cp's OEIS Frontend

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

Note: 1 might be a more natural starting offset for this sequence, although the identities concerning A053645 and A161511 would have to be changed. - Antti Karttunen, Oct 12 2009.

This can be regarded as an arrangement of the partitions, indexed by position in A125106. The partitions in a given row all have the same remaining partition when the largest part is removed; specifically, the partition indexed by the row number in A125106 (with row 0 having the empty partition remaining).

The first value on row n is A004760(n+1). The second value on each row is A004760(n+1) plus A062383(n); subsequent values increase by ever enlarging powers of two. Or equivalently, each subsequent value on the row after the first nonzero value is given by A004754(previous value on the same row).

A055941(r) tells how many terms the row r (>= 0) has been shifted rightward from its "natural position", i.e. with how many zeros that row has been prepended.

The number of (nonzero) entries in column k is A000041(k).

Multiplicative with a(2^e) = 2^e, a(p^e) = (-1)^((p^e-1)/2) otherwise. - Mitch Harris, May 17 2005

a(A091072(n)) > 0; a(A091067(n)) < 0. - Reinhard Zumkeller, Apr 28 2012

In the binary representation of n, clear everything left of the least significant 1 bit, and negate if the bit left of it was set originally. - Ralf Stephan, Aug 23 2013

This sequence is the trace of n in the minimal alternating binary representation of n (defined at A256696). - Clark Kimberling, Apr 07 2015

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)

"msb" = "most significant bit", A053644.

These encode lattice walks using steps (+1,+1) (= 1's in binary expansion) and (+1,-1) (= 0's in binary expansion) that start from the origin (0,0) and never "dive" under the "sea-level" y=0.

The number of such walks of length n (here: the terms of binary width n) is given by C(n,floor(n/2)) = A001405, which is based on the fact mentioned in Guy's article that the shallow diagonals of the Catalan triangle A009766 sum to A001405.

From Jason Kimberley, Feb 08 2013: (Start)

This sequence is a subsequence of A072601.

Define a map from this set onto the nonnegative integers as follows: set the output bit string to be empty, representing zero; process the input string from left to right; when 1 occurs, change the rightmost 0 in the output to 1; if there is no 0 in the output, prepend a 1; when 0 occurs in the input, change the rightmost 1 in the output to 1. The definition of this sequence ensures that we always have a 1 in the output when a 0 occurs in the input. We this map is onto by showing the restriction to the subset Asubsequence is onto. (End)

The binary representation of a(n) is the numeric representation of the left half of a symmetric balanced string of parentheses with "(" representing 1 and ")" representing 0 (see comments and examples in A001405). Some of the numbers in this sequence cannot be realized as the 1-0-pattern of the odd/even positions of 1's in any row n of A237048 that determines the parts and their widths in the symmetric representation of sigma(n), see A352696. - Hartmut F. W. Hoft, Mar 29 2022

A recurrence occurring in the analysis of a regular expression algorithm.

Define a sequence chf(n) of Christoffel words over an alphabet {-,+}:

chf(1) = '-',

chf(2*n+0) = negate(chf(n)),

chf(2*n+1) = negate(concatenate(chf(n),chf(n+1))).

Then the length of the chf(n) word is fusc(n) = A002487(n), the number of '-'-signs in the chf(n) word is c-fusc(n) = a(n) (the current sequence) and the number of '+'-signs in the chf(n) word is s-fusc(n) = A287730(n). See examples below.