cp's OEIS Frontend

Although this is a list, it has offset 0 for both historical and mathematical reasons.

Numbers whose set of base-4 digits is a subset of {0,1}. - Ray Chandler, Aug 03 2004, corrected by M. F. Hasler, Oct 16 2018

Numbers k such that the sum of the base-2 digits of k = sum of the base-4 digits of k. - Clark Kimberling

Numbers having the same representation in both binary and negabinary (A039724). - Eric W. Weisstein

This sequence has many other interesting and useful properties. Every term k corresponds to a unique pair i,j with k = a(i) + 2*a(j) (i=A059905(n), j=A059906(n)) -- see A126684. Every list of numbers L = [L1,L2,L3,...] can be encoded uniquely by "recursive binary interleaving", where f(L) = a(L1) + 2*a(f([L2,L3,...])) with f([])=0. - Marc LeBrun, Feb 07 2001

This may be described concisely using the "rebase" notation b[n]q, which means "replace b with q in the expansion of n", thus "rebasing" n from base b into base q. The present sequence is 2[n]4. Many interesting operations (e.g., 10[n](1/10) = digit reverse, shifted) are nicely expressible this way. Note that q[n]b is (roughly) inverse to b[n]q. It's also natural to generalize the idea of "basis" so as to cover the likes of F[n]2, the so-called "fibbinary" numbers (A003714) and provide standard ready-made images of entities obeying other arithmetics, say like GF2[n]2 (e.g., primes = A014580, squares = the present sequence, etc.). - Marc LeBrun, Mar 24 2005

a(n) is also equal to the product n X n formed using carryless binary multiplication (A059729, A063010). - Henry Bottomley, Jul 03 2001

Numbers k such that A004117(k) is odd. - Pontus von Brömssen, Nov 25 2008

Fixed point of the morphism: 0 -> 01; 1 -> 45; 2 -> 89; ...; n -> (4n)(4n+1), starting from a(0)=0. - Philippe Deléham, Oct 22 2011

If n is even and present, so is n+1. - Robert G. Wilson v, Oct 24 2014

Also: interleave binary digits of n with 0's. (Equivalent to the "rebase" interpretation above.) - M. F. Hasler, Oct 16 2018

Named after the Austrian-Canadian mathematician Leo Moser (1921-1970) and the Dutch mathematician Nicolaas Govert de Bruijn (1918-2012). - Amiram Eldar, Jun 19 2021

Conjecture: The sums of distinct powers of k > 2 can be constructed as the following (k-1)-ary rooted tree. For each n the tree grows and a(n) is then the total number of nodes. For n = 1, the root of the tree is added. For n > 1, if n is odd one leaf of depth n-2 grows one child. If n is even all leaves of depth >= (n - 1 - A000225(A001511(n/2))) grow the maximum number of children. An illustration is provided in the links. - John Tyler Rascoe, Oct 09 2022

A toothpick is a copy of the closed interval [-1,1]. (In the paper, we take it to be a copy of the unit interval [-1/2, 1/2].)

We start at stage 0 with no toothpicks.

At stage 1 we place a toothpick in the vertical direction, anywhere in the plane.

In general, given a configuration of toothpicks in the plane, at the next stage we add as many toothpicks as possible, subject to certain conditions:

- Each new toothpick must lie in the horizontal or vertical directions.

- Two toothpicks may never cross.

- Each new toothpick must have its midpoint touching the endpoint of exactly one existing toothpick.

The sequence gives the number of toothpicks after n stages. A139251 (the first differences) gives the number added at the n-th stage.

Call the endpoint of a toothpick "exposed" if it does not touch any other toothpick. The growth rule may be expressed as follows: at each stage, new toothpicks are placed so their midpoints touch every exposed endpoint.

This is equivalent to a two-dimensional cellular automaton. The animations show the fractal-like behavior.

After 2^k - 1 steps, there are 2^k exposed endpoints, all located on two lines perpendicular to the initial toothpick. At the next step, 2^k toothpicks are placed on these lines, leaving only 4 exposed endpoints, located at the corners of a square with side length 2^(k-1) times the length of a toothpick. - M. F. Hasler, Apr 14 2009 and others. For proof, see the Applegate-Pol-Sloane paper.

If the third condition in the definition is changed to "- Each new toothpick must have at exactly one of its endpoints touching the midpoint of an existing toothpick" then the same sequence is obtained. The configurations of toothpicks are of course different from those in the present sequence. But if we start with the configurations of the present sequence, rotate each toothpick a quarter-turn, and then rotate the whole configuration a quarter-turn, we obtain the other configuration.

If the third condition in the definition is changed to "- Each new toothpick must have at least one of its endpoints touching the midpoint of an existing toothpick" then the sequence n^2 - n + 1 is obtained, because there are no holes left in the grid.

A "toothpick" of length 2 can be regarded as a polyedge with 2 components, both on the same line. At stage n, the toothpick structure is a polyedge with 2*a(n) components.

