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%I #7 Dec 11 2015 11:52:11
%S 256,1140,1140,6028,5228,6028,31756,22624,22624,31756,171564,111552,
%T 88596,111552,171564,973388,588840,520800,520800,588840,973388,
%U 5538732,3048096,2408480,5048352,2408480,3048096,5538732,31668844,16072760
%N T(n,k) = number of (n+1) X (k+1) 0..3 arrays with every 2 X 3 or 3 X 2 subblock having exactly two counterclockwise and two clockwise edge increases.
%C Table starts
%C ......256.....1140......6028......31756.......171564........973388
%C .....1140.....5228.....22624.....111552.......588840.......3048096
%C .....6028....22624.....88596.....520800......2408480......10792800
%C ....31756...111552....520800....5048352.....21702240.....157206240
%C ...171564...588840...2408480...21702240....190349280....1990281312
%C ...973388..3048096..10792800..157206240...1990281312...24408220320
%C ..5538732.16072760..59892000.1542266208..18371031840..373319837280
%C .31668844.84203240.276975200.6705992160.163028308320.4839970998240
%H R. H. Hardin, <a href="/A206118/b206118.txt">Table of n, a(n) for n = 1..364</a>
%F Empirical for row k:
%F k=1: a(n) = 5*a(n-1) + 5*a(n-2) + 29*a(n-3) - 142*a(n-4) - 248*a(n-5) + 88*a(n-6) + 144*a(n-7).
%F k=2: a(n) = 2*a(n-1) + 19*a(n-2) + 6*a(n-3) - 90*a(n-4) + 19*a(n-5) + 45*a(n-6) for n>8.
%F k=3: a(n) = 115*a(n-3) for n>6.
%F k=4: a(n) = 309*a(n-3) for n>7.
%F k=5: a(n) = 859*a(n-3) for n>8.
%F k=6: a(n) = 2445*a(n-3) for n>9.
%F k=7: a(n) = 7075*a(n-3) for n>10.
%F k=8: a(n) = 20709*a(n-3) for n>11.
%F k=9: a(n) = 61099*a(n-3) for n>12.
%F k=10: a(n) = 181245*a(n-3) for n>13.
%F Apparently, a(n) = A183624(k)*a(n-3) for k>2 and n>k+3
%e Some solutions for n=4, k=3:
%e ..3..3..0..0....2..0..0..1....1..2..1..0....2..2..3..3....3..2..0..0
%e ..3..0..1..1....0..2..0..0....2..1..1..0....0..0..1..3....2..0..0..2
%e ..0..1..1..0....0..0..1..0....1..1..3..1....2..0..0..1....0..0..3..1
%e ..1..1..2..1....3..0..0..2....2..2..1..1....3..2..0..0....3..3..1..1
%e ..0..0..1..1....0..1..0..2....0..2..1..2....3..3..2..0....0..3..1..2
%K nonn,tabl
%O 1,1
%A _R. H. Hardin_, Feb 03 2012
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