A035486 -id:A035486 - cp's OEIS frontend

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

Clark Kimberling, Feb 13 2017

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,...).

			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

Cf. A035486, A282348 (diagonal).

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 *)