cp's OEIS Frontend

Might be called the "Lichtenberg sequence" after Georg Christoph Lichtenberg, who discussed it in 1769 in connection with the Chinese Rings puzzle (baguenaudier). - Andreas M. Hinz, Feb 15 2017

Number of steps to change from a binary string of n 0's to n 1's using a Gray code. - Jon Stadler (jstadler(AT)coastal.edu)

Popular puzzles such as Spin-Out and The Brain Puzzler are based on the Gray binary system and require a(n) steps to complete for some number n.

Conjecture: {a(n)} also gives all j for which A048702(j) = A000217(j); i.e., if we take the a(n)-th triangular number (a(n)^2 + a(n))/2 and multiply it by 3, we get a(n)-th even-length binary palindrome A048701(a(n)) concatenated from a(n) and its reverse. E.g., a(4) = 10, which is 1010 in binary; the tenth triangular number is 55, and 55*3 = 165 = 10100101 in binary. - Antti Karttunen, circa 1999. (This has been now proved by Paul K. Stockmeyer in his arXiv:1608.08245 paper.) - Antti Karttunen, Aug 31 2016

Number of ways to tie a tie of n or fewer half turns, excluding mirror images. Also number of walks of length n or less on a triangular lattice with the following restrictions; given l, r and c as the lattice axes. 1. All steps are taken in the positive axis direction. 2. No two consecutive steps are taken on the same axis. 3. All walks begin with l. 4. All walks end with rlc or lrc. - Bill Blewett, Dec 21 2000

a(n) is the minimal number of vertices to be selected in a vertex-cover of the balanced tree B_n. - Sen-peng Eu, Jun 15 2002

A087117(a(n)) = A038374(a(n)) = 1 for n > 1; see also A090050. - Reinhard Zumkeller, Nov 20 2003

Intersection of A003754 and A003714; complement of A107907. - Reinhard Zumkeller, May 28 2005

Equivalently, numbers m whose binary representation is effectively, for some number k, both the lazy Fibonacci and the Zeckendorf representation of k (in which case k = A022290(m)). - Peter Munn, Oct 06 2022

a(n+1) gives row sums of Riordan array (1/(1-x), x(1+2x)). - Paul Barry, Jul 18 2005

Total number of initial 01's in all binary words of length n+1. Example: a(3) = 5 because the binary words of length 4 that start with 01 are (01)00, (01)(01), (01)10 and (01)11 and the total number of initial 01's is 5 (shown between parentheses). a(n) = Sum_{k >= 0} k*A119440(n+1, k). - Emeric Deutsch, May 19 2006

In Norway we call the 10-ring puzzle "strikketoy" or "knitwear" (see link). It takes 682 moves to free the two parts. - Hans Isdahl, Jan 07 2008

Equals A002450 and A020988 interlaced. - Zak Seidov, Feb 10 2008

For n > 1, let B_n = the complete binary tree with vertex set V where |V| = 2^n - 1. If VC is a minimum-size vertex cover of B_n, Sen-Peng Eu points out that a(n) = |VC|. It also follows that if IS = V\VC, a(n+1) = |IS|. - K.V.Iyer, Apr 13 2009

Starting with 1 and convolved with [1, 2, 2, 2, ...] = A000295. - Gary W. Adamson, Jun 02 2009

a(n) written in base 2 is sequence A056830(n). - Jaroslav Krizek, Aug 05 2009

This is the sequence A(0, 1; 1, 2; 1) of the family of sequences [a,b:c,d:k] considered by G. Detlefs, and treated as A(a,b;c,d;k) in the W. Lang link given below. - Wolfdieter Lang, Oct 18 2010

From Vladimir Shevelev, Jan 30 2012, Feb 13 2012: (Start)

1) Denote by {n, k} the number of permutations of 1, ..., n with the up-down index k (for definition, see comment in A203827). Then max_k{n, k} = {n, a(n)} = A000111(n).

