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%I #8 Feb 15 2017 03:30:55
%S 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,
%T 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,
%U 11,11,11,9,14,13,21,22,28,12,12,12,12,12,6,12,17
%N Square array read by antidiagonals downwards (see Comments for definition).
%C 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,...).
%e The corner of the square array begins:
%e 1 2 3 4 5 6 7 8 9 10 11 12 13
%e 2 3 4 5 6 7 8 9 10 11 12 13 14
%e 4 2 5 6 7 8 9 10 11 12 13 14 15
%e 7 4 6 2 8 9 10 11 12 13 14 15 16
%e 10 7 9 4 8 6 11 12 13 14 15 16 17
%e 13 10 12 7 11 9 6 4 14 15 16 17 18
%e 16 13 15 10 14 12 4 7 6 11 17 18 19
%e 19 16 18 13 17 15 11 10 6 14 7 12 20
%t f[seq_] := Riffle[Take[Reverse[seq], #], Take[seq, #]] &[Floor[Length[seq]/2]];
%t rows = 200; row[1] = Table[n, {n, rows}];
%t Table[row[n + 1] = Flatten[{f[Take[row[n], 2 n - 1]], Drop[row[n], 2 n - 1]}], {n, Floor[(rows - 1)/3 + 1]}];
%t TableForm[Table[Take[row[n], 20], {n, 1, 20}]] (* A282347, array *)
%t Table[row[n][[n]], {n, 2 + Floor[(rows - 1)/3]}] (* A282347, sequence *)
%t (* _Peter J. C. Moses_, Feb 12 2017 *)
%Y Cf. A035486, A282348 (diagonal).
%K nonn,tabl,easy
%O 1,2
%A _Clark Kimberling_, Feb 13 2017
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