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.

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

Offset: 1

Author

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

Keywords

• nonn
• easy

Comments

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.

Examples

			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
		

Links

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

Crossrefs

Cf. A014062, A276449, A276451, A276454.

Programs

Mathematica

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 *)

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):
    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))

Formula

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