Polyomino question

[Jon Wild]

On Tue, 28 Feb 2006, Dr. Gordon Hamilton wrote:

> The best that can be done is 10. There are 11 different solutions. The 
> 9-solution below isn't right because it does not include the L 
> polyomino. I have been presenting this problem to elementary school 
> students in Calgary, Canada for the last year. You can see the problem 
> at: www.galileo.org/math/puzzles.html under the name "Babushka Squares".

Dear Gord - I hope they don't make you go back and visit all the primary 
schools to show the kids the solution with 9 squares!

I looked at the other problems you have for children at that site, and I 
think it's great if you can get kids to puzzle over these kinds of things 
- certainly more fun than anything I ever did at school in maths. The 
extension to that polyomino problem at your site contains another good 
sequence for the encyclopedia, if anyone would like to calculate it:

start with the set S0 of all (i.e. both) trominoes. Then find the set S1 
of all smallest n1-ominoes that contain any member of S0. Then find the 
set S2 of all smallest n2-ominoes that contain any member of S1. Iterate. 
This leads to two sequences: 3, n1, n2, n3...  and  2, #S1, #S2, #S3...

(From Gord's hint it seems the sequences start 3,4,5,7... and 2,2,4...)

I bet someone from the yahoo polyomino group can attack this with one of 
their solving problems and give us quite an extended answer in no time.

-Jon