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

Inverse binomial transform of powers of 5 (A000351) preceded by 0. - Paul Barry, Apr 02 2003

Number of walks of length n between any two distinct vertices of the complete graph K_5. Example: a(2)=3 because the walks of length 2 between the vertices A and B of the complete graph ABCDE are: ACB, ADB, AEB. - Emeric Deutsch, Apr 01 2004

The terms of the sequence are the number of segments (sides) per iteration of the space-filling Peano-Hilbert curve. - Giorgio Balzarotti, Mar 16 2006

General form: k=4^n-k. Also: A001045, A078008, A097073, A115341, A015518, A054878. - Vladimir Joseph Stephan Orlovsky, Dec 11 2008

A further inverse binomial transform generates A015441. - Paul Curtz, Nov 01 2009

For n >= 2, a(n) equals the permanent of the (n-1) X (n-1) tridiagonal matrix with 3's along the central diagonal, and 2's along the subdiagonal and the superdiagonal. - John M. Campbell, Jul 19 2011

Pisano period lengths: 1, 1, 2, 2, 10, 2, 6, 2, 6, 10, 10, 2, 6, 6, 10, 2, 4, 6, 18, 10, ... - R. J. Mathar, Aug 10 2012

Sum_{i=0..m} (-1)^(m+i)*4^i, for m >= 0, gives the terms after 0. - Bruno Berselli, Aug 28 2013

The ratio a(n+1)/a(n) converges to 4 as n approaches infinity. - Felix P. Muga II, Mar 09 2014

This is the Lucas sequence U(P=3,Q=-4), and hence for n>=0, a(n+2)/a(n+1) equals the continued fraction 3 + 4/(3 + 4/(3 + 4/(3 + ... + 4/3))) with n 4's. - Greg Dresden, Oct 07 2019

For n > 0, gcd(a(n), a(n+1)) = 1. - Kengbo Lu, Jul 27 2020

Inverse is Riordan array (1, xc(x)) (A106566).

Triangle T(n,k), 0 <= k <= n, read by rows, given by [0, -1, 1, 0, 0, 0, 0, 0, 0, ...] DELTA [1, 0, 0, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938.

Modulo 2, this sequence gives A106344. - Philippe Deléham, Dec 18 2008

Coefficient array of the polynomials Chebyshev_U(n, sqrt(x)/2)*(sqrt(x))^n. - Paul Barry, Sep 28 2009

The triangle pertaining to this sequence has the property that every row, every column and every diagonal contains a nontrivial geometric progression. More interestingly every line joining any two elements contains a nontrivial geometric progression. - Amarnath Murthy, Jan 02 2002

Kappraff states (pp. 148-149): "I shall refer to this as Nicomachus' table since an identical table of numbers appeared in the Arithmetic of Nicomachus of Gerasa (circa 150 A.D.)" The table was rediscovered during the Italian Renaissance by Leon Battista Alberti, who incorporated the numbers in dimensions of his buildings and in a system of musical proportions. Kappraff states "Therefore a room could exhibit a 4:6 or 6:9 ratio but not 4:9. This ensured that ratios of these lengths would embody musical ratios". - Gary W. Adamson, Aug 18 2003

After Nichomachus and Alberti several Renaissance authors described this table. See for instance Pierre de la Ramée in 1569 (facsimile of a page of his Arithmetic Treatise in Latin in the links section). - Olivier Gérard, Jul 04 2013

The triangle sums, see A180662 for their definitions, link Nicomachus's table with eleven different sequences, see the crossrefs. It is remarkable that these eleven sequences can be described with simple elegant formulas. The mirror of this triangle is A175840. - Johannes W. Meijer, Sep 22 2010

The diagonal sums Sum_{k} T(n - k, k) give A167762(n + 2). - Michael Somos, May 28 2012

Where d(n) is the divisor count function, then d(T(i,j)) = A003991, the rows of which sum to the tetrahedral numbers A000292(n+1). For example, the sum of the divisors of row 4 of this triangle (i = 4), gives d(16) + d(24) + d(36) + d(54) + d(81) = 5 + 8 + 9 + 8 + 5 = 35 = A000292(5). In fact, where p and q are distinct primes, the aforementioned relationship to the divisor function and tetrahedral numbers can be extended to any triangle of numbers in which the i-th row is of form {p^(i-j)*q^j, 0<=j<=i}; i >= 0 (e.g., A003593, A003595). - Raphie Frank, Nov 18 2012, corrected Dec 07 2012

Sequence (or tree) generated by these rules: 1 is in S, and if x is in S, then 2*x and 3*x are in S, and duplicates are deleted as they occur; see A232559. - Clark Kimberling, Nov 28 2013

