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A164056
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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).
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6
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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
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OFFSET
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0,1
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COMMENTS
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Complement of the sequence = A164057
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LINKS
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FORMULA
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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.
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EXAMPLE
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1;
1, 2;
1, 2, 3, 2;
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;
...
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MATHEMATICA
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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 *)
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CROSSREFS
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KEYWORD
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nonn,tabf
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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