cp's OEIS Frontend

Write n in base 2, n = sum b(i)*2^(i-1), then a(n) = sum b(i)*i. - Benoit Cloitre, Jun 09 2002

May be regarded as a triangular array read by rows, giving weighted sum of compositions in standard order. The standard order of compositions is given by A066099. - Franklin T. Adams-Watters, Nov 06 2006

Sum of all positive integer roots m_i of polynomial {m,k} - see link [Shevelev]; see also A264613. - Vladimir Shevelev, Dec 13 2015

Also the sum of binary indices of n, where a binary index of n (A048793) is any position of a 1 in its reversed binary expansion. For example, the binary indices of 11 are {1,2,4}, so a(11) = 7. - Gus Wiseman, May 22 2024

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

Numbers that are sums of distinct factorials (0! and 1! not treated as distinct).

Complement of A115945; A115944(a(n)) > 0; A115647 is a subsequence. - Reinhard Zumkeller, Feb 02 2006

A115944(a(n)) = 1. - Reinhard Zumkeller, Dec 04 2011

From Tilman Piesk, Jun 04 2012: (Start)

The inversion vector (compare A007623) of finite permutation a(n) (compare A055089, A195663) has only zeros and ones. Interpreted as a binary number it is 2*n (or n when the inversion vector is defined without the leading 0).

The inversion set of finite permutation a(n) interpreted as a binary number (compare A211362) is A211364(n).

(End)

The Zeckendorf expansion of n is obtained by repeatedly subtracting the largest Fibonacci number you can until nothing remains; for example, 100 = 89 + 8 + 3.

The Fibonacci successor to n is found by replacing each F_i in the Zeckendorf expansion by F_{i+1}; for example, the successor to 100 is 144 + 13 + 5 = 162.

If k appears, k + (rank of k) does not (10 is the 7th term in the sequence but 10 + 7 = 17 is not a term of the sequence). - Benoit Cloitre, Jun 18 2002

From Michele Dondi (bik.mido(AT)tiscalenet.it), Dec 30 2001: (Start)

a(n) = Sum_{k in A_n} F_{k+1}, where a(n)= Sum_{k in A_n} F_k is the (unique) expression of n as a sum of "noncontiguous" Fibonacci numbers (with index >= 2).

a(10^n) gives the first few digits of g = (sqrt(5)+1)/2.

The sequences given by b(n+1) = a(b(n)) obey the general recursion law of Fibonacci numbers. In particular the (sub)sequence (of a(-)) yielded by a starting value of 2=a(1), is the sequence of Fibonacci numbers >= 2. Starting points of all such subsequences are given by A035336.

a(n) = floor(phi*n+1/phi); phi = (sqrt(5)+1)/2. a(F_n)=F_{n+1} if F_n is the n-th Fibonacci number.

(End)

From Amiram Eldar, Sep 03 2022: (Start)

Numbers with an even number of trailing 1's in their dual Zeckendorf representation (A104326), i.e., numbers k such that A356749(k) is even.

The asymptotic density of this sequence is 1/phi (A094214). (End)

Replace 2^k with A000108(k+1) in binary expansion of n.

From Antti Karttunen, Jun 22 2014: (Start)

On the other hand, A244158 is similar, but replaces 10^k with A000108(k+1) in decimal expansion of n.

This sequence gives all k such that A014418(k) = A239903(k), which are precisely all nonnegative integers k whose representations in those two number systems contain no digits larger than 1. From this also follows that this is a subsequence of A244155.

(End)

The Zeckendorf and dual Zeckendorf representations both express a number n as a sum of distinct positive Fibonacci numbers; these distinct Fibonacci numbers can be encoded in binary (see A022290 for the decoding function):

- in the Zeckendorf representation (or greedy Fibonacci representation):

- Fibonacci numbers are as big as possible (see A035517),

- and the corresponding binary encoding, A003714(n),

cannot have two consecutive 1's;

- in the dual Zeckendorf representation (or lazy Fibonacci representation):

- Fibonacci numbers are as small as possible (see A112309),

- and the corresponding binary encoding, A003754(n+1),

cannot have two consecutive nonleading 0's.

See A356326 for a similar sequence.

The history of 1-D CA Rule 90 starting from the seed pattern 1 interpreted as Zeckendorffian expansion.

Also, product of distinct terms of A001566 and appropriate Fibonacci or Lucas numbers: a(n) = FL(n+2)Product(L(2^i)^bit(n,i),i=0..) Here L(2^i) = A001566 and FL(n) = n-th Fibonacci number if n even, n-th Lucas number if n odd. bit(n,i) is the i-th digit (0 or 1) in the binary expansion of n, with the least significant digit being bit(n,0).

This sequence looks on the Fibonacci sequence terms (from F_1 onwards) as an ordered set, considering F_1 and F_2 as distinct members mapped to the same integer value, namely 1. We run through all finite subsets, adding up the integer values from each subset (see table in the examples). In consequence, a number, k, occurs exactly A000121(k) times.

The definition is effectively an amendment of the definition of A022290 so that a(n) is a sum of Fibonacci terms starting with F_1 rather than F_2. If we took this a further step so that a(n) was a sum of Fibonacci terms starting with F_0, we would get this sequence with each term duplicated, that is 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, ... .

Doubling n increments the indices of the Fibonacci terms that sum to a(n), so a(2n) is in this sense a Fibonacci successor to a(n). When n is even, this sense matches the meaning of successor as used in A022342; however, for odd n, the generated successor of a(n) is A000201(a(n)) -- note, in this respect, that A000201 is a column in the extended Wythoff array (A287870).