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T(n,k) = number of n X k matrices containing a permutation of 1..n*k in increasing order rowwise, columnwise, diagonally and (downwards) antidiagonally.
10

%I #32 Jul 23 2021 06:55:31

%S 1,1,1,1,1,1,1,2,1,1,1,5,4,1,1,1,14,29,8,1,1,1,42,290,169,16,1,1,1,

%T 132,3532,6392,985,32,1,1,1,429,49100,352184,141696,5741,64,1,1,1,

%U 1430,750325,25097600,36372976,3142704,33461,128,1,1,1,4862,12310294

%N T(n,k) = number of n X k matrices containing a permutation of 1..n*k in increasing order rowwise, columnwise, diagonally and (downwards) antidiagonally.

%C Table starts:

%C .1.1...1......1..........1..............1...................1

%C .1.1...2......5.........14.............42.................132

%C .1.1...4.....29........290...........3532...............49100

%C .1.1...8....169.......6392.........352184............25097600

%C .1.1..16....985.....141696.......36372976.........14083834704

%C .1.1..32...5741....3142704.....3777546912.......8092149471168

%C .1.1..64..33461...69705920...392658046912....4673805856338368

%C .1.1.128.195025.1546100352.40820345224064.2702482348019033600

%H R. H. Hardin and Alois P. Heinz, <a href="/A181196/b181196.txt">Antidiagonals n = 1..30, flattened</a>

%H Brian T. Chan, <a href="https://arxiv.org/abs/1803.05594">Periodic P-Partitions</a>, arXiv:1803.05594 [math.CO], 2018-2020.

%H Ping Sun, <a href="https://doi.org/10.37236/6466">Enumeration of standard Young tableaux of shifted strips with constant width</a>, El. J. Comb., 24 (2017), #P2.41; arXiv:<a href="https://arxiv.org/abs/1506.07256">1506.07256</a> [math.CO], 2015.

%H Antonio Vera López, Luis Martínez, Antonio Vera Pérez, Beatriz Vera Pérez and Olga Basova, <a href="https://doi.org/10.1016/j.laa.2017.05.027">Combinatorics related to Higman's conjecture I: Parallelogramic digraphs and dispositions</a>, Linear Algebra and its Applications, Volume 530, 1 October 2017, p. 414-444. See Table 1.

%F Empirical column 1: a(n) = a(n-1).

%F Empirical column 2: a(n) = a(n-1).

%F Empirical column 3: a(n) = 2*a(n-1).

%F Empirical column 4: a(n) = 6*a(n-1)-a(n-2).

%F Empirical column 5: a(n) = 24*a(n-1)-40*a(n-2)-8*a(n-3).

%F Empirical column 6: a(n) = 120*a(n-1)-1672*a(n-2)+544*a(n-3)-6672*a(n-4) +256*a(n-5).

%F Empirical column 7: a(n) = 720*a(n-1) -84448*a(n-2) +1503360*a(n-3) -17912224*a(n-4) -318223104*a(n-5) +564996096*a(n-6) +270471168*a(n-7) -11373824*a(n-8) +65536*a(n-9).

%e All solutions for 3 X 4:

%e ..1..2..3..4....1..2..3..4....1..2..3..4....1..2..3..4....1..2..3..4

%e ..5..6..7..8....5..6..7..9....5..6..7.10....5..6..8..9....5..6..8.10

%e ..9.10.11.12....8.10.11.12....8..9.11.12....7.10.11.12....7..9.11.12

%e ...

%e ..1..2..3..6....1..2..3..6....1..2..3..6....1..2..3..6....1..2..3..6

%e ..4..5..7..8....4..5..7..9....4..5..7.10....4..5..8..9....4..5..8.10

%e ..9.10.11.12....8.10.11.12....8..9.11.12....7.10.11.12....7..9.11.12

%e ...

%e ..1..2..4..6....1..2..4..6....1..2..4..6....1..2..4..6....1..2..4..6

%e ..3..5..7..8....3..5..7..9....3..5..7.10....3..5..8..9....3..5..8.10

%e ..9.10.11.12....8.10.11.12....8..9.11.12....7.10.11.12....7..9.11.12

%e ...

%e ..1..2..3..5....1..2..3..5....1..2..3..5....1..2..3..5....1..2..3..5

%e ..4..6..7..8....4..6..7..9....4..6..7.10....4..6..8..9....4..6..8.10

%e ..9.10.11.12....8.10.11.12....8..9.11.12....7.10.11.12....7..9.11.12

%e ...

%e ..1..2..4..5....1..2..4..5....1..2..4..5....1..2..4..5....1..2..4..5

%e ..3..6..7..8....3..6..7..9....3..6..7.10....3..6..8..9....3..6..8.10

%e ..9.10.11.12....8.10.11.12....8..9.11.12....7.10.11.12....7..9.11.12

%e ...

%e ..1..2..3..7....1..2..3..7....1..2..4..7....1..2..4..7

%e ..4..5..8..9....4..5..8.10....3..5..8..9....3..5..8.10

%e ..6.10.11.12....6..9.11.12....6.10.11.12....6..9.11.12

%p b:= proc(l) option remember; local n; n:= nops(l);

%p `if`({l[]}={0}, 1, add(`if`((i=1 or l[i-1]<=l[i]) and l[i]>

%p `if`(i=n, 0, l[i+1]), b(subsop(i=l[i]-1, l)), 0), i=1..n))

%p end:

%p T:= (n,k)-> b([n$k]):

%p seq(seq(T(n, 1+d-n), n=1..d), d=1..12); # _Alois P. Heinz_, Jul 24 2012

%t b[l_List] := b[l] = With[{n = Length[l]}, If[Union[l] == {0}, 1, Sum[If[(i == 1 || l[[i-1]] <= l[[i]]) && l[[i]] > If[i == n, 0, l[[i+1]]], b[ReplacePart[l, i -> l[[i]]-1]], 0], {i, 1, n}]]]; T[n_, k_] := b[Array[n&, k]]; Table[Table[T[n, 1+d-n], {n, 1, d}], {d, 1, 12}] // Flatten (* _Jean-François Alcover_, Mar 06 2015, after _Alois P. Heinz_ *)

%Y Rows n=1-5 give: A000012, A000108, A181197, A181198, A181199.

%Y Columns 1+2, 3-8 give: A000012, A011782, A001653, A181192, A181193, A181194, A181195.

%Y A227578 is a similar but different array.

%K nonn,tabl

%O 1,8

%A _R. H. Hardin_, Oct 10 2010