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Triangle read by rows, giving number of partial quotients in continued fraction representation of terms in the left branch of the infinite Stern-Brocot tree.
8

%I #40 Jan 19 2023 22:35:51

%S 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,

%T 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,

%U 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

%N Triangle read by rows, giving number of partial quotients in continued fraction representation of terms in the left branch of the infinite Stern-Brocot tree.

%C Each next row is the last row concatenated with the last row reversed with elements incremented. A000120 is produced by a similar principle, omitting the reversal step. [Edited by _Andrey Zabolotskiy_, Mar 27 2020]

%C From _Gary W. Adamson_, Aug 08 2009: (Start)

%C 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/anti-codon; 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).

%C This produces the array:

%C 1 2 3 2 3 4 3 2

%C 2 1 2 3 4 3 2 3

%C 3 2 1 2 3 2 3 4

%C 2 3 2 1 2 3 2 3

%C 3 4 3 2 1 2 3 2

%C 4 3 2 3 2 1 4 3

%C 3 2 3 4 3 2 1 2

%C 2 3 4 3 2 3 2 1

%C ...

%C This produces a semi-magic 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/anti-codon shown in A147995. Top row of the hydrogen bond array as well as left column = (9, 8, 7, 8, 7, 6, 7, 8).

%C Alternatively, using the circulant rule for alternate rows and putting all 9's along the diagonal, we obtain the chart of hydrogen bonds. (End)

%C Rows tend to A088748 (which can also be generated from the dragon curve, A014577). - _Gary W. Adamson_, Aug 30 2009

%C Conjecture: positions of records are A081254. - _Andrey Zabolotskiy_, Mar 27 2020

%D R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, pp. 116-117.

%H Andrey Zabolotskiy, <a href="/A088696/b088696.txt">Table of n, a(n) for n = 1..8191</a>

%e Fractions in the left branch of the infinite Stern-Brocot tree (the fractions between 0 and 1), are:

%e 1/2;

%e 1/3, 2/3;

%e 1/4, 2/5, 3/5, 3/4;

%e 1/5, 2/7, 3/8, 3/7, 4/7, 5/8, 5/7, 4/5;

%e ...

%e and their corresponding continued fraction representations are:

%e [2]

%e [3] [1,2]

%e [4] [2,2] [1,1,2] [1,3]

%e [5] [3,2] [2,1,2] [2,3] [1,1,3] [1,1,1,2] [1,2,2] [1,4]

%e ...

%e with the number of terms in each continued fraction representation generating the present triangle:

%e 1

%e 1 2

%e 1 2 3 2

%e 1 2 3 2 3 4 3 2

%e ...

%t 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 *)

%t Flatten[NestList[Join[#, Reverse[#] + 1] &, {1}, 7]]; (* from A164738, _Jon Maiga_, Sep 26 2019 *)

%o (Haskell)

%o a088696 n = a088696_list !! (n-1)

%o a088696_list = f [1] where

%o f (x:xs) = x : f (xs ++ [x + 1 - x `mod` 2, x + x `mod` 2])

%o -- _Reinhard Zumkeller_, Mar 07 2011

%o (Python)

%o a = [[1]]

%o for n in range(6):

%o a.append(a[-1] + [x+1 for x in a[-1][::-1]])

%o print(sum(a, []))

%o # _Andrey Zabolotskiy_, Mar 27 2020, after _Jon Maiga_

%Y Cf. A000120, A007305, A007306.

%Y Cf. A147995.

%Y Cf. A088748, A014577.

%K nonn,tabf

%O 1,3

%A _Gary W. Adamson_, Oct 07 2003

%E Edited and extended by _Robert G. Wilson v_, Jun 09 2004