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A276449
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Number of 1-orbits of the cyclic group C_4 for a bi-colored square n X n grid with n squares of one color.
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4
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1, 0, 0, 4, 6, 0, 0, 120, 190, 0, 0, 7140, 11480, 0, 0, 635376, 1028790, 0, 0, 75287520, 122391522, 0, 0, 11143364232, 18161699556, 0, 0, 1978369382080, 3230129794320, 0, 0, 409663695276000, 669741609663270, 0, 0, 96930293990660064, 158625578809472060
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OFFSET
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1,4
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COMMENTS
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The old name was: Number of ways to choose n points from an n X n grid so that they have 90-degree rotational symmetry.
Consider a square n X n grid with n^2 squares. Each of the n^2 squares comes in two colors.
(E.g., an n X n chessboard with only two black fields, or a binary n X n matrix).
There are N(n) = binomial(n^2,n) = A014062(n) such 2-color grids. We are interested in configurations where n squares are colored in one way, say black, and the remaining ones stay white. Only colored grids modulo rotation around some axis perpendicular to the board through its center are of interest. These rotations represent the cyclic group C_4. Under C_4 operations R(90)^k, k=1..4, there will only be orbits of order 1 (colored grids invariant under R(90)^1, hence any rotation) order 2 (two different grids each not invariant under R(90)^1 but R(90)^2 operation, transforming into each other) and order 4 (four different grids each not invariant under R(90)^k for k=1,2,3, but under R(4)^4, transforming into each other). The orbit structure is denoted by 1^(e(n,1)) 2^(e(n,2)) 4^(e(n,4)) with e(n, 2^j) nonnegative integers for j=0,1,2. One has Sum_{j=0,1,2} 2^j*e(n,2^j) = N(n), and Sum_{j=0,1,2} e(n,2^j) which is the total number of orbits, given in A276454(n).
For example, one of the four 1-orbits of 4 X 4 board. (o) white, (+) black:
+ o o +
o o o o
o o o o
+ o o + ,
an example of a 2-orbit,
+ o + o o o o +
o o o o + o o o
o o o o o o o +
o + o + + o o o ,
an example of a 4-orbit,
+ + + + o o o + o o o o + o o o
o o o o o o o + o o o o + o o o
o o o o o o o + o o o o + o o o
o o o o o o o + + + + + + o o o .
The present sequence a(n) gives the number of 1-orbits of such 2-colored boards with n squares of one color under C_4.
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LINKS
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FORMULA
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a(n) = binomial((2*i)^2,i), for n = 4*i,
a(n) = binomial((2*i)*(2*i+1),i), for n = 4*i+1,
a(n) = 0, for others.
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EXAMPLE
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a(4) = 4, the arrangements are as follows:
+ o o + o + o o o o + o o o o o
o o o o o o o + + o o o o + + o
o o o o + o o o o o o + o + + o
+ o o + o o + o o + o o o o o o
a(5) = 6, the arrangements are as follows:
+ o o o + o + o o o o o + o o
o o o o o o o o o + o o o o o
o o + o o o o + o o + o + o +
o o o o o + o o o o o o o o o
+ o o o + o o o + o o o + o o
and
o o o + o o o o o o o o o o o
+ o o o o o + o + o o O + o o
O o + o o o o + o o o + + + o
o o o o + o + O + O o o + o o
o + o o o o o o o o o o o o o
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MAPLE
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seq(op([binomial(2*i*(2*i+1), i), 0, 0, binomial(4*(i+1)^2, i+1)]), i=0..30); # Robert Israel, Sep 05 2016
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MATHEMATICA
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Table[If[MemberQ[{2, 3}, #], 0, Function[i, Binomial[(2 i) (2 i + #), i]]@ Floor[n/4]] &@ Mod[n, 4], {n, 37}] (* Michael De Vlieger, Sep 07 2016 *)
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PROG
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(Python)
import math
def nCr(n, r):
f = math.factorial
return f(n) / f(r) / f(n-r)
# main program
for j in range(101):
if j%4 == 0:
a = nCr((j*j/4), (j/4))
elif j%4 == 1:
a = nCr(((j-1)/2)*((j-1)/2+1), ((j-1)/4))
else:
a = 0
print(str(j)+" "+str(a))
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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Edited: New name. Old name as a comment. Text substantially changed. Wolfdieter Lang, Oct 02 2016
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STATUS
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approved
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