[seqfan] Re: Counting polyominoes with peculiar symmetry

[Jon Wild]

I agree with Allan's count. The rule seems to be: a polyomino qualifies, 
in Allan's sense, if it meets at least one of these two conditions:

1) it has a mirror symmetry whose axis runs through the centre of a column 
or row of cells (i.e. not along a line that forms the cell edges)

2) it has a rotational symmetry about a fixed point that is at the 
midpoint of a cell edge (i.e. not at a cell corner, and not in the middle 
of a cell)

It looks to me as if there are 12 heptominoes that qualify. The sequence 
1,1,1,3,4,10,12 is not in the encyclopedia.

jon wild


On Wed, 22 Sep 2021, Allan Wechsler wrote:

> Take a polyomino and mark each of its square cells by drawing one of its
> two diagonals. But once you choose the first diagonal, all the others are
> forced because I require these diagonal markings to meet at their
> endpoints. (That is, /\ is fine, but // isn't allowed.)
>
> Depending on your choice of the first diagonal, you can mark up any
> polyomino in two ways, but sometimes these two ways turn out to be
> isomorphic. That only happens for polyominoes that have some symmetry, but
> it doesn't happen for all symmetric polyominoes, just some of them.
>
> By my hand count, the number of qualifying polyominoes of orders 1 through
> 6 are:
>
> 1,1,1,3,4,10
>
> But I am not sure of the hexomino number. None of the 11 matches at OEIS
> say anything about polyominoes.
>
> I hope some polyomino counters out there understand what I mean, and can
> verify or contradict my counts, and tell me whether the resulting sequence
> is already in OEIS or not.
>
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