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User: Jiangshan Sun

Jiangshan Sun has authored 3 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

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: Jiangshan Sun

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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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.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Extensions

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

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

Python