1 or 3 Rule

By inspection:
Q_N = 2Q_(N-1) + 8r_(N-2) + correction

By recognizing that by population count that r_N = Q_N (The alternating boundary acts as a row of zeros since the rule is 1 or 3, thus the growth of the r_N's is identical to q_N's on the NE half, and shifted by (-1,-1) on the SW half.) and finding correction through Excel we get
Q_N = 2Q_(N-1) + 8Q_(N-2) - 6

Upon examination the cells which are repeated are at locations:
(2^(N-1), 2^(N-1)) twice,
(2^(N-2), 2^(N-1)+2^(N-2)),
(2^(N-1)+2^(N-2), 2^(N-2)),
(2^(N-2)+1, 2^(N-1)+2^(N-2)-1),
(2^(N-1)+2^(N-2)-1, 2^(N-2)+1),

By subtracting by Q_(N-1) to remove the constant
Q_N = 3Q_(N-1) + 6Q_(N-2) - 8Q_(N-3)

which gives us the characteristic polynomial
x^3 - 3x^2 - 6x + 8 = 0

with roots 4, 1, -2. Thus:
Q_N = c_1*4^n + c_2*(-2)^n + c_3

Fitting this to data for Q_3, Q_4, Q_5, we get
Q_N = (4^n+2)/3.

Since the whole box B_N = 4Q_N - 3 (the origin is counted 3 times too many), B_N = 4/3*4^n - 1/3. Since this is the rotated version, it follows that the density of the 13 rule is 2/3.