A276449 -id:A276449 - cp's OEIS frontend

A276454 a(n) = A276452(n) + A276451(n) + A276449(n).

1, 2, 22, 464, 13302, 487152, 21475652, 1106550392, 65221981530, 4327577893800, 319187492622012, 25904823495240144, 2294089575287710984, 220132629099295901408, 22751391952803426496488, 2519687900505935894639088, 297684761086123702744203918

Offset: 1

Jason Y.S. Chiu, Hong-Chang Wang, 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 all orbits under C_4 of 2-colored n X n square grids with n squares of one color.
See A054772(n, k) for the table of these total C_4 orbit numbers for 2-colored grids with any number k from {0,1,...,n^2} of squares of one color. - Wolfdieter Lang, Oct 02 2016

			For n = 4 there are A276449(4) = 4 1-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  .
A276451(4) = 12 2-orbits: one of them is
   + 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 representative.
A276452(4) = 448 4-orbits: one of them is represented by
   + + + +
   o o o o
   o o o o
   o o o o .
The complete orbit structure for n=4 is 1^4 2^12 4^448, see A276449(4) = 4, A276451(4) = 12, A276452(4) = 448.
a(4) = 448 + 12 + 4 = 464.
A014062(4) = 448*4 + 12*2 + 4*1 = 1820.

Hong-Chang Wang, Table of n, a(n) for n = 1..70

Cf. A014062, A054772, A276451, A276452, 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 + (Function[j, Binomial[2 j (j + Boole@ OddQ@ n), j]]@ Floor[n/2] - f@ n)/2 + f@ n, {n, 17}] (* Michael De Vlieger, Sep 12 2016 *)

from math import comb as binomial
for j in range(1, 20):
    t = binomial(j * j, j)
    i = j // 2
    if j % 2 == 0:
        d = binomial(2 * i * i, i)
    else:
        d = binomial(2 * i * (i + 1), i)
    a = (t - d) // 4
    if j % 4 == 0:
        c = binomial((j * j // 4), (j // 4))
    elif j % 4 == 1:
        c = binomial(((j - 1) // 2) * ((j - 1) // 2 + 1), ((j - 1) // 4))
    else:
        c = 0
    b = (d - c) // 2
    print(str(j) + " " + str(a + b + c))

a(n) = A276452(n) + A276451(n) + A276449(n) for n = 1, 2, 3, ...,
A014062(n) = A276452(n)*4 + A276451(n)*2 + A276449(n).
a(n) = A054772(n, 2), n >= 1. - Wolfdieter Lang, Oct 02 2016

Edited by 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

A276454 a(n) = A276452(n) + A276451(n) + A276449(n).

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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.

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Author

Keywords

Comments

Examples

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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