Respuesta :
Answer:
[tex]a_n = 2^{\frac{n^2-1}{4} + 1} + \frac{2^{n^2} - \, 2^{\frac{n^2-1}{4} + 1}}{4}[/tex]
For n = 3, there are 134 possibilities
Step-by-step explanation:
First, lets calculate the generating function.
For each square we have 2 possibilities: red and blue. The Possibilities between n² squares multiply one with each other, giving you a total of 2^n² possibilities to fill the chessboard with the colors blue or red.
However, rotations are to be considered, then we should divide the result by 4, because there are 4 ways to flip the chessboard (including not moving it), that means that each configuration is equivalent to three other ones, so we are counting each configuration 4 times, with the exception of configurations that doesnt change with rotations.
A chessboard that doesnt change with rotations should have, in each position different from the center, the same colors than the other three positions it could be rotated into. As a result, we can define a symmetric by rotations chessboard with only (n²-1)/4 + 1 squares (the quarter part of the total of squares excluding the center plus the center).
We cocnlude that the total of configurations of symmetrical boards is [tex] 2^{\frac{n^2-1}{4} + 1} [/tex]
Since we have to divide by 4 the rest of configurations (because we are counted 4 times each one considering rotations), then the total number of configutations is
[tex]a_n = 2^{\frac{n^2-1}{4} + 1} + \frac{2^{n^2} - \, 2^{\frac{n^2-1}{4} + 1}}{4}[/tex]
If n = 3, then the total amount of possibilities are
[tex]a_3 = 2^{\frac{3^2-1}{4} + 1} + \frac{2^{3^2} - \, 2^{\frac{3^2-1}{4} + 1}}{4} = 134[/tex]
