Jason Y.S. Chiu - cp's OEIS frontend

User: Jason Y.S. Chiu

Jason Y.S. Chiu has authored 6 sequences.

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.

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

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

Original entry on oeis.org

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

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, ...

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.

Original entry on oeis.org

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

A273916 The Bingo-4 problem: minimal number of stones that must be placed on an infinite square grid to produce n groups of exactly 4 stones each. Groups consist of adjacent stones in a horizontal, vertical or diagonal line.

Original entry on oeis.org

0, 4, 7, 9, 11, 12, 12, 14, 15, 16, 16, 18, 19, 20, 22, 24

Offset: 0

Jiangshan Sun, Jason Y.S. Chiu, Hong-Chang Wang, Jun 03 2016

You are permitted to put 5 or more adjacent stones in a line, but cannot count them as a group.
Each pair of stones has at most one group that counts going through them. - David A. Corneth, Aug 01 2016
a(n) >= n and a(n+m) <= a(n) + a(m), e.g., a(16) <= a(10) + a(6) = 28. Placing stones in a 4 X k rectangular array shows that a(3k) <= 4(k+2). Fekete's subadditive lemma shows that 1 <= lim_{n->oo} a(n)/n <= 4/3 exists. - Chai Wah Wu, Jul 31 2016
Limit_{n->oo} a(n)/n = 1. See arXiv link. - Chai Wah Wu, Aug 25 2016

			From _M. F. Hasler_, Jul 30 2016: (Start)
One can get n=3 groups using a(3) = 9 stones (O) as follows:
   O O O O     The 3 groups are:
   . O O .     (1) the first line,
   . O . .     (2) the second column,
   O O . .     (3) the antidiagonal.
See the link for more examples. (End)

Hong-Chang Wang, Illustration of initial terms.
Chai Wah Wu, Minimal number of points on a grid forming patterns of blocks, arXiv:1608.07247 [math.CO], 2016.

See also the 4-trees-in-a-row orchard problem, A006065.

Edited by N. J. A. Sloane, Jul 29 2016

A273889 a(n) = ((4n-3)!! + (4n-2)!!) / (4n-1).

Original entry on oeis.org

1, 9, 435, 52017, 11592315, 4152126825, 2182133628675, 1581940549814625, 1512952069890336075, 1845586177840605209625, 2796710279417971723681875, 5153962250373844341910100625, 11351091844757135191108560046875, 29444207228221006416048397134215625, 88848552445321896564985597922269171875

Offset: 1

Jiangshan Sun, Brian Cheung, Jason Y.S. Chiu, Hong-Chang Wang, Jun 02 2016

nonn

			a(1) = (1 + 2)/3 = 1;
a(2) = (1*3*5 + 2*4*6)/7 = 9;
a(3) = (1*3*5*7*9 + 2*4*6*8*10)/11 = 435.

Hong-Chang Wang, Table of n, a(n) for n = 1..100
Brian Cheung, Proof that (4n-1) divides (4n-3)!! + (4n-2)!!
Hong-Chang Wang, Bridan's blog (in Chinese)

Cf. A000165, A001147, A006882, A004767, A103639.

B[n_, k_] := (Product[k (i - 1) + 1, {i, 2 n - 1}] + Product[k (i - 1) + 2, {i, 2 n - 1}])/(2 k (n - 1) + 3); Table[B[n, 2], {n, 15}] (* Michael De Vlieger, Jun 10 2016 *)
Table[((4n-3)!!+(4n-2)!!)/(4n-1),{n,20}] (* Harvey P. Dale, Mar 08 2018 *)

for n in range(1,101):
    if n == 1:
        a = 1
        b = 2
    else:
        a = a*(4*n-5)*(4*n-3)
        b = b*(4*n-4)*(4*n-2)
    c = (a+b)/(4*n-1)
    print(str(n)+" "+str(c))

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User: Jason Y.S. Chiu

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

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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.

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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.

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A273916 The Bingo-4 problem: minimal number of stones that must be placed on an infinite square grid to produce n groups of exactly 4 stones each. Groups consist of adjacent stones in a horizontal, vertical or diagonal line.

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A273889 a(n) = ((4n-3)!! + (4n-2)!!) / (4n-1).

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