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%I #17 Oct 05 2019 04:26:36
%S 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,
%T 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,
%U 1,0,0,1,1,1,0,0,1,0,0,1,1,1,0,1,1,0,0
%N 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).
%C Complement of the sequence = A164057
%H Jon Maiga, <a href="/A164056/b164056.txt">Table of n, a(n) for n = 0..1022</a> (Rows 0..9)
%F 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.
%e A088696 begins:
%e 1;
%e 1, 2;
%e 1, 2, 3, 2;
%e 1, 2, 3, 2, 3, 4, 3, 2;
%e ...
%e Triangle A164056 =
%e 0;
%e 0, 1;
%e 0, 1, 1, 0;
%e 0, 1, 1, 0, 1, 1, 0, 0;
%e 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0;
%e ...
%t 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 *)
%Y Cf. A088696, A164057
%K nonn,tabf
%O 0,1
%A _Gary W. Adamson_, Aug 08 2009
%E More terms from _Jon Maiga_, Sep 30 2019
%E Keyword tabf from _Michel Marcus_, Sep 30 2019
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