2) a(n) is the minimal number > a(n-1) with the Hamming distance d_H(a(n-1), a(n)) = n. Thus this sequence is the Hamming analog of triangular numbers 0, 1, 3, 6, 10, ... (End)

From Hieronymus Fischer, Nov 22 2012: (Start)

Represented in binary form each term after the second one contains every previous term as a substring.

The terms a(2) = 2 and a(3) = 5 are the only primes. Proof: For even n we get a(n) = 2*(2^(2*n) - 1)/3, which shows that a(n) is even, too, and obviously a(n) > 2 for all even n > 2. For odd n we have a(n) = (2^(n+1) - 1)/3 = (2^((n+1)/2) - 1) * (2^((n+1)/2) + 1)/3. Evidently, at least one of these factors is divisible by 3, both are greater than 6, provided n > 3. Hence it follows that a(n) is composite for all odd n > 3.

Represented as a binary number, a(n+1) has exactly n prime substrings. Proof: Evidently, a(1) = 1_2 has zero and a(2) = 10_2 has 1 prime substring. Let n > 1. Written in binary, a(n+1) is 101010101...01 (if n + 1 is odd) and is 101010101...10 (if n + 1 is even) with n + 1 digits. Only the 2- and 3-digits substrings 10_2 (=2) and 101_2 (=5) are prime substrings. All the other substrings are nonprime since every substring is a previous term and all terms unequal to 2 and 5 are nonprime. For even n + 1, the number of prime substrings equal to 2 = 10_2 is (n+1)/2, and the number of prime substrings equal to 5 = 101_2 is (n-1)/2, makes a sum of n. For odd n + 1 we get n/2 for both, the number of 2's and 5's prime substrings, in any case, the sum is n. (End)

Number of different 3-colorings for the vertices of all triangulated planar polygons on a base with n+2 vertices if the colors of the two base vertices are fixed. - Patrick Labarque, Feb 09 2013

A090079(a(n)) = a(n) and A090079(m) <> a(n) for m < a(n). - Reinhard Zumkeller, Feb 16 2013

a(n) is the number of length n binary words containing at least one 1 and ending in an even number (possibly zero) of 0's. a(3) = 5 because we have: 001, 011, 100, 101, 111. - Geoffrey Critzer, Dec 15 2013

a(n) is the number of permutations of length n+1 having exactly one descent such that the first element of the permutation is an even number. - Ran Pan, Apr 18 2015

a(n) is the sequence of the last row of the Hadamard matrix H(2^n) obtained via Sylvester's construction: H(2) = [1,1;1,-1], H(2^n) = H(2^(n-1))*H(2), where * is the Kronecker product. - William P. Orrick, Jun 28 2015

Conjectured record values of A264784: a(n) = A264784(A155051(n-1)). - Reinhard Zumkeller, Dec 04 2015. (This is proved by Paul K. Stockmeyer in his arXiv:1608.08245 paper.) - Antti Karttunen, Aug 31 2016

Decimal representation of the x-axis, from the origin to the right edge, of the n-th stage of growth of the two-dimensional cellular automaton defined by "Rule 131", based on the 5-celled von Neumann neighborhood. See A279053 for references and links. - Robert Price, Dec 05 2016

For n > 4, a(n-2) is the second-largest number in row n of A127824. - Dmitry Kamenetsky, Feb 11 2017

Conjecture: a(n+1) is the number of compositions of n with two kinds of parts, n and n', where the order of the 1 and 1' does not matter. For n=2, a(3) = 5 compositions, enumerated as follows: 2; 2'; 1,1; 1',1 = 1',1; 1',1'. - Gregory L. Simay, Sep 02 2017

