%I #6 Mar 31 2012 12:36:42
%S 5,10,10,10,10,10,5,20,20,5,1,79,92,79,1,5,21,537,537,21,5,10,226,140,
%T 1225,140,226,10,10,157,3157,1095,1095,3157,157,10,5,227,3604,15023,
%U 4919,15023,3604,227,5,1,678,6692,95845,27000,27000,95845,6692,678,1,5,120
%N T(n,k)=Number of nXk 0..4 arrays with every row and column nondecreasing rightwards and downwards, and the number of instances of each value within one of each other
%C Table starts
%C ..5..10....10......5.......1........5.........10..........10...........5
%C .10..10....20.....79......21......226........157.........227.........678
%C .10..20....92....537.....140.....3157.......3604........6692.......26168
%C ..5..79...537...1225....1095....15023......95845......268375......325061
%C ..1..21...140...1095....4919....27000......97341......383172.....1188521
%C ..5.226..3157..15023...27000...752271....8434277....39361323....80878071
%C .10.157..3604..95845...97341..8434277...29555346...171588826..1801771085
%C .10.227..6692.268375..383172.39361323..171588826..1278883650.16557969007
%C ..5.678.26168.325061.1188521.80878071.1801771085.16557969007.65182166083
%C ..1.120..5408.168262.3704820.64001290..853004113..9528745293.89750034774
%H R. H. Hardin, <a href="/A200990/b200990.txt">Table of n, a(n) for n = 1..199</a>
%F T(n,1) = binomial(5,n modulo 5). For a 0..z array, T(n,1) = binomial(z+1, n modulo (z+1)).
%e Some solutions for n=5 k=3
%e ..0..1..2....0..0..2....0..1..2....0..0..1....0..0..0....0..1..1....0..0..0
%e ..0..1..3....0..1..2....0..1..2....0..1..3....1..1..2....0..1..2....1..1..1
%e ..0..1..4....1..3..4....0..1..2....1..2..3....1..2..3....0..3..3....2..3..4
%e ..2..2..4....1..3..4....3..3..4....2..2..3....2..3..3....2..3..4....2..3..4
%e ..3..3..4....2..3..4....3..4..4....4..4..4....4..4..4....2..4..4....2..3..4
%Y Column 1 for a 0..z array is binomial(z+1,n modulo (z+1))
%K nonn,tabl
%O 1,1
%A _R. H. Hardin_ Nov 25 2011