A276454 -id:A276454 - cp's OEIS frontend

A276449 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.

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

Offset: 1

Jason Y.S. Chiu, Chiang, Tung-Ying, Hsiang-An Wang, Hong-Chang Wang, Sep 02 2016

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.

			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
reformatted - _Wolfdieter Lang_, Oct 02 2016

Hong-Chang Wang, Table of n, a(n) for n = 1..100

Cf. A014062, A276451, A276452, A276454.

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

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 *)

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))

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.

Edited: New name. Old name as a comment. Text substantially changed. Wolfdieter Lang, Oct 02 2016

A276451 Number of 2-orbits of the cyclic group C_4 for a bi-colored square n X n grid with n squares of one color.

Original entry on oeis.org

0, 1, 2, 12, 30, 408, 1012, 17920, 45600, 1059380, 2730756, 78115884, 203235032, 6917206576, 18113945256, 714851008512, 1881039165696, 84449819514060, 223049005408900, 11225502116862880, 29736777118603962, 1658138369930988088, 4403069737450280832

Offset: 1

Chiang, Tung-Ying, Jason Y.S. Chiu, Hong-Chang Wang, Jiangshan Sun, Sep 03 2016

For a definition and examples of this problem see the comment section of A276449.

The present sequence a(n) gives the number of 2-orbits of such 2-color boards with n squares of one color under C_4.

			n = 4: one of the two 2-orbits is (o white, + black)
+ 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 one can take the first one as a representative.
For n = 3 there are a(3) = 2 2-orbits, represented by
+ o o       o o o
o + o  and  + + +
o o +       o o o.
The orbit structure for n=3 is 1^0 2^2 4^20; see A276449(3) = 0, a(3) = 2, A276452(3) = 20.
For the 12 2-orbits for n=4, see the representatives given in the link.

Hong-Chang Wang, Table of n, a(n) for n = 1..100
Hong-Chang Wang, Example for n = 4

Cf. A014062, A276449, A276452, A276454.

Table[(Function[j, Binomial[2 j (j + Boole@ OddQ@ n), j]]@ Floor[n/2] - If[MemberQ[{2, 3}, #], 0, Function[i, Binomial[(2 i) (2 i + #), i]]@ Floor[n/4]] &@ Mod[n, 4])/2, {n, 23}] (* Michael De Vlieger, Sep 07 2016 *)

import math
def nCr(n,r):
    f = math.factorial
    return f(n) / f(r) / f(n-r)
# main program
for j in range(101):
    i = j/2
    if j%2==0:
        b = nCr(2*i*i,i)
    else:
        b = nCr(2*i*(i+1),i)
    if j%4==0:
        c = nCr((j*j/4),(j/4))
    elif j%4==1:
        c = nCr(((j-1)/2)*((j-1)/2+1),((j-1)/4))
    else:
        c = 0
    print(str(j)+" "+str((b-c)/2))

a(n) = (binomial(2*i*i,i) - A276449(n))/2, for n = 2*i.

a(n) = (binomial(2*i*(i+1),i) - A276449(n))/2, for n = 2*i+1.

Edited: Wolfdieter Lang, Oct 02 2016

A276452 Number of 4-orbits of the cyclic group C_4 for a bi-colored square n X n grid with n squares of one color.

Original entry on oeis.org

0, 1, 20, 448, 13266, 486744, 21474640, 1106532352, 65221935740, 4327576834420, 319187489891256, 25904823417117120, 2294089575084464472, 220132629092378694832, 22751391952785312551232, 2519687900505221042995200, 297684761086121821704009432, 37370623083548749203599933004

Offset: 1

Hong-Chang Wang, Jason Y.S. Chiu, Chiang, Tung-Ying, Sep 03 2016

For a definition and examples of this problem see the comment section of A276449. The present sequence a(n) gives the number of 4-orbits under C_4 of such 2-colored n X n grids with n squares of one color.

			a(2) = 1: the 4-orbit is
+ +   o +   o o   + o
o o   o +   + +   + o  ,
and one can take the first one as representative.
For n = 3 there are a(3) = 20 4-orbits, represented by
+ + +   + + 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
--------------------------------------
+ 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 .
--------------------------------------
The complete orbit structure for n=3 is 1^0 2^2 4^20, see A276449(3) = 0, A276451(3) = 2, a(3) = 20

Hong-Chang Wang, Table of n, a(n) for n = 1..69

Cf. A014062, A276449, A276451, A276454.

f[n_] := If[MemberQ[{2, 3}, #], 0, Function[i, Binomial[(2 i) (2 i + #), i]]@ Floor[n/4]] &@ Mod[n, 4]; g[n_] := (Function[j, Binomial[2 j (j + Boole@ OddQ@ n), j]]@ Floor[n/2] - f@ n)/2; Table[(Binomial[n^2, n] - 2 g@ n - f@ n)/4, {n, 18}] (* Michael De Vlieger, Sep 07 2016 *)

import math
def nCr(n,r):
    f = math.factorial
    return f(n) / f(r) / f(n-r)
# main program
for j in range(101):
    a = nCr(j*j,j)
    i = j/2
    if j%2==0:
        b = nCr(2*i*i,i)
    else:
        b = nCr(2*i*(i+1),i)
    print(str(j)+" "+str((a-b)/4))

a(n) = (A014062(n) - A276451(n)*2 - A276449(n))/4 for n = 1, 2, 3, ...

cp's OEIS Frontend

A276449 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.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Python

Formula

Extensions

A276451 Number of 2-orbits of the cyclic group C_4 for a bi-colored square n X n grid with n squares of one color.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Python

Formula

Extensions

A276452 Number of 4-orbits of the cyclic group C_4 for a bi-colored square n X n grid with n squares of one color.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Python

Formula