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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.

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 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.

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.

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

Sequence is inspired by A273889, A274119 and A273983.

Since Product_{i=0..n}(i*k+a) - Product_{i=0..n}(-i*k-b) ≡ 0 mod (n*k+a+b), then define B(n,k,a,b) = (Product_{i=0..n}(i*k+a) - Product_{i=0..n}(-i*k-b))/(n*k+a+b), with n*k+a+b <> 0, n >= 0 and k,a,b are integers, such that B(1,n,2,1) = (Product_{i=0..1}(i*n+2) - Product_{i=0..1}(-i*n-1))/(n+3) = A000012(n) with n >= 0;

B(3,n,2,1) = (Product_{i=0..3}(i*n+2) - Product_{i=0..3}(-i*n-1))/(3*n+3) = A100040(n+2) with n >= 0;

B(2*n+1,0,2,1) = (Product_{i=0..2*n+1}(2) - Product_{i=0..2*n+1}(-1))/3 = A002450(n+1) with n >= 0;

B(2*n+1,2,2,1) = (Product_{i=0..2*n+1}(2*i+2) - Product_{i=0..2*n+1}(-2*i-1))/(4*n+5) = A273983(n+1) with n >= 0;

and a(n) is B(2*n+1,1,2,1). - Hong-Chang Wang, Jun 17 2016

Sequence is inspired by A273983. The same argument as in A273889 can be used here to prove the expression evaluates to an integer.

Since Product_{i=0..n}(i*k+a) - Product_{i=0..n}(-i*k-b) ≡ 0 mod (n*k+a+b), then define B(n,k,a,b) = (Product_{i=0..n}(i*k+a) - Product_{i=0..n}(-i*k-b))/(n*k+a+b), with n*k+a+b <> 0, n >= 0 and k,a,b are integers, such that B(2*n,2,2,1) = (Product_{i=0..2*n}(2*i+2) - Product_{i=0..2*n}(-2*i-1))/ (4*n+3) = A273889(n+1), n >= 0; B(2*n,3,2,1) = (Product_{i=0..2*n}(3*i+2) - Product_{i=0..2*n}(-3*i-1))/(6*n+3) = A274117(n+1), n >= 0; B(2,n,2,1) = (Product_{i=0..2}(i*n+2) - Product_{i=0..2}(-i*n-1))/(2*n+3) = A008585(n+1), n >= 0; and a(n) is B(4,n,2,1). - Hong-Chang Wang, Jun 17 2016

Sequence is similar to A273889, with a similar proof of divisibility.

Apply the P card and the B card, the two cards on the bottom of the stack, to do shuffling a deck of cards. The T cards are on the top of cards and the M cards are between the T cards and the P card.

The shuffling steps are like this: (T)(M) P B -> B (T)(M) P -> B (T) P (M).

Let n = T + M +2, (n >= 2). New P = ceiling((n+1)/2).

Count cycles to return original order for the deck of size n.

Order (4k+1)th and (4k+2)th items, k >= 0, can get a Latin square.