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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]

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/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).

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 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).

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

Positions of records are A081254. - Andrey Zabolotskiy, Mar 27 2020

If the terms (n>0) are written as an array (in a left-aligned fashion) with rows of length 2^m, m >= 0:

2,

3, 3,

5, 4, 4, 5,

8, 7, 5, 7, 7, 5, 7, 8,

13,11, 9,12, 9, 6,10,11,11,10,6, 9,12, 9,11,13,

21,18,14,19,16,11,17,19,14,13,7,11,17,13,15,18,18,15,13,17,11,7,13,14,19,17,11,16, ...

a(n) is palindromic in each level m >= 0 (ranks between 2^m and 2^(m+1)-1), because in each level m >= 0 A162910 is the reverse of A162909:

a(2^m + k) = a(2^(m+1) - 1 - k), m >= 0, 0 <= k < 2^m.

All columns have the Fibonacci sequence property: a(2^(m+2) + k) = a(2^(m+1) + k) + a(2^m + k), m >= 0, 0 <= k < 2^m (empirical observations).

a(2^m + k) = A162909(2^(m+2) + k), a(2^m + k) = A162909(2^(m+1)+ 2^m + k), a(2^m + k) = A162910(2^(m+1) + k), m >= 0, 0 <= k < 2^m (empirical observations).

a(n) = A162911(n) + A162912(n), where A162911(n)/A162912(n) is the bit reversal permutation of A162909(n)/A162910(n) in each level m >= 0 (empirical observations).

a(n) = A162911(2n+1), a(n) = A162912(2n) for n > 0 (empirical observations). n > 1 occurs in this sequence phi(n) = A000010(n) times, as it occurs in A007306 (Franklin T. Adams-Watters's comment), which is the sequence obtained by adding numerator and denominator in the Calkin-Wilf enumeration system of positive rationals. A162909(n)/A162910(n) is also an enumeration system of all positive rationals (Bird system), and in each level m >= 0 (ranks between 2^m and 2^(m+1)-1) rationals are the same in both systems. Thus a(n) has the same terms in each level as A007306.

The same property occurs in all numerator+denominator sequences of enumeration systems of positive rationals, as, for example, A007306 (A007305+A047679), A071585 (A229742+A071766), and A086592 (A020650+A020651).

Define a sequence chf(n) of Christoffel words over an alphabet {-,+}:

chf(1) = '-',

chf(2*n+0) = negate(chf(n)),

chf(2*n+1) = negate(concatenate(chf(n),chf(n+1))).

Each chf(n) word has the length fusc(n) = A002487(n) and splits uniquely into two parent Christoffel words - the left Christoffel word lef(n) of the length l-fusc(n) = a(n) and the right Christoffel word rig(n) of the length r-fusc(n) = A288003(n). See the example below.

The Markov tree is a complete, infinite binary tree. Vertices are labeled by triples. The root vertex is (1, 5, 2). The left child of (a, b, c) is (a, 3*a*b - c, b); its right child is (b, 3*b*c - a, c). The sequence is a triangle read by rows consisting of the middle element of each triple, which is always the largest element of the triple. Row r contains 2^r elements.

The tree contains contains exactly one representative of each class of permutation equivalent nonsingular solutions to Markov's equation, a^2 + b^2 + c^2 = 3 * a * b * c. Nonsingular solutions are those in which a, b, and c are three distinct numbers. The two singular triples (1, 1, 1) and (1, 2, 1) are omitted in this sequence.

A consequence of Markov's equation is that the recurrence for the tree may be reformulated as follows: the left child of (a, b, c) is (a, (a^2 + b^2) / c, b); its right child is (b, (b^2 + c^2) / a, c).

An open problem is to prove the uniqueness conjecture, which asserts that the largest element of a triple determines the other two.

Frobenius proposed assigning a rational number index in (0,1) to each vertex of the tree, and hence to each term in this sequence. This is the Farey index, obtained by assigning the triple (0/1, 1/2, 1/1) to the root vertex and using the following rules to assign triples to the rest of the tree: the vertex labeled (u/v, w/x, y/z) with w = u + u and x = v + z has left child (u/v, (u+w)/(v+x), w/x) and right child (w/x, (w+y)/(x+z), y/z). The Farey index is the center element of the triple. Each rational number in (0, 1) appears as the Farey index of exactly one vertex of the tree. The index of a(n) is A007305(n+2) / A007306(n+2).

A sequence of leftward steps in the tree produces odd-indexed Fibonacci numbers, A001519, which have Farey indices of the form 1 / n. A sequence of rightward steps in the tree produces odd-indexed Pell numbers, A001653, which have Farey indices of the form (n - 1) / n. A sequence of leftward steps followed by a single rightward step produces A350922, corresponding to Farey indices of the form 2 / (2 * n + 1). Alternating steps right, left, right, left, right, ... produces A064098, which corresponds to Farey indices of the form F(n) / F(n + 1), where F(n) is the n-th Fibonacci number.

A variant of Stern's diatomic series A002487. See the link [Luschny] and the Maple function below for the classification by types which is based on a generalization of Dijkstra's fusc function.

a(n) is also the number of superduperbinary integer partitions of n.

It appears that a(n) is equal to the multiplicative inverse of A002487(n+2) mod A002487(n+1). - Gary W. Adamson, Dec 23 2023

Number of 1's on row 2n+1 of table A125184.