A368148 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).
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, 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, 2, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1
Offset: 1
Examples
Array A(n, k) begins:
n\k | 1 2 3 4 5 6 7 8 9 10
----+-----------------------------
1 | 1 2 2 1 2 2 1 2 2 1
2 | 2 1 3 2 1 3 2 1 3 2
3 | 2 3 3 2 3 3 2 3 3 2
4 | 1 2 2 1 2 2 1 2 2 1
5 | 2 1 3 2 1 3 2 1 3 2
6 | 2 3 3 2 3 3 2 3 3 2
7 | 1 2 2 1 2 2 1 2 2 1
8 | 2 1 3 2 1 3 2 1 3 2
9 | 2 3 3 2 3 3 2 3 3 2
10 | 1 2 2 1 2 2 1 2 2 1
.
We can chose A(1, 1) = 1.
A(2, 1) cannot equal 1; we chose A(2, 1) = 2.
Likewise we chose A(1, 2).
A(2, 2) cannot equal 2 as this would imply a component with 3 or more 2's.
So, by necessity, we chose A(3, 1) = A(1, 3) = 2.
We chose A(2, 2) = 1.
We chose A(4, 1) = 1.
A(3, 2) cannot equal 1 or 2; we chose A(3, 2) = 3.
Likewise we chose A(2, 3) = 3.
We chose A(1, 4) = 1.
A(5, 1) cannot equal 1; we chose A(5, 1) = 2.
A(4, 2) cannot equal 1 (or 3); we chose A(4, 2) = 2.
By necessity, A(3, 3) = 3.
etc.
Crossrefs
Cf. A130196 (one-dimensional variant).
Programs
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PARI
A(n,k) = { [1,2,2; 2,1,3; 2,3,3][1+(n-1)%3, 1+(k-1)%3] }
Formula
A(n+3, k) = A(n, k+3) = A(n, k).
A(n, k) = A(k, n).
Comments