A282347 Square array read by antidiagonals downwards (see Comments for definition).

1, 2, 2, 3, 3, 4, 4, 4, 2, 7, 5, 5, 5, 4, 10, 6, 6, 6, 6, 7, 13, 7, 7, 7, 2, 9, 10, 16, 8, 8, 8, 8, 4, 12, 13, 19, 9, 9, 9, 9, 8, 7, 15, 16, 22, 10, 10, 10, 10, 6, 11, 10, 18, 19, 25, 11, 11, 11, 11, 11, 9, 14, 13, 21, 22, 28, 12, 12, 12, 12, 12, 6, 12, 17

Offset: 1

Author

Clark Kimberling, Feb 13 2017

Keywords

• nonn
• tabl
• easy

Comments

Define f(x(1),x(2),...,x(2k)) = (a(2k),x(1),x(2k-1),x(2),a(2k-1),a(3),...a(k-1)). The array is defined by rows as follows. row 1 = (1,2,3,4,5,...) = A000027. To get from (row n) = (r(1),r(2),r(3),...) to (row n+1), the first 2n-2 terms are f(r(1),r(2),...,r(n-1),r(n+1),...,r(2n-1)), where r(n) is skipped, followed by (r(2n),r(2n+1),...) = (3n-1, 3n, 3n+1,...).

Examples

			The corner of the square array begins:
1    2    3    4    5    6    7    8    9    10   11   12   13
2    3    4    5    6    7    8    9    10   11   12   13   14
4    2    5    6    7    8    9    10   11   12   13   14   15
7    4    6    2    8    9    10   11   12   13   14   15   16
10   7    9    4    8    6    11   12   13   14   15   16   17
13   10   12   7    11   9    6    4    14   15   16   17   18
16   13   15   10   14   12   4    7    6    11   17   18   19
19   16   18   13   17   15   11   10   6    14   7    12   20
		

Crossrefs

Cf. A035486, A282348 (diagonal).

Programs

Mathematica

f[seq_] := Riffle[Take[Reverse[seq], #], Take[seq, #]] &[Floor[Length[seq]/2]];
rows = 200; row[1] = Table[n, {n, rows}];
Table[row[n + 1] = Flatten[{f[Take[row[n], 2 n - 1]], Drop[row[n], 2 n - 1]}], {n,    Floor[(rows - 1)/3 + 1]}];
TableForm[Table[Take[row[n], 20], {n, 1, 20}]]   (* A282347, array *)
Table[row[n][[n]], {n, 2 + Floor[(rows - 1)/3]}] (* A282347, sequence *)
(* Peter J. C. Moses, Feb 12 2017 *)