A229637 T(n,k) = number of defective 3-colorings of an n X k 0..2 array connected horizontally, diagonally and antidiagonally with exactly two mistakes, and colors introduced in row-major 0..2 order.
0, 0, 0, 1, 6, 0, 3, 40, 39, 0, 12, 122, 244, 202, 0, 40, 488, 1109, 1496, 925, 0, 120, 1608, 6031, 10227, 8800, 3924, 0, 336, 5392, 28448, 77620, 89331, 50084, 15795, 0, 896, 17368, 136778, 535671, 960325, 747299, 277996, 61182, 0, 2304, 55232, 633328
Offset: 1
Examples
Some solutions for n=3, k=4: 0 1 0 2 0 1 0 1 0 1 0 2 0 1 0 0 0 1 1 2 2 1 0 2 2 1 0 1 2 2 0 1 0 2 1 2 0 1 0 2 2 1 2 0 1 2 0 1 1 1 0 1 0 2 1 0 0 1 0 1
Links
- R. H. Hardin, Table of n, a(n) for n = 1..287
Formula
Empirical for column k:
k=1: a(n) = a(n-1)
k=2: a(n) = 9*a(n-1) - 27*a(n-2) + 27*a(n-3) for n > 5
k=3: a(n) = 15*a(n-1) - 81*a(n-2) + 185*a(n-3) - 162*a(n-4) + 60*a(n-5) - 8*a(n-6) for n > 7.
k=4: [order 6] for n > 9.
k=5: [order 18] for n > 20.
k=6: [order 27] for n > 30.
k=7: [order 57] for n > 60.
Empirical for row n:
n=1: a(n) = 6*a(n-1) - 12*a(n-2) + 8*a(n-3) for n > 6.
n=2: a(n) = 6*a(n-1) - 6*a(n-2) - 16*a(n-3) + 12*a(n-4) + 24*a(n-5) + 8*a(n-6).
n=3: [order 9] for n > 12.
n=4: [order 18] for n > 21.
n=5: [order 30] for n > 33.
n=6: [order 69] for n > 72.
Comments