cp's OEIS Frontend

a(n) is the complement of the bit to the left of the least significant "1" in the binary expansion of n+1. E.g., n = 3, n+1 = 4 = 100_2, so a(3) = (complement of bit to left of 1) = 1. - Robert L. Brown, Nov 28 2001 [Adjusted to match offset by N. J. A. Sloane, Apr 15 2021]

To construct the sequence: start from 1,(..),0,(..),1,(..),0,(..),1,(..),0,(..),1,(..),0,... and fill undefined places with the sequence itself. - Benoit Cloitre, Jul 08 2007

This is the characteristic function of A091072 - 1. - Gary W. Adamson, Apr 11 2010

Turns (by 90 degrees) of the Heighway dragon which can be rendered as follows: [Init] Set n=0 and direction=0. [Draw] Draw a unit line (in the current direction). Turn left/right if a(n) is zero/nonzero respectively. [Next] Set n=n+1 and goto (draw). See fxtbook link below. - Joerg Arndt, Apr 15 2010

Sequence can be obtained by the L-system with rules L->L1R, R->L0R, 1->1, 0->0, starting with L, and then dropping all L's and R's (see example). - Joerg Arndt, Aug 28 2011

From Gary W. Adamson, Jun 20 2012: (Start)

One half of the infinite Farey tree can be mapped one-to-one onto A014577 since both sequences can be derived directly from the binary. First few terms are

1, 1, 0, 1, 1, 0, 0, 1, 1, 1, ...

1/2 2/3 1/3 3/4 3/5 2/5 1/4 4/5 5/7 5/8, ...

Infinite Farey tree fractions can be derived from the binary by appending a repeat of rightmost binary term to the right, then recording the number of runs to obtain the continued fraction representation. Example: 9 = 1001 which becomes 10011 which becomes [1,2,2] = 5/7. (End)

The sequence can be considered as a binomial transform operator for a target sequence S(n). Replace the first 1 in A014577 with the first term in S(n), then for successive "1" term in A014577, map the next higher term in S(n). If "0" in A014577, map the next lower term in S(n). Using the sequence S(n) = (1, 3, 5, 7, ...), we obtain (1), (3, 1), (3, 5, 3, 1), (3, 5, 7, 5, 3, 5, 3, 1), .... Then parse the terms into subsequences of 2^k terms, adding the terms in each string. We obtain (1, 4, 12, 32, 80, ...), the binomial transform of (1, 3, 5, 7, ...). The 8-bit string has one 1, three 5's, three 7's and one 1) as expected, or (1, 3, 3, 1) dot (1, 3, 5, 7). - Gary W. Adamson, Jun 24 2012

From Gary W. Adamson, May 29 2013: (Start)

The sequence can be generated directly from the lengths of continued fraction representations of fractions in one half of the Stern-Brocot tree (fractions between 0 and 1):

1/2

1/3 2/3

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

1/5 2/7 3/8 3/7 4/7 5/8 5/7 4/5

...

and their corresponding continued fraction representations are:

[2]

[3] [1,2]

[4] [2,2] [1,1,2] [1,3]

[5] [3,2] [2,1,2] [2,3] [1,1,3] [1,1,1,2] [1,2,2] [1,4]

...

Record the lengths by rows then reverse rows, getting:

1,

2, 1,

2, 3, 2, 1,

2, 3, 4, 3, 2, 3, 2, 1,

...

start with "1" and if the next term is greater than the current term, record a 1, otherwise 0; getting the present sequence, the Harter-Heighway dragon curve. (End)

The paper-folding word "110110011100100111011000..." can be created by concatenating the terms of a fixed point of the morphism or string substitution rule: 00 -> 1000, 01 -> 1001, 10 -> 1100 & 11 -> 1101, beginning with "11". - Robert G. Wilson v, Jun 11 2015

From Gary W. Adamson, Jun 04 2021: (Start)

Since the Heighway dragon is composed of right angles, it can be mapped with unit trajectories (Right = 1, Left = (-1), Up = i and Down = -i) on the complex plane where i = sqrt(-1). The initial (0th) iterate is chosen in this case to be the unit line from (0,0) to (-1,0). Then follow the directions below as indicated, resulting in a reflected variant of the dragon curve shown at the Eric Weisstein link. The conjectured system of complex plane trajectories is:

0 -1

1 -1, i

2 -1, i, 1, i

3 -1, i, 1, i, 1, -i, 1, i

4 -1, i, 1, i, 1, -i, 1, i, 1, -i, -1, -i, 1, -i, 1, i

...

