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

1,2

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,...).

Links

Table of n, a(n) for n=1..74.

Example

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

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 *)

Crossrefs

Cf. A035486, A282348 (diagonal).

Sequence in context: A074719 A079730 A035486 * A172397 A237815 A238701

Adjacent sequences: A282344 A282345 A282346 * A282348 A282349 A282350

Keyword

nonn,tabl,easy

Author

Clark Kimberling, Feb 13 2017

Status

approved