

A088696


Triangle read by rows, giving number of partial quotients in continued fraction representation of terms in the left branch of the infinite SternBrocot tree.


7



1, 1, 2, 1, 2, 3, 2, 1, 2, 3, 2, 3, 4, 3, 2, 1, 2, 3, 2, 3, 4, 3, 2, 3, 4, 5, 4, 3, 4, 3, 2, 1, 2, 3, 2, 3, 4, 3, 2, 3, 4, 5, 4, 3, 4, 3, 2, 3, 4, 5, 4, 5, 6, 5, 4, 3, 4, 5, 4, 3, 4, 3, 2, 1, 2, 3, 2, 3, 4, 3, 2, 3, 4, 5, 4, 3, 4, 3, 2, 3, 4, 5, 4, 5, 6, 5, 4, 3, 4, 5, 4, 3, 4, 3, 2, 3, 4, 5, 4, 5, 6, 5, 4, 5, 6
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OFFSET

1,3


COMMENTS

A000120 is produced by following each row of this triangle by its reversal.
From Gary W. Adamson, Aug 08 2009: (Start)
The row with 8 terms: (1, 2, 3, 2, 3, 4, 3, 2); can be used to generate the numbers of hydrogen bonds per codon/anticodon; superimposed on the DNA codon array of A147995 as follows: top row and left column of an 8 X 8 array is composed of the 8 terms (1, 2, 3, 2, 3, 4, 3, 2). If rows and columns have an offset of "1", then odd rows circulate downward starting from the position (n,n). Even rows circulate in the opposite direction starting from position (n,n).
This produces the array:
1 2 3 2 3 4 3 2
2 1 2 3 4 3 2 3
3 2 1 2 3 2 3 4
2 3 2 1 2 3 2 3
3 4 3 2 1 2 3 2
4 3 2 3 2 1 4 3
3 2 3 4 3 2 1 2
2 3 4 3 2 3 2 1
...
This produces a semimagic square with a diagonal of (1,1,1,...). Using the simple replacement rule ("complement to 10"): (1>9); (2>8); (3>7); (4>6) we obtain the chart of DNA hydrogen bonds per codon/anticodon shown in A147995. Top row of the hydrogen bond array as well as left column = (9, 8, 7, 8, 7, 6, 7, 8).
Alternatively, using the circulant rule for alternate rows and putting all 9's along the diagonal, we obtain the chart of hydrogen bonds. (End)
Rows tend to A088748 (which can also be generated from the dragon curve, A014577).  Gary W. Adamson, Aug 30 2009


REFERENCES

R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. AddisonWesley, Reading, MA, 1990, pp. 116117.


LINKS

Table of n, a(n) for n=1..105.


EXAMPLE

Fractions in the left branch of the infinite SternBrocot tree (the fractions between 0 and 1), are:
1/2;
1/3, 2/3;
1/4, 2/5, 3/5, 3/4;
1/5, 2/7, 3/8, 3/7, 4/7, 5/8, 5/7, 4/5;
...
and their corresponding continued fraction representations are:
[2]
[3] [1,2]
[4] [2,2] [1,1,2] [1,3]
[5] [3,2] [2,1,2] [2,3] [1,1,3] [1,1,1,2] [1,2,2] [1,4]
...
with the number of terms in each continued fraction representation generating A088696:
1
1 2
1 2 3 2
1 2 3 2 3 4 3 2
...


MATHEMATICA

sb[n_List] := Block[{k = l = Length[n], a = n}, While[k > 1, a = Insert[ a, (Numerator[ a[[k]]] + Numerator[ a[[k  1]]]) / (Denominator[ a[[k]]] + Denominator[ a[[k  1]]]), k]; k ]; a]; sbn[n_] := Complement[ Nest[ sb, {0, 1}, n], Nest[ sb, {0, 1}, n  1]]; f[n_] := Length /@ (ContinuedFraction /@ sbn[n])  1; Flatten[ Table[ f[n], {n, 7}]] (* Robert G. Wilson v, Jun 09 2004 *)
Flatten[NestList[Join[#, Reverse[#] + 1] &, {1}, 7]]; (* from A164738, Jon Maiga, Sep 26 2019 *)


PROG

(Haskell)
a088696 n = a088696_list !! (n1)
a088696_list = f [1] where
f (x:xs) = x : f (xs ++ [x + 1  x `mod` 2, x + x `mod` 2])
 Reinhard Zumkeller, Mar 07 2011


CROSSREFS

Cf. A000120, A007305, A007306.
Cf. A147995.  Gary W. Adamson, Aug 08 2009
Cf. A088748, A014577.  Gary W. Adamson, Aug 30 2009
Sequence in context: A192099 A193101 A100661 * A257249 A267108 A004738
Adjacent sequences: A088693 A088694 A088695 * A088697 A088698 A088699


KEYWORD

nonn,tabf


AUTHOR

Gary W. Adamson, Oct 07 2003


EXTENSIONS

Edited and extended by Robert G. Wilson v, Jun 09 2004


STATUS

approved



