%I #10 Dec 16 2023 08:59:40

%S 1,2,2,2,1,2,1,3,3,1,2,2,3,2,2,2,1,2,2,1,2,1,3,3,1,3,3,1,2,2,3,2,2,3,

%T 2,2,2,1,2,2,1,2,2,1,2,1,3,3,1,3,3,1,3,3,1,2,2,3,2,2,3,2,2,3,2,2,2,1,

%U 2,2,1,2,2,1,2,2,1,2,1,3,3,1,3,3,1,3,3,1,3,3,1

%N Square array A(n, k), n, k > 0, read and filled in the greedy way by upwards antidiagonals such that A(n, k) corresponds to the size of the connected component (relative to the Von Neumann neighborhood) of terms equal to A(n, k) including the position (n, k).

%C The array is doubly periodic (see first formula) and consists of the following repeating 3 X 3 pattern with two components of 1 1's, two components of 2 2's and one component of 3 3's:

%C +---+-------+

%C | 1 | 2 2 |

%C +---+---+---+

%C | 2 | 1 | 3 |

%C | +---+ |

%C | 2 | 3 3 |

%C +---+-------+

%F A(n+3, k) = A(n, k+3) = A(n, k).

%F A(n, k) = A(k, n).

%e Array A(n, k) begins:

%e n\k | 1 2 3 4 5 6 7 8 9 10

%e ----+-----------------------------

%e 1 | 1 2 2 1 2 2 1 2 2 1

%e 2 | 2 1 3 2 1 3 2 1 3 2

%e 3 | 2 3 3 2 3 3 2 3 3 2

%e 4 | 1 2 2 1 2 2 1 2 2 1

%e 5 | 2 1 3 2 1 3 2 1 3 2

%e 6 | 2 3 3 2 3 3 2 3 3 2

%e 7 | 1 2 2 1 2 2 1 2 2 1

%e 8 | 2 1 3 2 1 3 2 1 3 2

%e 9 | 2 3 3 2 3 3 2 3 3 2

%e 10 | 1 2 2 1 2 2 1 2 2 1

%e .

%e We can chose A(1, 1) = 1.

%e A(2, 1) cannot equal 1; we chose A(2, 1) = 2.

%e Likewise we chose A(1, 2).

%e A(2, 2) cannot equal 2 as this would imply a component with 3 or more 2's.

%e So, by necessity, we chose A(3, 1) = A(1, 3) = 2.

%e We chose A(2, 2) = 1.

%e We chose A(4, 1) = 1.

%e A(3, 2) cannot equal 1 or 2; we chose A(3, 2) = 3.

%e Likewise we chose A(2, 3) = 3.

%e We chose A(1, 4) = 1.

%e A(5, 1) cannot equal 1; we chose A(5, 1) = 2.

%e A(4, 2) cannot equal 1 (or 3); we chose A(4, 2) = 2.

%e By necessity, A(3, 3) = 3.

%e etc.

%o (PARI) A(n,k) = { [1,2,2; 2,1,3; 2,3,3][1+(n-1)%3, 1+(k-1)%3] }

%Y Cf. A130196 (one-dimensional variant).

%K nonn,tabl,easy

%O 1,2

%A _Rémy Sigrist_, Dec 13 2023