Proof that a(n) = 4*4^n + 1530*3^n + 37320*2^n + 223290.

Encode each row of the array as a 9-tuple (a, b, c, d, e, f, g, h, i) with entries in {0,1,...,3}. There are 4^9 = 262144 such 9-tuples. Define the 262144 X 262144 transfer matrix T by T[(a, b, c, d, e, f, g, h, i),(j, k, l, m, n, o, p, q, r)] = 1 if consecutive rows are compatible (every 2 X 2 subblock sums to 6), and 0 otherwise. Compatibility forces k = 6-a-b-j; l = 6-b-c-k; m = 6-c-d-l; n = 6-d-e-m; o = 6-e-f-n; p = 6-f-g-o; q = 6-g-h-p; r = 6-h-i-q; the entry is 1 iff k, l, m, n, o, p, q, r all lie in {0,1,...,3}. Then a(n) = 1^T * T^n * 1, where 1 denotes the all-ones vector of length 262144, since summing T^n over all start and end rows counts all valid (n+1) X 9 arrays. (The computations below use the row-statistics quotient — encoding each row by its first entry and the min/max of its even- and odd-indexed entries — which reduces T to a much smaller matrix with the same action on constant-lifted functions, making the exact integer arithmetic feasible.)

Let p(x) = (x-1)(x-2)(x-3)(x-4). An exact integer computation (applying each factor (T-k*I) in turn to the vector 1) shows that p(T)*1 = 0. Hence the sequence satisfies a linear recurrence whose characteristic polynomial divides p(x).

Since p has 4 distinct roots, the p-null space of T decomposes as a direct sum of eigenspaces for eigenvalues 1, 2, 3, 4. Applying the annihilator product for each eigenvalue k (the product of all factors (T-k'*I) for k' != k) to 1 yields a nonzero vector in the k-eigenspace, confirming all 4 eigenvalues contribute. Writing 1 = v_1 + ... + v_4 with T*v_k = k*v_k gives T^n * 1 = sum_k k^n * v_k, so a(n) = sum_k alpha_k * k^n where alpha_k = 1^T * v_k.

The coefficients alpha_k are computed by dividing the dot product 1^T * (annihilator_k * 1) by the scalar prod_{k' != k}(k - k'):

  k=1: alpha=223290,  k=2: alpha=37320,  k=3: alpha=1530,  k=4: alpha=4

This gives a(n) = 4*4^n + 1530*3^n + 37320*2^n + 223290, which matches all terms in the sequence. This also proves the existing empirical formulas: the recurrence follows because each term k^n satisfies (x-k), making p(x) the characteristic polynomial; the g.f. follows by summing alpha_k * k*x / (1-k*x) over k = 1..4.