cp's OEIS Frontend

If c(2) is even then c(k) = 1 for k >= 2*c(2), hence there is no even value in the sequence. If n is in the sequence, there exist an integer k(n) and an integer m(n) such that, for any k >= k(n), c(2k) - c(2k-1) = 2*m(n) and c(2k+1) - c(2k) = -m(n). Sometimes m(n) = (n-1)/2 but not always. If B(n) = a(n+1) - a(n) then B(n) = 2 or 4, but B(n) does not seem to follow any pattern.

Conjecture: a(n) = A036554(n)+1. - Vladeta Jovovic, Apr 01 2003

a(n) = A036554(n)+1 = A079523(n)+2. - Ralf Stephan, Jun 09 2003

Conjecture: this sequence gives the positions of 0's in the limiting 0-word of the morphism 0->11, 1->10, A285384. - Clark Kimberling, Apr 26 2017

Conjecture: This also gives the positions of the 1's in A328979. - N. J. A. Sloane, Nov 05 2019

Also union of numbers of the form 8*A121539(n)+k, 0<=k<8.

Generalization: the numbers n such that A010060(n)+A010060(n+2^m)=1 constitute the union of sequences {2^m*A121539(n)+k}, k=0,1,...,2^m-1.

Let A=Axxxxxx be any sequence. Denote by A^* the intersection of A and the union of sequences {4*A(n)+k}, k=-1,0,1,2. Then the present sequence is the union of A079523^* and A121539^*.

Conjecture. In every sequence of numbers n such that A010060(n)=A010060(n+k) for fixed odd k, the odious (A000069) and evil (A001969) terms alternate. [Vladimir Shevelev, Jul 31 2009]

Let A=Axxxxxx be any sequence from OEIS. Denote by A^* the intersection of the union of sequences {2*A(n)+j}, j=0,1, and the union of sequences {4*A(n)+k}, k=-2,-1,0,1. Then the sequence is the union of (A079523)^* and (A121539)^*.

Or union of intersection of A161639 and {A079523(n)-8} and intersection of A161673 and {A121539(n)-8}. In general, for a>=1, consider equations A010060(x+a)+A010060(x)=1, A010060(x+a)=A010060(x). Denote via B_a (C_a) the sequence of nonnegative solutions of the first (second) equation. Then we have recursions: B_(a+1) is the union of transactions 1) C_a and {A121539(n)-a}, 2) B_a and {A079523(n)-a}; C_(a+1) is the union of transactions 1) C_a and {A079523(n)-a}, 2) B_a and {A121539(n)-a}.

Conjecture. In every sequence of numbers n, such that A010060(n)=A010060(n+k), for fixed odd k, the odious (A000069) and evil (A001969) terms alternate. - Vladimir Shevelev, Jul 31 2009

This conjecture was actually proved in a later version of the Shevelev arxiv article cited below, and it can also easily be proved by the Walnut prover. - Jeffrey Shallit, Oct 12 2022

To construct the sequence: start from the Feigenbaum sequence A035263 = 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, ..., then change 1 -> 0, 2 and 0 -> 1, 1. - Philippe Deléham, Apr 18 2004

This Feigenbaum interpretation is equivalent to writing n+1 = binary "...1 00..00 x" where x is the least significant bit and zero or more 0's. If an odd number of 0's then a(n) = 1, otherwise a(n) = 2*x. In a similar way, if n-1 = binary "...0 11..11 x" with an odd number of 1's then a(n)=1 and otherwise a(n) = 2*x. - Kevin Ryde, Oct 17 2020

From Mikhail Kurkov, Mar 25 2021: (Start)

This sequence can be represented as a binary tree. Each child to the right is obtained by applying mex to the parent, and each child to the left is obtained by applying mex to the set formed by the parent and its second child:

( )

|

...................0...................

2 1

1......../ \........0 2......../ \........0

/ \ / \ / \ / \

/ \ / \ / \ / \

/ \ / \ / \ / \

2 0 2 1 1 0 2 1

1 0 2 1 1 0 2 0 2 0 2 1 1 0 2 0

etc.

Here mex means smallest nonnegative missing number.

Each parent and its two children form a set {0,1,2}. (End)

Each entry is 1, 2 or 3, associated with positions registered in A079523, A081706, and A161579, respectively.

Write the sequence of odious odd numbers above the sequence of evil odd numbers connecting all that are 2 apart:

1 7 11-13 19-21 25 31 35-37 41 47-49 55 59-61 67-69 73 79-81 87 91-93 97

3-5 9 15-17 23 27-29 33 39 43-45 51-53 57 63-65 71 75-77 83-85 89 95 99-

Remove the connected numbers:

1 7 25 31 41 55 73 87 97

9 23 33 39 57 71 89 95

Define these as "isolated".

The sequence is the smaller of the remaining pairs that are 2 apart.

The 1 is not a member since there is no change in parity between 1 and 7.

All of the differences between adjacent numbers in both the evil and odious sequences are either 2, 4 or 6, with 4 being the indicator that a transition in parity occurs. The program provided utilizes that fact to produce the sequence.

The sequence that includes all numbers along this path is A047522 (numbers congruent to {1,7} mod 8). This is also the same as the odd terms of A199398 (XOR of the first n odd numbers).

This sequence is similar to A044449 (numbers n such that string 1,3 occurs in the base 4 representation of n but not of n+1), but it contains additional terms. An example is 119. Its base 4 representation is 1313 while the base 4 representation of 120 is 1320. It may be that another workable definition of the sequence is -- numbers n in base 4 representation such that string 1,3 occurs one less time in n+1 than n, but I have not been able to check this.

The difference between the numbers in the sequence is always either 8 or 16, however there appears to be no recurring repetitions in it. Writing the 8 as a 0 and the 16 as a 1 (or dividing the difference pattern by 2 and subtracting a 1) produces a difference pattern of: 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 1... which is an infinite word.

A similar process writing Even Odious over Even Evils produces 6, 22, 30, 38, 54, 70... which is twice A131323 (Odd numbers n such that the binary expansion ends in an even number of 1's), with all numbers along the path given by A047451 (numbers congruent to {0,6} mod 8) and yields the same difference pattern which produces the same infinite word.