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T(n,k) = number of permutations of 0..floor((n*k-2)/2) on odd squares of an n X k array such that each row and column of odd squares is increasing.
5

%I #21 Mar 28 2019 10:45:50

%S 1,1,1,1,2,1,1,3,3,1,1,6,6,6,1,1,10,30,30,10,1,1,20,70,280,70,20,1,1,

%T 35,420,2100,2100,420,35,1,1,70,1050,23100,23100,23100,1050,70,1,1,

%U 126,6930,210210,1051050,1051050,210210,6930,126,1,1,252,18018,2522520,14294280

%N T(n,k) = number of permutations of 0..floor((n*k-2)/2) on odd squares of an n X k array such that each row and column of odd squares is increasing.

%C Table starts

%C .1...1.....1........1...........1..............1.................1

%C .1...2.....3........6..........10.............20................35

%C .1...3.....6.......30..........70............420..............1050

%C .1...6....30......280........2100..........23100............210210

%C .1..10....70.....2100.......23100........1051050..........14294280

%C .1..20...420....23100.....1051050.......85765680........5703417720

%C .1..35..1050...210210....14294280.....5703417720......577185873264

%C .1..70..6930..2522520...814773960...577185873264...337653735859440

%C .1.126.18018.25729704.12547518984.48236247979920.43364386933948080

%C Even columns match A215292.

%C The first column is number of symmetric standard Young tableaux of shape (n), the second column is number of symmetric standard Young tableaux of shape (n,n) and the third column is number of symmetric standard Young tableaux of shape (n,n,n). - _Ran Pan_, May 21 2015

%H R. H. Hardin, <a href="/A215297/b215297.txt">Table of n, a(n) for n = 1..1000</a>

%H Hodge, Jonathan K.; Krines, Mark; Lahr, Jennifer, <a href="https://doi.org/10.1007/s11083-009-9112-1">Preseparable extensions of multidimensional preferences</a>, Order 26, No. 2, 125-147 (2009), Table 1.

%H Ran Pan, <a href="http://www.math.ucsd.edu/~projectp/warmups/eP.html">Exercise P</a>, <a href="http://www.math.ucsd.edu/~projectp/problems/p4.html">Problem 4</a>, Project P.

%F f1=floor(k/2), f2=floor((k+1)/2), f3=floor((n+1)/2), f4=floor(n/2);

%F T(n,k) = A060854(f1,f3)*A060854(f2,f4)*binomial(f1*f3+f2*f4,f1*f3).

%e Some solutions for n=5, k=4:

%e ..x..0..x..4....x..0..x..1....x..1..x..3....x..0..x..6....x..0..x..1

%e ..1..x..2..x....4..x..7..x....0..x..8..x....3..x..5..x....3..x..7..x

%e ..x..3..x..8....x..2..x..3....x..2..x..5....x..1..x..7....x..2..x..5

%e ..6..x..7..x....5..x..9..x....4..x..9..x....4..x..9..x....6..x..8..x

%e ..x..5..x..9....x..6..x..8....x..6..x..7....x..2..x..8....x..4..x..9

%Y Column 2 is A001405. Column 4 is A215288. Column 6 is A215290.

%K nonn,tabl

%O 1,5

%A _R. H. Hardin_, Aug 07 2012