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

0,1

Comments

Complement of the sequence = A164057

Links

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

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.

Example

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;
...

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 *)

Crossrefs

Cf. A088696, A164057

Sequence in context: A072784 A092436 A288306 * A286059 A163539 A143538

Adjacent sequences: A164053 A164054 A164055 * A164057 A164058 A164059

Keyword

nonn,tabf

Author

Gary W. Adamson, Aug 08 2009

Extensions

More terms from Jon Maiga, Sep 30 2019

Keyword tabf from Michel Marcus, Sep 30 2019

Status

approved