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The P-pentomino is rep-n² for all n. The diagram below shows solutions for n = 2, 3, 5 and a method of extending the rep-5² to rep-7², rep-9², rep11² and rep13² which can be extended to give solutions for all odd n. Since any even n is composite this shows that the P is rep-n² for all n.
The L-pentomino forms a 2x5 rectangle and so can also form a 10x10 square and is, therefore, rep-10². It it is also rep-n² for all n>4. Below are example for n = 4, 5, 6, 7, 8.
The diagram below shows how a rep-n² can be extended to a rep-(n+5)².
The Y-pentomino forms 10x10 and 15x15 squares and so is rep-(10k)² and rep-(15k)². Mike Reid reports that it is also rep-n² for n = 9, 11 and 16. These are shown below - the rep-16² and the rep-19² use a number of Y-pentomino rectangles - details of these can be found in Michael Reid's Rectifiable polyomino page or Torsten Sillke's Y pentomino page.
Below are general constructions for rep-(5n)² and rep-(5n+1)². (If m>2 then a 5m x k rectangle exists for all k > 19 and so all the rectangles exist for n > 4- see possible rectangles) This shows that the Y-pentomino is rep-n² for all n = 0, 1 (mod 5).
The next image shows how to create a rep-54
from three rep-9
The constructions below, made from I- and Y-pentominoes, can be expanded by a factor of 9 to give solutions to the rep-63
The next diagram shows how a rep-n
Mike Reid has also found the following replications. Each has a part made up of Y-pentominoes and the figures are then completed with various rectangles which can be made with the Y-pentominoes.
The above shows that the Y is rep-n
Patricvk Hamlyn has now found the rep-12
The above are the only pentominoes which are reptiles but we could, as with rectifiable polyominoes, consider pairs of pentominoes. The problem here would be to create n-fold copies of the pair of pentominoes. Three examples are given below - V and Z (rep-5
