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A216675
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Number of ways one can draw arrows between adjacent nodes of an n X n grid such that each node has one outgoing and one incoming arrow.
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2
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0, 4, 0, 1296, 0, 45265984, 0, 168709341081856, 0, 66865709036047973991424, 0, 2815414274858422422282241600000000, 0, 12589335654221209921194197564847684000000000000, 0, 5977481098898922857923760209743284068237948337696882106105856, 0
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OFFSET
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1,2
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COMMENTS
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"Adjacent" is meant in the sense of von Neumann neighborhoods (4 neighbors for "interior" nodes, 3 resp. 2 for nodes on the borders resp. in the corners).
Alternate definition: Number of permutations of an n X n array with each element moving exactly one step horizontally or vertically. (Suggested by R. H. Hardin.)
Also the permanent of the adjacency matrix of the n X n grid graph, which is the determinant of the modified adjacency matrix where vertical and horizontal edges have weights of 1 and i, respectively.
Also the square of the number of domino tilings of an n X n chessboard.
(End)
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LINKS
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FORMULA
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EXAMPLE
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For a 1 X 1 grid, there is no such possibility.
For a 2 X 2 grid, on can draw arrows between 2 pairs of nodes in horizontal or vertical sense, and the clockwise and counterclockwise cyclic "permutation" of the 4 nodes.
For a 3 X 3 grid, there is no possibility, neither for a 5 X 5 grid.
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MATHEMATICA
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Table[If[Mod[n, 2]==0, Det[MapIndexed[(#1 I^Mod[Total[#2], 2])&, Normal[AdjacencyMatrix[GridGraph[{n, n}]]], {2}]], 0], {n, 1, 20}] (* Adam P. Goucher, Aug 01 2013 *)
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PROG
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(Python)
from sympy.abc import x
from sympy import resultant, chebyshevu, I
def A216675(n): return 0 if n&1 else resultant(chebyshevu(n, x/2), chebyshevu(n, I*x/2)) # Chai Wah Wu, Nov 07 2023
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CROSSREFS
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See A216678 for the same problem with an additional constraint ("no 2-loops").
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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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