A201072 - OEIS

%I #5 Mar 31 2012 12:36:42

%S 7,21,21,35,70,35,35,35,35,35,21,77,514,77,21,7,749,2611,2611,749,7,1,

%T 972,3937,22440,3937,972,1,7,127,50334,43308,43308,50334,127,7,21,

%U 3034,4448,1127514,2250982,1127514,4448,3034,21,35,7161,381982,175865

%N T(n,k)=Number of nXk 0..6 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 ..7...21......35........35.........21.............7.............1

%C .21...70......35........77........749...........972...........127

%C .35...35.....514......2611.......3937.........50334..........4448

%C .35...77....2611.....22440......43308.......1127514........175865

%C .21..749....3937.....43308....2250982......11512566.......3558888

%C ..7..972...50334...1127514...11512566......45244488......76714325

%C ..1..127....4448....175865....3558888......76714325....1001060834

%C ..7.3034..381982..20393189..473983329....4070916182...13053459189

%C .21.7161..206357..14766300.4102010820...93433967419..126679336122

%C .35.2170.1377351.158314010.3773723044.1001816580924.1146503112430

%H R. H. Hardin, <a href="/A201072/b201072.txt">Table of n, a(n) for n = 1..143</a>

%F T(n,1) = binomial(7,n modulo 7). For a 0..z array, T(n,1) = binomial(z+1, n modulo (z+1)).

%e Some solutions for n=3 k=7

%e ..0..0..1..1..3..4..5....0..0..1..1..3..4..4....0..0..1..1..2..4..4

%e ..0..2..2..3..4..5..6....0..2..2..3..3..5..6....0..1..3..3..4..5..5

%e ..1..2..3..4..5..6..6....1..2..4..5..5..6..6....2..2..3..5..6..6..6

%K nonn,tabl

%O 1,1

%A _R. H. Hardin_ Nov 26 2011