mersenneforum.org July 2016
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 2016-07-02, 13:30 #1 Xyzzy     "Mike" Aug 2002 32·907 Posts July 2016
 2016-07-04, 01:02 #2 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 2·3·1,571 Posts A wonderful problem, indeed!
 2016-07-04, 18:25 #3 R. Gerbicz     "Robert Gerbicz" Oct 2005 Hungary 2×733 Posts There is an update: Code: "Update (3/7): Your pentominos should be able to tile all the 4^N*4^N boards with any possible missing square. You can use free pentominos (rotation and reflections are allowed). Solving with less than three pentominos types will earn you a '*'."
 2016-07-04, 19:12 #4 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 2×3×1,571 Posts I guess that there is a trade-off there: if we consider solution using (non-free) one-sided pentominos, then the solution(s?) exists with three, but two of them [I]can[/I] be reflections of one free pentomino. Makes for an interesting follow-up to enumerate all 3-{one-sided pentomino} meta-solutions.
 2016-08-06, 22:51 #5 R. Gerbicz     "Robert Gerbicz" Oct 2005 Hungary 2×733 Posts My sent solution also used P and L (but named P as B), [I would say my solution has a nicer tiling of the larger version of P and L] : I'll use only two types of pentominos, (see the attached pento.jpg file [[used Windows paint]] for the pictures and further tilings), so using B and L pentominos. The case N=1 is trivial, using symmetry there is only 3 cases, see the tilings (S1,S2,S3) for these cases using only B and L.. For larger N values we use induction in the following way: suppose that the tiling for smaller K

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