Conjecture proved by recognizing the appropriate g.f. is x/(1 - x)(1 - x)(1 - 2*x^2 - 2x^3 - ...) = x/(1 - 2*x - x^2 + 2x^3). - Gregory L. Simay, Sep 10 2017

a(n) = 2^(n-1) + 2^(n-3) + 2^(n-5) + ... a(2*k -1) = A002450(k) is the sum of the powers of 4. a(2*k) = 2*A002450(k). - Gregory L. Simay, Sep 27 2017

a(2*n) = n times the string [10] in binary representation, a(2*n+1) = n times the string [10] followed with [1] in binary representation. Example: a(7) = 85 = (1010101) in binary, a(8) = 170 = (10101010) in binary. - Ctibor O. Zizka, Nov 06 2018

Except for 0, these are the positive integers whose binary expansion has cuts-resistance 1. For the operation of shortening all runs by 1, cuts-resistance is the number of applications required to reach an empty word. Cuts-resistance 2 is A329862. - Gus Wiseman, Nov 27 2019

From Markus Sigg, Sep 14 2020: (Start)

Let s(k) be the length of the Collatz orbit of k, e.g. s(1) = 1, s(2) = 2, s(3) = 5. Then s(a(n)) = n+3 for n >= 3. Proof by induction: s(a(3)) = s(5) = 6 = 3+3. For odd n >= 5 we have s(a(n)) = s(4*a(n-2)+1) = s(12*a(n-2)+4)+1 = s(3*a(n-2)+1)+3 = s(a(n-2))+2 = (n-2)+3+2 = n+3, and for even n >= 4 this gives s(a(n)) = s(2*a(n-1)) = s(a(n-1))+1 = (n-1)+3+1 = n+3.

Conjecture: For n >= 3, a(n) is the second largest natural number whose Collatz orbit has length n+3. (End)

From Gary W. Adamson, May 14 2021: (Start)

With offset 1 the sequence equals the numbers of 1's from n = 1 to 3, 3 to 7, 7 to 15, ...; of A035263; as shown below:

..1 3 7 15...

..1 0 1 1 1 0 1 0 1 0 1 1 1 0 1...

..1.....2...........5......................10...; a(n) = Sum_(k=1..2n-1)A035263(k)

.....1...........2.......................5...; as to zeros.

..1's in the Tower of Hanoi game represent CW moves On disks in the pattern:

..0, 1, 2, 0, 1, 2, ... whereas even numbered disks move in the pattern:

..0, 2, 1, 0, 2, 1, ... (End)

Except for 0, numbers that are repunits in Gray-code representation (A014550). - Amiram Eldar, May 21 2021

From Gus Wiseman, Apr 20 2023: (Start)

Also the number of nonempty subsets of {1..n} with integer median, where the median of a multiset is the middle part in the odd-length case, and the average of the two middle parts in the even-length case. For example, the a(1) = 1 through a(4) = 10 subsets are:

{1} {1} {1} {1}

{2} {2} {2}

{3} {3}

{1,3} {4}

{1,2,3} {1,3}

{2,4}

{1,2,3}

{1,2,4}

{1,3,4}

{2,3,4}

The complement is counted by A005578.

For mean instead of median we have A051293, counting empty sets A327475.

For normal multisets we have A056450, strongly normal A361202.

For partitions we have A325347, strict A359907, complement A307683.

(End)

In binary form, because the sequence 10101_2... keeps repeating, one can represent a(n) = floor(0.(10)2 * 2^n). Because 0.(10)_2 = 0.(10)_2 * 4 - 2, then 0.(10)_2 = 2/3. Substituting this value in the first expression, we obtain a(n) = floor(2^(n+1)/3). This also agrees with the formula of Paul Barry, Oct 08 2005. - _Giovanni Ciriani, Mar 31 2025

Following Lenormand (2003), we define the "runs-resistance" of a finite list L to be the number of times the RUNS transformation must be applied to L in order to reduce L to a list with a single element.

Here it is immaterial whether we read the binary representation of n from left to right or right to left.

