cp's OEIS Frontend

a(k) is the distance between k and the largest power of 2 not exceeding k, where k = n + 1. [Consider the sequence of even numbers <= k; after sending the first term to the last position delete all odd-indexed terms; the final term that remains after iterating the process is the a(k)-th even number.] - Lekraj Beedassy, May 26 2005

Triangle read by rows in which row n lists the first 2^(n-1) positive integers, n >= 1; see the example. - Omar E. Pol, Sep 10 2013

This sequence and A341392 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]

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]

From Yosu Yurramendi, Mar 21 2017: (Start)

This sequence is a self-inverse permutation of the integers. Except for fixed points 0, 1, it consists completely of 2-cycles: (2^(m+1)+k, 2^(m+1)+2^m+k), m >= 0, 0 <= k < 2^m.

A071766(a(n)) = A229742(n), A229742(a(n)) = A071766(n), n > 0.

A245325(a(n)) = A245326(n), A245326(a(n)) = A245325(n), n > 0.

A065190(a(n)) = a(A065190(n)), n > 0.

A054429(a(n)) = a(A054429(n)) = A117120(n), n > 0.

A258746(a(n)) = a(A258746(n)), n > 0.

A258996(a(n)) = a(A258996(n)), n > 0. (End)

A324337(a(n)) = A324338(n), A324338(a(n)) = A324337(n), n > 0. - Yosu Yurramendi, Nov 04 2019

The (n+1)-th row has A001316(n) terms and sums to n * A001316(n) / 2.

For any n >= 0 and k such that 0 <= k < A001316(n):

- if A000120(n) > 0 then T(n, 1) = A006519(n),

- if A000120(n) > 1 then T(n, 2) = 2^A285099(n),

- if A000120(n) > 0 then T(n, A001316(n)/2 - 1) = A053645(n),

- if A000120(n) > 0 then T(n, A001316(n)/2) = 2^A000523(n),

- if A000120(n) > 0 then T(n, A001316(n) - 2) = A129760(n),

- T(n, A001316(n) - 1) = n,

- the six previous relations correspond respectively (when applicable) to the second term, the third term, the pair of central terms, the penultimate term and the last term of a row,

- T(n, k) AND T(n, A001316(n) - k - 1) = 0,

- T(n, k) + T(n, A001316(n) - k - 1) = n,

- T(n, k) = k for any k < A006519(n+1),

- A000120(T(n, k)) = A000120(k).

If we plot (n, T(n,k)) then we obtain a skewed Sierpinski triangle (see Links section).

If interpreted as a flat sequence a(n) for n >= 0:

- a(n) = 0 iff n = A006046(k) for some k >= 0,

- a(n) = 1 iff n = A006046(2*k + 1) + 1 for some k >= 0,

- a(A006046(k) - 1) = k - 1 for any k > 0.

Permutation of natural numbers: let r(n,0)=n, r(n,k)=a(r(n,k-1)) for k>0, then r(n,floor(log_2(n))) = n and for n>1: r(n,floor(log_2(n))-1) = A080542(n).

Discarding their most significant bit, binary representations of numbers present in each cycle of this permutation form a distinct equivalence class of binary necklaces, thus there are A000031(n) separate cycles in each range [2^n .. (2^(n+1))-1] (for n >= 0) of this permutation. A256999 gives the largest number present in n's cycle. - Antti Karttunen, May 16 2015

Permutation of natural numbers with inverse = A080541: A080541(a(n)) = a(A080541(n)) = n;

let r(n,0)=n, r(n,k)=a(r(n,k-1)) for k>0, then r(n,floor(log_2(n))) = n and for n>1: r(n,floor(log_2(n))-1) = A080541(n).

Discarding their most significant bit, binary representations of numbers present in each cycle of this permutation form a distinct equivalence class of binary necklaces, thus there are A000031(n) separate cycles in each range [2^n .. (2^(n+1))-1] (for n >= 0) of this permutation. A256999 gives the largest number present in n's cycle. - Antti Karttunen, May 16 2015

These correspond to the maximal (lexicographically largest) representatives selected from each equivalence class of binary necklaces. See the last example.

