Lattice Animals with self-avoiding perimeters

[franktaw at netscape.net]

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).

Franklin T. Adams-Watters


-----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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