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

Also numbers k such that the binary digital mean dm(2, k) = (Sum_{i=1..d} 2*d_i - 1) / (2*d) = 0, where d is the number of digits in the binary representation of k and d_i the individual digits. - Reikku Kulon, Sep 21 2008

From Reikku Kulon, Sep 29 2008: (Start)

Each run of values begins with 2^(2k + 1) + 2^(k + 1) - 2^k - 1. The initial values increase according to the sequence {2^(k - 1), 2^(k - 2), 2^(k - 3), ..., 2^(k - k)}.

After this, the values follow a periodic sequence of increases by successive powers of two with single odd values interspersed.

Each run ends with an odd increase followed by increases of {2^(k - k), ..., 2^(k - 2), 2^(k - 1), 2^k}, finally reaching 2^(2k + 2) - 2^(k + 1).

Similar behavior occurs in other bases. (End)

Numbers k such that A000120(k)/A070939(k) = 1/2. - Ctibor O. Zizka, Oct 15 2008

Subsequence of A053754; A179888 is a subsequence. - Reinhard Zumkeller, Jul 31 2010

A000120(a(n)) = A023416(a(n)); A037861(a(n)) = 0.

A001700 gives number of terms having length 2*n in binary representation: A001700(n-1) = #{m: A070939(a(m))=2*n}. - Reinhard Zumkeller, Jun 08 2011

The number of terms below 2^k is A079309(floor(k/2)) for k > 1. - Amiram Eldar, Nov 21 2020

The expression a(n) = B^n*Sum_{ k=0..n } binomial(2*k,k)/B^k gives A006134 for B=1, A082590 (B=2), A132310 (B=3), A002457 (B=4), A144635 (B=5). - N. J. A. Sloane, Jan 21 2009

T(n+1,1) from table A045912 of characteristic polynomial of negative Pascal matrix. - Michael Somos, Jul 24 2002

p divides a((p-3)/2) for p=11, 13, 23, 37, 47, 59, 61, 71, 73, 83, 97, 107, 109, 131, 157, 167, ...: A097933. Also primes congruent to {1, 2, 3, 11} mod 12 or primes p such that 3 is a square mod p (excluding 2 and 3) A038874. - Alexander Adamchuk, Jul 05 2006

Partial sums of the even central binomial coefficients. For p prime >=5, a(p-1) = 1 or -1 (mod p) according as p = 1 or -1 (mod 3) (see Pan and Sun link). - David Callan, Nov 29 2007

First column of triangle A187887. - Michel Marcus, Jun 23 2013

From Gus Wiseman, Apr 20 2023: (Start)

Also the number of nonempty subsets of {1,...,2n+1} with median n+1, where the median of a multiset is either the middle part (for odd length), or the average of the two middle parts (for even length). The odd/even-length cases are A000984 and A006134(n-1). For example, the a(0) = 1 through a(2) = 9 subsets are:

{1} {2} {3}

{1,3} {1,5}

{1,2,3} {2,4}

{1,3,4}

{1,3,5}

{2,3,4}

{2,3,5}

{1,2,4,5}

{1,2,3,4,5}

Alternatively, a(n-1) is the number of nonempty subsets of {1,...,2n-1} with median n.

(End)

The rank of a partition is its largest part minus the number of parts.

Number of partitions of n such that 2*(least part) > greatest part. - Clark Kimberling, Feb 16 2014

Also the number of partitions of n with the same median as maximum. These are conjugate to the partitions described above. For minimum instead of maximum we have A361860. - Gus Wiseman, Apr 23 2023

Total number of leaves in all rooted ordered trees with at most n edges. - Michael Somos, Feb 14 2006

Also: Number of UH-free Schroeder paths of semilength n with horizontal steps only at level less than two [see Yan]. - R. J. Mathar, May 24 2008

Hankel transform is A010892. - Paul Barry, Apr 28 2009

Binomial transform of A005773. - Philippe Deléham, Dec 13 2009

Number of vertices all of whose children are leaves in all ordered trees with n+1 edges. Example: a(3) = 15; for an explanation see David Callan's comment in A001519. - Emeric Deutsch, Feb 12 2015

The 4-term sequence 2,4,8,20 is the answer to the "Solitaire Army" problem, or checker-jumping puzzle. It is too short to have its own entry. See Conway et a;., Winning Ways, Vol. 2, pp. 715-717. - N. J. A. Sloane, Mar 01 2018

Number of subsets of {-n..n} with sum 0. Also the number of subsets of {0..2n} that are empty or have mean n. For median instead of mean we have twice A024718. - Gus Wiseman, Apr 23 2023

Comments from Alexander Adamchuk, Jul 02 2006: (Start)

Every a(n) is divisible by prime 2, a(n)/2 = A079309(n).

a(n) is divisible by prime 3 only for n=12,30,36,84,90,108,120,... A083096.

a(p) is divisible by p^2 for primes p=5,11,17,23,29,41,47,... Primes of form 6n-1. A007528.

a(p-1) is divisible by p^2 for primes p=7,13,19,31,37,43,... Primes of form 6n+1. A002476.

