Lattice Animals with self-avoiding perimeters

franktaw at franktaw at
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).

Franklin T. Adams-Watters

-----Original Message-----
From: jvospost3 at

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

 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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