating diagonal 2s?

Regarding further consequences of your insight:

First of all, obviously this pattern is rotationally symmetric.

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Now for the following, imagine a clue to the far right, followed by an arbitrarily long row of directly adjacent clues to its left.

If the left half (side) of the first clue has an odd number of lines (1 or 3), the right side of the adjacent clue must also have an odd number of lines (1 or 3).

If that second clue is even, then its left side must also have an odd number of lines and the pattern propagates further to *its* left side, since odd + odd = even. This logic extends to an arbitrary number of adjacent even clues.

Once an adjacent clue is odd, then its left side has an even number of lines (0 or 2). This breaks the odd-number-of-lines pattern, but if there's a *second* consecutive odd clue, then its right side will also be even (0 or 2), but its left side will once again have an odd number of lines, which restores the odd-number-of-lines pattern.

What about an arbitrary number of odd clues? An even number of odd clues with an odd number of lines on its right also has an odd number of lines on its left; whereas the opposite is true for an odd number of odd clues.

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To generalize this: Adjacent even clues preserve parity. (i.e. odd in, odd out; even in, even out) Adjacent odd clues reverse parity. (i.e. odd in, even out; even in, odd out)

From this, it also follows that the two ends of an arbitrarily long chain of adjacent clues preserve parity if its sum is even (i.e. if it contains an even number of odd clues), and reverses parity if its sum is odd. Or more concisely: An even chain preserves parity. An odd chain reverses it.

Good point! I think there's a lot of potentially interesting logic here. It is definitely similar to diagonal chains of clues in a square board.

You can also use this logic to help understand patterns like "1 next to 5", "1 next to 4", and "5 next to 5". Also, on any 5, one half is a "1 and 3" pattern and the other half is "2 and 2", maybe that will be useful in certain situations.