Partial sums of rows produce Stirling numbers of the 2nd kind: A000392(n+2) = Sum_{m=1..(n^2+n)/2} a(m). - Fred Daniel Kline, Sep 22 2014

A permutation of A003586. - L. Edson Jeffery, Sep 22 2014

Form a word of length i by choosing a (possibly empty) word on alphabet {0,1} then concatenating a word of length j on alphabet {2,3,4}. T(i,j) is the number of such words. - Geoffrey Critzer, Jun 23 2016

Form of Zorach additive triangle (see A035312) where each number is sum of west and northwest numbers, with the additional condition that each number is GCD of the two numbers immediately below it. - Michel Lagneau, Dec 27 2018

Construct a graph as follows: form the graph whose adjacency matrix is the tensor product of that of P_3 and [1,1;1,1], then add a loop at each of the extremity nodes. a(n-1) counts walks of length n between adjacent nodes. - Paul Barry, Nov 12 2004

The compositions of n in which each natural number is colored by one of p different colors are called p-colored compositions of n. For n >= 2, 9*a(n-2) equals the number of 9-colored compositions of n with all parts >= 2, such that no adjacent parts have the same color. - Milan Janjic, Nov 26 2011

Pisano period lengths: 1, 1, 6, 1, 24, 6, 16, 1, 6, 24, 110, 6, 56, 16, 24, 2, 16, 6, 60, 24, ... - R. J. Mathar, Aug 10 2012

Original name: Generalized Fibonacci numbers.

The compositions of n in which each natural number is colored by one of p different colors are called p-colored compositions of n. For n >= 2, 6*a(n-2) equals the number of 6-colored compositions of n with all parts >= 2, such that no adjacent parts have the same color. - Milan Janjic, Nov 26 2011

Pisano period lengths: 1, 3, 6, 6, 1, 6, 21, 12, 18, 3, 40, 6, 56, 21, 6, 24, 16, 18, 360, 6, .... - R. J. Mathar, Aug 10 2012

From Wolfdieter Lang, Jan 02 2024: (Start)

This sequence {a(n-1)}, with a(-1) = 0, appears in the formula for powers of phi21 := (1 + sqrt(21))/2 = A222134 = 2.791287..., together with A(n) = A365824(n), as phi21^n = A(n) + a(n-1)*phi21(n), for n >= 0.

Limit_{n->oo} a(n)/a(n-1) = phi21. (End)

One obtains A015523 through a binomial transform, and A197189 by shifting one place left (starting 1,1,8 with offset 0) followed by a binomial transform. - R. J. Mathar, Oct 11 2011

The compositions of n in which each positive integer is colored by one of p different colors are called p-colored compositions of n. For n>=2, 8*a(n-1) equals the number of 8-colored compositions of n, with all parts >=2, such that no adjacent parts have the same color. - Milan Janjic, Nov 26 2011

a(n+1) is the number of compositions (ordered partitions) of n into parts 1 and 2, where there are 7 sorts of part 2. - Joerg Arndt, Jan 16 2024

Pisano period lengths: 1, 3, 8, 6, 4, 24, 1, 6, 24, 12, 60, 24, 12, 3, 8, 6, 288, 24, 120, 12, ... - R. J. Mathar, Aug 10 2012

a(n) gives the length of the word obtained after n steps with the substitution rule 0->1^6, 1->(1^6)0, starting from 0. The number of 1's and 0's of this word is 6*a(n-1) and 6*a(n-2), resp.

Triangle T(n,k), 0 <= k <= n, read by rows, given by [0, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...] DELTA [1, 1, -1, 0, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938.

Row sums are the Jacobsthal numbers A001045(n+1) and column sums form Pell numbers A000129.

Maximal column entries: A038149 = {1, 1, 2, 5, 10, 22, ...}.

T(n,k) gives a convolved Fibonacci sequence (A001629, A001872, ...).

Triangle read by rows: T(n,n-k) is the number of ways to tile a 2 X n rectangle with k pieces of 2 X 2 tiles and n-2k pieces of 1 X 2 tiles (0 <= k <= floor(n/2)). - Philippe Deléham, Feb 17 2014

Diagonal sums are A013979(n). - Philippe Deléham, Feb 17 2014

T(n,k) is the number of ways to tile a 2 X n rectangle with k pieces of 2 X 2 tiles and 1 X 2 tiles. - Emeric Deutsch, Aug 14 2014

a(n) is also the coefficient of x^(n-1) in the bivariate Fibonacci polynomials F(n)(x,y)=xF(n-1)(x,y)+yF(n-2)(x,y),F(0)(x,y)=0,F(1)(x,y)=1, when we write 13x for x and -36x^2 for y. - Mario Catalani (mario.catalani(AT)unito.it), Dec 09 2002