cp's OEIS Frontend

From Antti Karttunen, Feb 10 2021: (Start)

This sequence can be represented as a binary tree. Each child to the left is obtained by multiplying its parent with (1+{binary weight of its breadth-first-wise index in the tree}), while each child to the right is just a clone of its parent:

1

|

...................1...................

2 1

4......../ \........2 3......../ \........1

/ \ / \ / \ / \

/ \ / \ / \ / \

/ \ / \ / \ / \

8 4 6 2 9 3 4 1

16 8 12 4 18 6 8 2 27 9 12 3 16 4 5 1

etc.

(End)

This sequence and A243499 have the same set of values on intervals from 2^m to 2^(m+1) - 1 for m >= 0. - Mikhail Kurkov, Jun 18 2021 [verification needed]

FindStat provides a sequence of mappings between this sequence and A000110 starting from collection [Set partitions] (see Links section for illustration). - Mikhail Kurkov, May 20 2023 [verification needed]

Without the initial 0, a(n) is the lexicographically minimal sequence of distinct positive integers such that all values of a(n) mod n are distinct and nonnegative. - Ivan Neretin, Apr 27 2015

A002487(n)/A002487(n+1), n > 0, runs through all the reduced nonnegative rationals exactly once. A002487 is the Stern's sequence. Permutation from denominators (A002487(n+1))

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

where labels are

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

to numerators (A002487(n))

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

where changed labels are

1 3 2 7 4 5 6 15 8 9 10 11 12 13 14

Thus, b(n) = A002487(n+1), b(a(n)) = A002487(n), n>0. - Yosu Yurramendi, Jul 07 2016

Scatter plot might be called "Cypress forest on a windy day". - Antti Karttunen, Nov 30 2021

A002487(n)/A002487(n+1), n > 0, runs through all the reduced nonnegative rationals exactly once. A002487 is the Stern's sequence. A002487(a(n)) = A002487(n+1) n>0 . - Yosu Yurramendi, Jul 07 2016

Let gpf(m) = A006530(m) and let phi(m) = A000010(m) for m in A005117.

Row n contains m in A005117 such that A000720(A006530(m)) = n, sorted such that phi(m)/m increases as k increases.

Let m be the squarefree kernel A007947(m') of m'. We only consider squarefree m since phi(m)/m = phi(m')/m'. Let prime p | n and prime q be a nondivisor of n.

Since m is squarefree, we might encode the multiplicities of its prime divisors in a positional notation M that is finite at n significant digits. For example, m = 42 can be encoded reverse(A067255(42)) = 1,0,1,1 = 7^1 * 5^0 * 3^1 * 2^1. It is necessary to reverse row m of A067255 (hereinafter simply A067255(m)) so as to preserve zeros in M = A067255(m) pertaining to small nondivisor primes q < p. The code M is a series of 0's and 1's since m is squarefree. Then it is clear that row n contains all m such that A067255(m) has n terms, and there are 2^(n - 1) possible terms for n >= 1.

We may use an approach that generates the binary expansion of the range 2^(n - 1) < M < 2^n - 1, or we may append 1 to the reversed (n - 1)-tuples of {1, 0} to achieve codes M -> m for each row n, which is tantamount to ordering according to A059894.

Originally it was thought that the codes M were in order of the latter algorithm, and we could avoid sorting. Observation shows that the m still require sorting by the function phi(m)/m indeed to be in increasing order in row n. Still, the latter approach is slightly more efficient than the former in generating the sequence.

The terms (n>0) may be written as a left-justified array with rows of length 2^m, m >= 0:

2,

3, 3,

5, 4, 5, 4,

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

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

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

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).

The terms (n>0) may also be written as a right-justified array with rows of length 2^m, m >= 0:

2,

3, 3,

5, 4, 5, 4,

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

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

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

Each column is an arithmetic sequence. The differences of the arithmetic sequences give the sequence A071585: a(2^(m+1)-1-k) - a(2^m-1-k) = A071585(k), m >= 0, 0 <= k < 2^m.

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. A245325(n)/A245326(n) is also an enumeration system of all positive rationals, 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), A086592 (A020650+A020651), A268087 (A162909+A162910).

A cell is colored white if the binary digit is 0 and a cell is colored black if the binary digit is 1. A biased rook on a white cell moves only to the left and otherwise moves only to the right.

For the definition of viabin number see comment in A290253.

The binary representation of a(n) is obtained by concatenating the binary representation of (n-1) and the reversed bit-flipped binary representation of (n-1) and then dropping the last bit. This suggests that it would have been more natural to index from 0. - Peter J. Taylor, Sep 24 2021