The conjecture succeeds through the 4th iterate. It appears that to generate the (n+1)-th row, bring down the n-th row as the left half of row (n+1). For the right half of row (n+1), bring down the n-th row but change the signs of the first half of row n. For example, to get the complex plane instructions for the third iterate of the dragon curve, bring down (-1, i, 1, i) as the left half, and the right half is (1, -i, 1, i). (End)

From Gary W. Adamson, Jun 09 2021: (Start)

Partial sums of the iterate trajectories produce a sequence of complex addresses for unit segments. Partial sums of row 4 are: -1, (-1+i), i, 2i, (1+2i), (1+i), (2+i), (2+2i), (3+2i), (3+i), (2+i), 2, 3, (3-i), (4-i), 4. (zeros are omitted with terms of the form a+0i). The reflected variant of the dragon curve has the 0th iterate from (0,0) to (1,0), and the respective addresses simply change the signs of the real terms. (End)

The n-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of n, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.

The Stern polynomials B(n,t) are defined by B(0,t)=0, B(1,t)=1, B(2n,t)=tB(n,t), B(2n+1,t)=B(n+1,t)+B(n,t) for n>=1 (see S. Klavzar et al.).

Also number of hyperbinary representations of n-1 containing exactly k digits 1. A hyperbinary representation of a nonnegative integer n is a representation of n as a sum of powers of 2, each power being used at most twice. Example: row 9 of the triangle is 1,2,1; indeed the hyperbinary representations of 8 are 200 (2*2^2+0*2^1+0*2^0), 120 (1*2^2+2*2^1+0*2^0), 1000 (1*2^3+0*2^2+0*2^1+0*2^0) and 112 (1*2^2+1*2^1+2*1^0), having 0,1,1 and 2 digits 1, respectively (see S. Klavzar et al. Corollary 3).

Number of terms in row n is A277329(n) (= 1+A057526(n) for n >= 1).

Row sums yield A002487 (Stern's diatomic series).

T(2n+1,1) = A005811(n) = number of 1's in the standard Gray code of n (S. Klavzar et al. Theorem 8). T(4n+1,1)=1, T(4n+3,1)=0 (S. Klavzar et al., Lemma 5).

From Antti Karttunen, Oct 27 2016: (Start)

Number of nonzero terms on row n is A277314(n).

Number of odd terms on row n is A277700(n).

Maximal term on row n is A277315(n).

Product of nonzero terms on row n is A277325(n).

Number of times where row n and n+1 both contain nonzero term in the same position is A277327(n).

(End)

Every sequence can be uniquely split into a sequence of non-overlapping runs. For example, the runs of (2,2,1,1,1,3,2,2) are ((2,2),(1,1,1),(3),(2,2)), with sums (4,3,3,4).

Every sequence can be uniquely split into non-overlapping runs, read left-to-right. For example, the runs of (2,2,1,1,1,3,2,2) are ((2,2),(1,1,1),(3),(2,2)), with sums (4,3,3,4).

The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.

A044813 contains those positive integers that when written in binary, have all run-lengths (of both 0's and 1's) distinct.

A175414 contains those positive integers in A175413 that are not in A044813. (A175414 contains those positive integers that when written in binary, at least one run of 0's is the same length as one run of 1's, even though all run of 0 are of distinct length and all runs of 1's are of distinct length.)

Also numbers whose binary expansion has all distinct runs (not necessarily run-lengths). - Gus Wiseman, Feb 21 2022

Every sequence can be uniquely split into a sequence of non-overlapping runs. For example, the runs of (2,2,1,1,1,3,2,2) are ((2,2),(1,1,1),(3),(2,2)), with sums (4,3,3,4).

The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.

The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.

Every sequence can be uniquely split into a sequence of non-overlapping runs. For example, the runs of (2,2,1,1,1,3,2,2) are ((2,2),(1,1,1),(3),(2,2)), with sums (4,3,3,4).