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

1

|

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

4 3

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

/ \ / \ / \ / \

/ \ / \ / \ / \

/ \ / \ / \ / \

16 12 18 10 27 15 25 7

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

etc.

Sequence A253565 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) tells distance of n from 1 in all these trees. Of these four trees, this is the one where the left child is always larger than the right child.

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

a(n) (n>=1) can be obtained by the composition of a bijection between {1,2,3,4,...} and the set of integer partitions and a bijection between the set of integer partitions and {2,3,4,...}. Explanation on the example n=10. Write 2*n = 20 as a binary number: 10100. Consider a Ferrers board whose southeast border is obtained by replacing each 1 by an east step and each 0 by a north step. We obtain the Ferrers board of the partition p = (2,2,1). Finally, a(10) = 2'*2'*1', where m' = m-th prime. Thus, a(10)= 3*3*2 = 18. - Emeric Deutsch, Sep 17 2016

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

Let's define "property S" for sequences as follows: If s is any sequence of positive natural numbers, normalized to begin with offset 1, then it satisfies the S-property if LCM-transform(s) is equal to the sequence obtained by applying A014963 to sequence s, or in other words, when for all n >= 1, lcm {s(1)..s(n)} / lcm {s(1)..s(n-1)} = A014963(s(n)). This holds if and only if, for all n >= 1, when, either (case A): s(n) is of the form p^k, p prime, then gcd(s(n), lcm {s(1)..s(n-1)}) must be equal to p^(k-1), or (case B): when s(n) is not a prime power, then gcd(s(n), lcm {s(1)..s(n-1)}) must be equal to s(n). Together the cases (A) and (B) reduce to the condition that each prime power should appear in s before any of its multiples do.

Clearly the Doudna-sequence satisfies the property by the way of its construction, as do many of its variants like A356867 (see A369060).

Also, for any base-2 related permutation b that keeps all the numbers of range [2^k, 2^(1+k)[ in the same range, i.e., if for all n >= 1, A000523(b(n)) = A000523(n), then the above property is automatically satisfied.

Furthermore, because in Doudna-sequence no multiple of any term is located on the same row as the term itself (see the tree-illustration in A005940), it follows that any composition of A005940 with any such base-2 related permutation as mentioned above also automatically satisfies the S-property, for example, the permutations A163511, A243353, A253563, A253565, A366260, A366263 and A366275.

Note: Like A005940 itself, also this sequence might be more logical with the starting offset 0 instead of 1, to better align with the underlying mapping from the binary expansion of n to the prime factorization. - Antti Karttunen, Jan 24 2024

This can be calculated using the binary expansion of n; see the PARI program.

The n-th row consists of all partitions with hook size (maximum + number of parts - 1) equal to n.

The partitions in row n of this sequence are the conjugates of the partitions in row n of A125106 taken in reverse order.

Row n is also the reversed partial sums plus one of the n-th composition in standard order (A066099) minus one. - Gus Wiseman, Nov 07 2022

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.

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.

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.

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.

Note the indexing: domain starts from one, while the range includes also zero. See also comments in A253564.

The a(n)-th composition in standard order (graded reverse-lexicographic, A066099) is one plus the first differences of the weakly increasing sequence of prime indices of n with 1 prepended. See formula for a simplification. The triangular form is A358169. The inverse is A253565. Not prepending 1 gives A358171. For Heinz numbers instead of standard compositions we have A325351 (without prepending A325352). - Gus Wiseman, Dec 23 2022