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Sep 12 at 6:21 history edited Laska CC BY-SA 4.0
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Jun 30 at 1:29 comment added alices_and_bobs Thanks for the suggestion! I looked it up in OEIS and it doesn't seem to match anything. You are welcome to submit it there, it's your question after all and you got the correct answer first. Maybe you can also work out the number of distinct units for nxn chessboard that might be interesting too.
Jun 28 at 3:43 comment added Laska Thanks: good coding sir. I think the series begins 1, 2, before the 5. Worth sticking it into OEIS for the grins. Now maybe Mr Gilman who likes naming fairy pieces can come up with a trillion names for this lot!
Jun 28 at 3:33 comment added alices_and_bobs It's impressive you got the right answer (hopefully ;) by analyzing. I first used a semi-automated list, but apparently missed a few repeated cases somewhere. I counted it again using a script and it also gives 329 unique lines but because of the poor formatting I'm not sure where's my mistake. Do you know any better explanation of these numbers? 5, 11, 28, 58, 161, 329
Jun 28 at 3:32 history edited Laska CC BY-SA 4.0
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Jun 28 at 3:26 history edited Laska CC BY-SA 4.0
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Jun 28 at 2:27 history edited Laska CC BY-SA 4.0
Correcter
Jun 28 at 1:26 history edited Laska CC BY-SA 4.0
Correct
Jun 27 at 23:36 comment added alices_and_bobs I think (0,1) rider +(0,2) rider is not the same as (0,1) rider alone, or even (0,1) rider+(0,2) leaper, because (0,1) rider +(0,2) rider can not only leap over a piece at (0,1) but also changes the dependence squares of (0,4) from (0,1/2/3) to (0,2). I believe it's very tricky to calculate without counting all cases explicitly that's why I did that, but it would be great if there's a way to calculate more easily or even get a recurrence relation or something. It's one of the more stupid things I've counted by hand lately :)
Jun 27 at 23:29 history edited Laska CC BY-SA 4.0
Reorder
Jun 27 at 23:24 history answered Laska CC BY-SA 4.0