cp's OEIS Frontend

Each n > 1 occurs A025147(n) times in the sequence.

Numerators from the ordering of positive rational numbers by increasing balanced ternary representation of the "factorization" of positive rational numbers into terms of A186285 (prime powers with a power of three as exponent).

Each a(n) is always either a divisor or a multiple of a(n+1).

Consider A052330. Imagine that it is an automatic piano that "plays sequences" when an appropriate punched "tape" is fed to it (as its input), i.e., when it is composed from the right with an appropriate sequence p, as A019565(p(n)). The 1-bits in the binary expansion of each p(n) are the "holes" in the tape, and they determine which "tunes" are present on beat n. The "tunes" are actually "Fermi-Dirac primes" (A050376) that are multiplied together.

If the tape is constructed in such a way that between the successive beats (when moving from p(n) to p(n+1)), either a subset of 0-bits are toggled on (changed to 1's), or a subset of 1-bits are toggled off (changed to 0's), but no both kind of changes occur simultaneously, then when fed as an input to this piano, the resulting sequence "s" (the output) is guaranteed to satisfy the condition that s(n+1) is either a multiple or a divisor of s(n). Furthermore, if the given sequence p is itself a permutation of natural numbers, then also the produced sequence is. For example, Gray code A003188 and its inverse A006068 are such sequences, and when given as an "input tape" for A052330, they produce permutations A207901 and A302783.

There is a simpler instrument, called "squarefree piano" (A019565), with which it is possible to produce similar divisor-or-multiple sequences, but that contain only squarefree numbers. Given A003188 or A006068 as an input tape for it produces correspondingly sequences A302033 and A284003.

This sequence is obtained by playing "squarefree piano" with the same tape which yields A304531 when "Fermi-Dirac piano" is played with it. However, in this case the sequence A304531 is produced by a greedy algorithm, and thus its tape (A304533) is actually a back-formation, obtained from the "music" (A304531) by applying "tape-recorder" (A052331) to it. Note that this in not a subsequence of A304531, as the terms occur in different order than the squarefree terms of A304531.

See also Peter Munn's Apr 11 2018 message on SeqFan-mailing list.

This is GF(2)[X] analog of A005940, but note the indexing: here the domain starts from 0, although the range excludes zero.

This sequence can be represented as a binary tree. Each child to the left is obtained by applying A305421 to the parent, and each child to the right is obtained by doubling the parent:

1

|

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

3 4

7......../ \........6 5......../ \........8

/ \ / \ / \ / \

/ \ / \ / \ / \

/ \ / \ / \ / \

11 14 9 12 21 10 15 16

13 22 29 28 49 18 27 24 69 42 63 20 107 30 17 32

Sequence A305427 is obtained by scanning the same tree level by level from right to left.

For all i, j: A323074(i) = A323074(j) => a(i) = a(j).

Like the related A322822 also this filter sequence satisfies the following two implications, for all i, j >= 1:

a(i) = a(j) => A322356(i) = A322356(j),

a(i) = a(j) => A290105(i) = A290105(j).

Multiplicative with a(A050376(m)) = Prime(m) = A000040(m). If k = 2^{i_1} + ... + 2^{i_j} is the binary representation of k, a(p^k) = a(p^2^{i_1}) * ... * a(p^2^{i_j}). [edited by Peter Munn, Jan 07 2020]

Equivalently, a(A050376(m)) = A000040(m); a(A059897(n,k)) = A059897(a(n), a(k)). - Peter Munn, Dec 30 2019

For Fermi-Dirac representation of n see A182979. - N. J. A. Sloane, Jul 21 2018

For any n > 0, either a(n)/a(n+1) or a(n+1)/a(n) belongs to A050376.

This sequence has similarities with A282291; in both sequences, each pair of consecutive terms contains a term that divides the other.

This is a left inverse of A302853, and also the right inverse if A282291 (and thus also A302853) is surjective (a permutation of natural numbers), in which case the fallback-clause is unnecessary.

For all i, j: a(i) = a(j) => A005811(i) = A005811(j).

For n > 1, a(n) gives the number of edges needed to traverse from n to reach the leftmost branch (where the terms of A050376 are located) in the binary tree illustrated in A052330.