rep-n for n = 7k, 7k+1, 7k+6
The diagram below shows an extension from rep-(2k) to rep-(2k+7) and from rep-(2k+1) to rep-(2k+8). In fact with k=0 the first one gives another solution for rep-7.
rep-n for all n = 4 and all n >= 6
In these constructions we use the fact that 14xn rectangles exist for all n other than 1, 3 and 5.
The diagram above shows an extension from rep-n to rep-(n+14). This immediately gives solutions for rep-14 and rep-15 with n = 0, 1.
Below are diagrams for n = 4, 6, 8, 10, and 12.
The diagrams below shows rep-n for n = 11, 13, 17 and 21.
The final diagrams give solutions for n = 19 and 23.
rep-n for n = 11, 12 and all n >= 14
Since a 14x14 and 21x21 squares are possible with this heptomino it is rep-14 and rep-21. The diagram below shows how a rep-k can be extended to a rep-(k+14). (see possible rectangles).
n = 7k with k = 4, 6, 9,..+3r; 10, 15,...+5r; 10, 12, ... +2r n = 7k+1 with k = 12, 14, .. +2r; and probably others
These are based on the construction of rectangles as shown below (see possible rectangles). The construction on the right works for k = 10+2k (k>0). The lower construction gives solutions for n = 7k+1 with k>10 and even.
rep-n for n = 7k (k>1), 7k+1 (k>1) and probably others
The prime rectangles for this are 2x7, 11x49, 13x35 and 19x21. Thus a 14x14 square is possible and the 19x21 can be extended to form a 21x21 square. This shows that the pice is rep-n for n = 7k (k>0).
21xn rectangles exist for all even n and for odd n>=19. This leads to the extension constructions shown below.