A164056 Triangle of 2^n terms by rows, derived from A088696 as to length of continued fractions, lengths increase = 1, decrease = 0. A088696 can be generated using the following algorithm: Rows 0 and 1 begin 1; 1,2; then for all further rows, bring down current row then append to the right: (1 added to each term in current row). Row 2 (1, 2, 3, 2) then becomes: (1, 2, 3, 2, 3, 4, 3, 2).

0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 1, 1, 0, 0

Offset: 0

Author

Gary W. Adamson, Aug 08 2009

Keywords

• nonn
• tabf

Comments

Complement of the sequence = A164057

Examples

			A088696 begins:
1;
1, 2;
1, 2, 3, 2;
1, 2, 3, 2, 3, 4, 3, 2;
...
Triangle A164056 =
0;
0, 1;
0, 1, 1, 0;
0, 1, 1, 0, 1, 1, 0, 0;
0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0;
...
		

Links

• Jon Maiga, Table of n, a(n) for n = 0..1022 (Rows 0..9)

Crossrefs

Cf. A088696, A164057

Programs

Mathematica

A088696[n_]:=A088696[n]=Flatten[NestList[Join[#,Reverse[#]+1]&,{1},15]][[n]]; Table[If[n>1,If[A088696[n+1]>A088696[n], 1, 0], 0], {n,0,100}] (* Jon Maiga, Sep 30 2019 *)

Formula

Given number of terms in half of the Stern-Brocot infinite Farey tree (cf. A088696); left the leftmost term in each row = 0, then the next term = 1 if the next corresponding positional term in A088696 increases; otherwise 0.

Extensions

More terms from Jon Maiga, Sep 30 2019

Keyword tabf from Michel Marcus, Sep 30 2019