The RUNS transformation must be applied at least once, in order to obtain a list, so a(n) >= 1.

If the binary expansion of n is 1^b 0^c 1^d 0^e ..., then a(n) is the number whose binary expansion is 1^(b-1) 0^(c-1) 1^(d-1) 0^(e-1) .... Leading 0's are omitted, and if the result is the empty string, here we set a(n) = 0. See A319419 for a version which represents the empty string by -1.

Lenormand refers to this operation as planing ("raboter") the runs (or blocks) of the binary expansion.

A175046 expands the runs in a similar way, and a(A175046(n)) = A001477(n). - Andrew Weimholt, Sep 08 2018. In other words, this is a left inverse to A175046: A318921 o A175046 = A001477 = id on [0..oo). - M. F. Hasler, Sep 10 2018

Conjecture: For n in the range 2^k, ..., 2^(k+1)-1, the total value of a(n) appears to be 2*3^(k-1) - 2^(k-1) (see A027649), and so the average value of a(n) appears to be (3/2)^(k-1) - 1/2. - N. J. A. Sloane, Sep 25 2018

The above conjecture was proved by Doron Zeilberger on Nov 16 2018 (see link) and independently by Chai Wah Wu on Nov 18 2018 (see below). - N. J. A. Sloane, Nov 20 2018

From Chai Wah Wu, Nov 18 2018: (Start)

Conjecture is correct for k > 0. Proof: looking at the least significant 2 bits of n, it is easy to see that a(4n) = 2a(2n), a(4n+1) = a(2n), a(4n+2) = a(2n+1) and a(4n+3) = 2a(2n+1)+1. Define f(k) = Sum_{i=2^k..2^(k+1)-1} a(i), i.e. the sum ranges over all numbers with a (k+1)-bit binary expansion. Thus f(0) = a(1) = 0 and f(1) = a(2)+a(3) = 1. By summing over the recurrence relations for a(n), we get f(k+2) = Sum_{i=2^k..2^(k+1)-1} (f(4i) + f(4i+1) + f(4i+2) + f(4i+3)) = Sum_{i=2^k..2^(k+1)-1} (3a(2i) + 3a(2i+1) + 1) = 3*f(k+1) + 2^k. Solving this first order recurrence relation with the initial condition f(1) = 1 shows that f(k) = 2*3^(k-1)-2^(k-1) for k > 0.

(End)

For the operation of shortening all runs by 1, cuts-resistance is defined to be the number of applications required to reach an empty word.

A composition of n is a finite sequence of positive integers summing to n.

For the operation of shortening all runs by 1, cuts-resistance is defined as the number of applications required to reach an empty word.

Cuts-resistance is defined in A319416.

This triangle summarizes the data shown in A319420.

Conjecture (Sloane): Sum_{i = 1..n} i * T(n,i) = A189391(n).

This is the Riordan transform of A000217 (triangular numbers) with the Riordan matrix (of the Bell type) A053121 (inverse of the Chebyshev S Bell matrix). See the resulting formulae below. - Wolfdieter Lang, Feb 18 2017.

Conjecture: a(n) is also half the sum of the "cuts-resistance" (see A319416, A319420, A319421) of all binary vectors of length n (see Lenormand, page 4). - N. J. A. Sloane, Sep 20 2018

For the operation of taking the sequence of run-lengths of a finite sequence, runs-resistance is defined to be the number of applications required to reach a singleton.

For the operation of shortening all runs by 1, cuts-resistance is defined to be the number of applications required to reach an empty word.

The cuts-resistance of a vector is defined in A319416. The 2^n vectors of length n are taken in lexicographic order.

Note that here the vectors can begin with either 0 or 1, whereas in A319416 only vectors beginning with 1 are considered (since there we are considering binary representations of numbers).

Conjecture: The row sums, halved, appear to match A189391.

For the operation of shortening all runs by 1, cuts-resistance is defined to be the number of applications required to reach an empty word.