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A181198
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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.
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2
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1, 1, 8, 169, 6392, 352184, 25097600, 2152061145, 212012802584, 23263015359672, 2781709560836960, 356806123331844056, 48516442013911012288, 6930091952294051922080, 1032505514388962439665280, 159544871422153344631037625, 25451354639006231998529405016
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OFFSET
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1,3
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LINKS
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FORMULA
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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
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.
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EXAMPLE
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Some solutions for 4 X 4:
1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4
5 6 7 8 5 6 7 8 5 6 7 8 5 6 7 8 5 6 7 8
9 10 11 12 9 10 11 13 9 10 11 14 9 10 12 13 9 10 12 14
13 14 15 16 12 14 15 16 12 13 15 16 11 14 15 16 11 13 15 16
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MATHEMATICA
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Table[
NextPartitions[n1_, n2_, n3_, n4_] :=
If[n1 < n, f[n1 + 1, n2, n3, n4], 0] +
If[n2 < n1 - 1 || n2 === n - 1, f[n1, n2 + 1, n3, n4], 0] +
If[n3 < n2 - 1 || n3 === n - 1 === n2 - 1, f[n1, n2, n3 + 1, n4], 0] +
If[n4 < n3 - 1, f[n1, n2, n3, n4 + 1], 0];
pp = f[1, 0, 0, 0];
Do[pp = Expand[pp /. f[ns__] :> NextPartitions[ns]], {4 n - 2}];
pp /. f[n, n, n, n - 1] -> 1,
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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