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This sequence can be represented as a binary tree. Each child to the left is obtained by applying A253550 to the parent, and each child to the right is obtained by applying A253560 to the parent:

1

|

...................2...................

3 4

5......../ \........9 6......../ \........8

/ \ / \ / \ / \

/ \ / \ / \ / \

/ \ / \ / \ / \

7 25 15 27 10 18 12 16

11 49 35 125 21 75 45 81 14 50 30 54 20 36 24 32

etc.

Sequence A253563 is the mirror image of the same tree. Also in binary trees A005940 and A163511 the terms on level of the tree are some permutation of the terms present on the level n of this tree. A252464(n) gives the distance of n from 1 in all these trees. Of these four trees, this is the one where the left child is always smaller than the right child.

Note that the indexing of sequence starts from 0, although its range starts from one.

The term a(n) is the Heinz number of the adjusted partial sums of the n-th composition in standard order, where (1) the k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again, (2) the Heinz number of a partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k), and (3) we define the adjusted partial sums of a composition to be obtained by subtracting one from all parts, taking partial sums, and adding one back to all parts. See formula for a simplification. A triangular form is A242628. The inverse is A253566. The non-adjusted version is A358170. - Gus Wiseman, Dec 17 2022

Let (10...0)_i (i>=0) denote 2^i in binary. Under (10...0)_i^k we understand a concatenation of (10...0)_i k times.

If n=Product_{i=1..m} p_i^t_i is the prime power factorization of n, then in the name r(n)=concatenation{i=1..m} ((10...0_(i-1)^t_i).

Numbers q and s are called c-equivalent if their binary expansions contain the same set of parts of the form 10...0. For example, 14=(1)(1)(10)~(10)(1)(1)=11.

Conversely, if n~n_1 such that n_1 is in A114994 and has c-factorization: n_1 = concatenation{i=m,...,0} ((10...0)i^t_i), one can consider "converse" sequence {s(n)}, where s(n) = Product{i=m..0} p_(i+1)^t_i.

For example, for n=22, n_1=21=((10)^2)(1), and s(22)=3^2*2=18.

The k-th composition in standard order (row k of A066099) is obtained by taking the set of positions of 1's in the reversed binary binary expansion of k, prepending 0, taking first differences, and reversing again. Then a(n) is the number k such that the k-th composition in standard order consists of the prime indices of n in weakly decreasing order (the partition with Heinz number n). - Gus Wiseman, Apr 02 2020

A self-inverse permutation. Also a(n) = A054429(A059893(n)) = A059893(A054429(n)).

a(n) is the viabin number of the integer partition that is conjugate to the integer partition with viabin number n. Example: a(9) = 11. Indeed, 9 and 11 are the viabin numbers of the conjugate partitions [2,1,1] and [3,1], respectively. For the definition of viabin number see comment in A290253. - Emeric Deutsch, Aug 23 2017

Fixed points union { 0 } are in A290254. - Alois P. Heinz, Aug 24 2017

Row (partition) sums of A125106.

a(n) is also the weight (= sum of parts) of the integer partition having viabin number n. The viabin number of an integer partition is defined in the following way. Consider the southeast border of the Ferrers board of the integer partition and consider the binary number obtained by replacing each east step with 1 and each north step, except the last one, with 0. The corresponding decimal form is, by definition, the viabin number of the given integer partition. "Viabin" is coined from "via binary". For example, consider the integer partition [2,2,2,1]. The southeast border of its Ferrers board yields 10100, leading to the viabin number 20. - Emeric Deutsch, Jul 24 2017

If the terms (n>0) are written as an array:

1,

1, 2,

1, 2, 3, 3,

1, 2, 3, 3, 4, 5, 4, 5,

1, 2, 3, 3, 4, 5, 4, 5, 5, 7, 7, 8, 5, 7, 7, 8,

1, 2, 3, 3, 4, 5, 4, 5, 5, 7, 7, 8, 5, 7, 7, 8, 6, 9, 10, 11, 9, 12, 11, 13, 6, 9,

then the sum of the m-th row is 3^m (m = 0,1,2,3,...), each column is constant and the terms are from A071585 (a(2^m+k) = A071585(k), k = 0,1,2,...).

If the rows are written in a right-aligned fashion:

1,

1, 2,

1, 2, 3, 3,

1, 2, 3, 3, 4, 5, 4, 5,

1, 2, 3, 3, 4, 5, 4, 5, 5, 7, 7, 8, 5, 7, 7, 8,

..., 7, 7, 8, 5, 7, 7, 8, 6, 9, 10, 11, 9, 12, 11, 13, 6, 9, 10, 11, 9, 12, 11, 13,

then each column is a Fibonacci sequence (a(2^(m+2)+k) = a(2^(m+1)+k) + a(2^m+k), m = 0,1,2,..., k = 0,1,2,...,2^m-1 with a_k(1) = A071766(k) and a_k(2) = A086593(k) being the first two terms of each column sequence). - Yosu Yurramendi, Jun 23 2014

Conjecture: For all n, a(A054429(n)) = A054429(a(n)), i.e. A054429 acts as a homomorphism (automorphism) of the cyclic group generated by this permutation. This implies also a weaker conjecture given in A209860.

From Gus Wiseman, Aug 24 2021: (Start)

As a triangle with row lengths 2^n, T(n,k) for n > 0 appears (verified up to n = 2^15) to be the unique nonnegative integer whose binary indices are the k-th subset of {1..n} containing n. Here, a binary index of n (row n of A048793) is any position of a 1 in its reversed binary expansion, and sets are sorted first by length, then lexicographically. For example, the triangle begins:

1

2 3

4 5 6 7

8 9 10 12 11 13 14 15

16 17 18 20 24 19 21 25 22 26 28 23 27 29 30 31

Mathematica: Table[Total[2^(Append[#,n]-1)]&/@Subsets[Range[n-1]],{n,5}]

Row lengths are A000079 (shifted right). Also Column k = 1.

Row sums are A010036.

Using reverse-lexicographic order gives A059893.

Using lexicographic order gives A059894.

Taking binary indices to prime indices gives A339195 (or A019565).

The ordering of sets is A344084.

A version using Heinz numbers is A344085.

(End)

The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions. See also A000120, A059893, A070939, A114994, A225620.

A fixed point of composition c is an index i such that c_i = i.

Another way to describe this: starting with the binary representation and a counter set at one, count the 0's from right to left. Write a term equal to the counter for each "1" encountered.

A101211 is a similar sequence, with A005811 elements per row which maps natural numbers to compositions (ordered partitions).

There are two ways to consider this as a table: taking each partition as a row, or taking the partitions generated by 2^(n-1) through 2^n-1 as a row.

Taking the n-th row as multiple partitions, it consists of those partitions with the first hook size (largest part plus number of parts minus 1) equal to n. The number of integers in this n-th row is A001792(n-1), and the row sum is A049611.

Taking each partition as a separate row, the row lengths are A000120, and the row sums are A161511.

Heinz numbers of the rows are A005940. - Gus Wiseman, Jan 17 2023

The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions. See also A000120, A059893, A070939, A114994, A225620.

A nonfixed point in a composition c is an index i such that c_i != i.

The "unbalancing operation" used here is what is usually called "rotation of binary trees" (e.g. in Lucas, Ruskey et al. article)