Indexing starts from zero, because a(0) = 0 is a special case.

If k is a member then so also is 2*k, i.e., k with 0 appended to the end of its binary representation.

If k is a member then so also is A004755(k), i.e., k with 1 prepended to the front of its binary representation.

One obtains A065609 if one erases the most significant bit of each term [as A053645(a(n))] and then discards any zero-terms produced from the terms that originally were powers of two (A000079).

First differs from A328607 in lacking 108, with binary expansion 1101100. If we define a dual-necklace to be a finite sequence that is lexicographically maximal (not minimal) among all of its cyclic rotations, these are numbers whose binary expansion, without the most significant digit, is a dual-necklace. - Gus Wiseman, Nov 04 2019

From Davis Smith, May 09 2021: (Start)

For n > 2, a(n) cannot be a power of 2.

If A007088(n) (the binary expansion of n) contains a string of k zeros, then it contains A007088(2^m), where 0 <= m <= k, as a substring. Similarly, if A007088(n) contains a string of k ones, then it contains A007088(2^m - 1), where 1 <= m <= k. Strings of zeros and ones are the most compact way to have powers of 2 and powers of 2 minus 1 (respectively) as substrings in a binary expansion. This means that A007088(a(n)) will contain a string of A000523(n) ones and a string of A000523(n) zeros. The binary expansion of a(2^k - 1) will contain a string of k ones and a string of k - 1 zeros.

Conjecture: a(n) == 0 (mod A053644(n)), i.e., A007088(a(n)) ends with the longest string of zeros. It follows from this that a(2^k) = 2*a(2^k - 1). A conjecture related to this is that a(2^k - 1) = 2*a(2^k - 2) + 2^(k - 1), i.e., A007088(a(2^k - 1)) ends with the longest string of ones followed by the longest string of zeros. Ending with the longest string of ones followed by the longest string of zeros is not true for all A007088(a(n)), as some have a hiccup before starting their string of zeros, e.g., a(10), a(18), a(22), and a(34).

Conjecture: a(2^k + 1) = 2^(k + floor(log_2(a(2^k)))) + a(2^k), i.e., concatenate the binary expansion of 2^(k - 1) to the front of the binary expansion of a(2^k) in order to get the binary expansion of a(2^k + 1).

(End)

All terms belong to A261467. - Rémy Sigrist, May 11 2021

From Jon E. Schoenfield, Jun 03 2021: (Start)

Conjecture: the binary expansion of a(n) contains exactly ceiling(n/2) 1's iff 2^m - 7 <= n <= 2^m + 6 for some integer m >= 3. (See Links.)

Conjecture: for n > 1, the binary expansion of a(n) begins with that of 2^floor(log_2(n-1)) + 1.(End)

From Davis Smith, Jun 05 2021: (Start)

For a proof that a(n) == 2^floor(log_2(n)) (mod 2^(floor(log_2(n)) + 1)), see my second link (not the b-file). This also proves the conjecture from May 09 2021 which states that it is congruent to 0 (mod A053644(n)). A proof for the related conjecture would likely rely on an explanation of values of n such that a(n) is not congruent to (2^floor(log_2(n)) - 1)*2^floor(log_2(n)) (mod 2^(2*floor(log_2(n)))), i.e. the values of n such that A007088(a(n)) does not end with a string of floor(log_2(n)) ones followed immediately by a string of floor(log_2(n)) zeros. A proof for Jon E. Schoenfield's second conjecture on Jun 03 2021 would satisfy my more restricted second conjecture and it may follow necessarily from my proof, assuming that A007088(a(n)) must begin with either A007088(2^floor(log_2(n - 1)) + 1) or A007088(2^floor(log_2(n))). (End)