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Number of 4 X n matrices containing a permutation of 1..4*n in increasing order rowwise, columnwise, diagonally and (downwards) antidiagonally.
2

%I #39 Jan 01 2024 08:01:56

%S 1,1,8,169,6392,352184,25097600,2152061145,212012802584,

%T 23263015359672,2781709560836960,356806123331844056,

%U 48516442013911012288,6930091952294051922080,1032505514388962439665280,159544871422153344631037625,25451354639006231998529405016

%N Number of 4 X n matrices containing a permutation of 1..4*n in increasing order rowwise, columnwise, diagonally and (downwards) antidiagonally.

%H Christoph Koutschan, <a href="/A181198/b181198.txt">Table of n, a(n) for n = 1..100</a> (terms 1..27 from Alois P. Heinz)

%H Joerg Arndt, <a href="/A181198/a181198.txt">The a(4)=199 shifted standard Young tableaux of shape [4,4,4,4]</a>

%H Manuel Kauers and Christoph Koutschan, <a href="https://arxiv.org/abs/2303.02793">Some D-finite and some Possibly D-finite Sequences in the OEIS</a>, arXiv:2303.02793 [cs.SC], 2023, pp. 38-40.

%F Conjectured recurrence of order 2 and degree 9: 3*(n + 1)*(2*n + 3)*(3*n + 4)*(3*n + 5)*(7*n^2 - 1)*(n + 2)^3*a(n + 2) - 8*(n + 1)*(2*n + 1)*(4*n + 3)*(4*n + 5)*(364*n^5 + 84*n^4 - 1025*n^3 - 534*n^2 + 157*n + 54)*a(n + 1) - 64*(2*n - 1)^2*(2*n + 1)*(4*n - 1)*(4*n + 1)*(4*n + 3)*(4*n + 5)*(7*n^2 + 14*n + 6)*a(n) = 0. - _Christoph Koutschan_, Feb 26 2023

%F Conjectured formula, solution to the above recurrence, for n > 1: a(n) = (-64)^n * (n-1) * (-1/2)_{2*n} * (1/2)_{n} / (4*(3*n)!) * (-1 + 3*Sum_{k=2..n-1} (-4)^k * (7*k^2-1) / ((k-1) * k * (k+1)^2 * (2*k-1)^2 * (2*k+1)^3) * binomial(3*k,2*k) * binomial(k+1/2,k)), where (a)_{n} is the Pochhammer symbol.

%e Some solutions for 4 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 8 5 6 7 8 5 6 7 8 5 6 7 8

%e 9 10 11 12 9 10 11 13 9 10 11 14 9 10 12 13 9 10 12 14

%e 13 14 15 16 12 14 15 16 12 13 15 16 11 14 15 16 11 13 15 16

%t Table[

%t NextPartitions[n1_, n2_, n3_, n4_] :=

%t If[n1 < n, f[n1 + 1, n2, n3, n4], 0] +

%t If[n2 < n1 - 1 || n2 === n - 1, f[n1, n2 + 1, n3, n4], 0] +

%t If[n3 < n2 - 1 || n3 === n - 1 === n2 - 1, f[n1, n2, n3 + 1, n4], 0] +

%t If[n4 < n3 - 1, f[n1, n2, n3, n4 + 1], 0];

%t pp = f[1, 0, 0, 0];

%t Do[pp = Expand[pp /. f[ns__] :> NextPartitions[ns]], {4 n - 2}];

%t pp /. f[n, n, n, n - 1] -> 1,

%t {n, 20}] (* _Christoph Koutschan_, Feb 26 2023 *)

%Y Row 4 of A181196.

%K nonn

%O 1,3

%A _R. H. Hardin_, Oct 10 2010

%E a(12)-a(27) from _Alois P. Heinz_, Jul 24 2012

%E a(28)-a(100) from _Christoph Koutschan_, Feb 26 2023