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Triangle with 2*n+1 terms per row, read by rows: the first row is 1 (by decree), following rows contain 0 to 2n+1 but omitting 2n.
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%I #20 Apr 29 2016 09:20:00

%S 1,0,1,3,0,1,2,3,5,0,1,2,3,4,5,7,0,1,2,3,4,5,6,7,9,0,1,2,3,4,5,6,7,8,

%T 9,11,0,1,2,3,4,5,6,7,8,9,10,11,13,0,1,2,3,4,5,6,7,8,9,10,11,12,13,15,

%U 0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,17

%N Triangle with 2*n+1 terms per row, read by rows: the first row is 1 (by decree), following rows contain 0 to 2n+1 but omitting 2n.

%C Row n is row 2n+1 of A128138, a bisection.

%C The second bisection by rows

%C 0, 2,

%C 0, 1, 2, 4,

%C 0, 1, 2, 3, 4, 6,

%C 0, 1, 2, 3, 4, 5, 6, 8,

%C etc

%C is the basis of

%C 0, 2, 4, 6, 8, 10, 12, ... the even numbers A005843(n)

%C 0, 1, 2, 4, 3, 6, 8, 5, 10, ... a permutation of the nonnegative integers A265667(n).

%C 0, 1, 2, 3, 4, 6, 5, 8, 7, 10, 12, ... a permutation of the nonnegative integers A265734(n)

%C etc.

%C A005843(n) - A005843(n-1) = 2, for n>0.

%C A265667(n) - A265667(n-3) = 4, 2, 4 (period 3), for n>2.

%C A265734(n) - A265734(n-5) = 6, 4, 6, 4, 6 (period 5), for n>4.

%C See A267654.

%C For

%C 1, 3, 5, 7, 9, 11, 13 ... the odd numbers A005408(n),

%C 0, 1, 3, 2, 5, 7, 4, 9, 11, ... a permutation of the nonnegative numbers A006369,

%C 0, 1, 2, 3, 5, 4, 7, 6, 9, 11, 8, 13, 10, 15, ... another permutation,

%C a(n) must be extended with one term by row:

%C 1, 3,

%C 0, 1, 3, 2,

%C 0, 1, 2, 3, 5, 4,

%e Irregular triangle:

%e 1,

%e 0, 1, 3,

%e 0, 1, 2, 3, 5,

%e 0, 1, 2, 3, 4, 5, 7,

%e 0, 1, 2, 3, 4, 5, 6, 7, 9,

%e 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11,

%e 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13,

%e etc.

%t Table[Delete[Range[0, 2 n + 1], 2 n + 1], {n, 0, 8}] // Flatten (* _Michael De Vlieger_, Apr 25 2016 *)

%Y Cf. A001477, A005408, A005843, A006369, A028310 (main diagonal), A053186, A265667, A265734, A267654.

%K nonn,tabf,easy

%O 0,4

%A _Paul Curtz_, Apr 22 2016