Lattice Animals with self-avoiding perimeters

franktaw at netscape.net franktaw at netscape.net
Thu Feb 22 10:56:49 CET 2007

```I think this variation of the question has a much simpler answer:
arrange n size 1 holes in a pattern as close to a square as possible
(fit in the smallest grid k x k or k x (k+1) - or actually 2k+1 x 2k+1
or 2k+1 x 2k+3, in order to leave space between the holes), and
surround them.  Thus if k^2 < n <= k^2+k, the solution has a perimeter
of 4n+2*(2k+1+2k+3) = 4n+8k+8 , and if k^2+k < n <= (k+1)^2, the
perimeter is  4n+2*2*(2k+3) = 4n+8k+12.  This is 4*A121150(n).

-----Original Message-----
From: jvospost3 at gmail.com

Following up on Frank TAW's comment about holes in A057730, we may
define:

...
Frank TAW's hole question is simplified here, in that we exclude the
smallest holey polyomino, as that heptomino does not have a
self-avoiding perimeter. But we do include the holey octomino with a
monomino square. Note that it's perimeter is a 3x3 square with a
smaller 1x1 inside, which is the same length but distinct from 3x3 +
1x1 (the square nonomino with disjoint monomino).
...
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