Every a(n) from a((p-1)/2) to a(p-1) is divisible by prime p for p=7,13,19,31,37,43,... Primes of form 6n+1. A002476.

Every a(n) from a((p^2-1)/2) to a(p^2-1) is divisible by prime p>3.

a(p^2-1), a(p^2-2) and a(p^2-3) are divisible by p^2 for prime p>3.

a(p^2-4) is divisible by p^2 for prime p>5.

a(p^2-5) is divisible by p^2 for prime p>7.

a(p^2-6) is divisible by p^2 for prime p>7.

a(p^2-7) is divisible by p^2 for prime p>11.

a(p^2-8) is divisible by p^2 for prime p>13.

a(p^3) is divisible by p^2 for prime 2 and prime p=5,11,... Primes of form 6n-1. A007528.

a(p^3-1) is divisible by p^2 for prime p=7,13,... Primes of form 6n+1. A002476.

a(p^4-1) is divisible by p^2 for prime p>3. (End)

Mod[ a(3^k), 9 ] = 1 for integer k>0. Smallest number k such that 2^n divides a(k) is k(n) = {1,2,2,11,11,46,46,707,707,707,...}. Smallest number k such that 3^n divides a(k) is k(n) = {12,822,2466,...}. a(2(p-1)/3) is divisible by p^2 for prime p = {7,13,19,31,37,43,61,...} = A002476 Primes of form 6n+1. Every a(n) from a(p^2-(p+1)/2) to a(p^2-1) is divisible by p^2 for prime p>3. Every a(n) from a((4p+3)(p-1)/6) to a((2p+3)(p-1)/3) is divisible by p^2 for prime p = {7,13,19,31,37,43,61,...} = A002476 Primes of form 6n+1. - Alexander Adamchuk, Jan 04 2007

From Gus Wiseman, Apr 19 2023: (Start)

Appears to be the number of nonempty subsets of {1,...,n} with median k, where the median of a multiset is either the middle part (for odd length), or the average of the two middle parts (for even length). For example, row n = 5 counts the following subsets:

{1} {2} {3} {4} {5}

{1,3} {1,5} {3,5}

{1,2,3} {2,4} {1,4,5}

{1,2,4} {1,3,4} {2,4,5}

{1,2,5} {1,3,5} {3,4,5}

{2,3,4}

{2,3,5}

{1,2,4,5}

{1,2,3,4,5}

Including half-steps gives A231147.

For mean instead of median we have A327481.

(End)

Or, constant term in expansion of Product_{k=1..n} (x^k + 1/x^k)^2. - N. J. A. Sloane, Jul 09 2008

Or, maximal coefficient of the polynomial (1+x)^2 * (1+x^2)^2 *...* (1+x^n)^2.

a(n) = A000302(n) - A181765(n).

From Gus Wiseman, Apr 18 2023: (Start)

Also the number of subsets of {1..2n} that are empty or have mean n. The a(0) = 1 through a(3) = 10 subsets are:

{} {} {} {}

{1} {2} {3}

{1,3} {1,5}

{1,2,3} {2,4}

{1,2,6}

{1,3,5}

{2,3,4}

{1,2,3,6}

{1,2,4,5}

{1,2,3,4,5}

Also the number of subsets of {-n..n} with no 0's but with sum 0. The a(0) = 1 through a(3) = 10 subsets are:

{} {} {} {}

{-1,1} {-1,1} {-1,1}

{-2,2} {-2,2}

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

{-3,1,2}

{-2,-1,3}

{-2,-1,1,2}

{-3,-1,1,3}

{-3,-2,2,3}

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

(End)

The Berndt-type sequence number 4 for the argument 2Pi/7 - see also A215007, A215008, A215143 and A215494.

We have a(n)=A079309(n) for n=1..6, and A079309(7)-a(7)=1.