Conjecture: Consider the rectangles in the sieve (including the squares). The area of each rectangle (A = b*c) and the edges (b and c) are powers of 2, but at least one of the edges (b or c) is <= 2.

In the toothpick structure, if n >> 1, we can see some patterns that look like "canals" and "diffraction patterns". For example, see the Applegate link "A139250: the movie version", then enter n=1008 and click "Update". See also "T-square (fractal)" in the Links section. - Omar E. Pol, May 19 2009, Oct 01 2011

From Benoit Jubin, May 20 2009: The web page "Gallery" of Chris Moore (see link) has some nice pictures that are somewhat similar to the pictures of the present sequence. What sequences do they correspond to?

For a connection to Sierpiński triangle and Gould's sequence A001316, see the leftist toothpick triangle A151566.

Eric Rowland comments on Mar 15 2010 that this toothpick structure can be represented as a 5-state CA on the square grid. On Mar 18 2010, David Applegate showed that three states are enough. See links.

Equals row sums of triangle A160570 starting with offset 1; equivalent to convolving A160552: (1, 1, 3, 1, 3, 5, 7, ...) with (1, 2, 2, 2, ...). Equals A160762: (1, 0, 2, -2, 2, 2, 2, -6, ...) convolved with 2*n - 1: (1, 3, 5, 7, ...). Starting with offset 1 equals A151548: [1, 3, 5, 7, 5, 11, 17, 15, ...] convolved with A078008 signed (A151575): [1, 0, 2, -2, 6, -10, 22, -42, 86, -170, 342, ...]. - Gary W. Adamson, May 19 2009, May 25 2009

For a three-dimensional version of the toothpick structure, see A160160. - Omar E. Pol, Dec 06 2009

From Omar E. Pol, May 20 2010: (Start)

Observation about the arrangement of rectangles:

It appears there is a nice pattern formed by distinct modular substructures: a central cross surrounded by asymmetrical crosses (or "hidden crosses") of distinct sizes and also by "nuclei" of crosses.

Conjectures: after 2^k stages, for k >= 2, and for m = 1 to k - 1, there are 4^(m-1) substructures of size s = k - m, where every substructure has 4*s rectangles. The total number of substructures is equal to (4^(k-1)-1)/3 = A002450(k-1). For example: If k = 5 (after 32 stages) we can see that:

a) There is a central cross, of size 4, with 16 rectangles.

b) There are four hidden crosses, of size 3, where every cross has 12 rectangles.

c) There are 16 hidden crosses, of size 2, where every cross has 8 rectangles.

d) There are 64 nuclei of crosses, of size 1, where every nucleus has 4 rectangles.

Hence the total number of substructures after 32 stages is equal to 85. Note that in every arm of every substructure, in the potential growth direction, the length of the rectangles are the powers of 2. (See illustrations in the links. See also A160124.) (End)

It appears that the number of grid points that are covered after n-th stage of the toothpick structure, assuming the toothpicks have length 2*k, is equal to (2*k-2)*a(n) + A147614(n), k > 0. See the formulas of A160420 and A160422. - Omar E. Pol, Nov 13 2010

Version "Gullwing": on the semi-infinite square grid, at stage 1, we place a horizontal "gull" with its vertices at [(-1, 2), (0, 1), (1, 2)]. At stage 2, we place two vertical gulls. At stage 3, we place four horizontal gulls. a(n) is also the number of gulls after n-th stage. For more information about the growth of gulls see A187220. - Omar E. Pol, Mar 10 2011

From Omar E. Pol, Mar 12 2011: (Start)

Version "I-toothpick": we define an "I-toothpick" to consist of two connected toothpicks, as a bar of length 2. An I-toothpick with length 2 is formed by two toothpicks with length 1. The midpoint of an I-toothpick is touched by its two toothpicks. a(n) is also the number of I-toothpicks after n-th stage in the I-toothpick structure. The I-toothpick structure is essentially the original toothpick structure in which every toothpick is replaced by an I-toothpick. Note that in the physical model of the original toothpick structure the midpoint of a wooden toothpick of the new generation is superimposed on the endpoint of a wooden toothpick of the old generation. However, in the physical model of the I-toothpick structure the wooden toothpicks are not overlapping because all wooden toothpicks are connected by their endpoints. For the number of toothpicks in the I-toothpick structure see A160164 which also gives the number of gullwing in a gullwing structure because the gullwing structure of A160164 is equivalent to the I-toothpick structure. It also appears that the gullwing sequence A187220 is a supersequence of the original toothpick sequence A139250 (this sequence).

For the connection with the Ulam-Warburton cellular automaton see the Applegate-Pol-Sloane paper and see also A160164 and A187220.

(End)

A version in which the toothpicks are connected by their endpoints: on the semi-infinite square grid, at stage 1, we place a vertical toothpick of length 1 from (0, 0). At stage 2, we place two horizontal toothpicks from (0,1), and so on. The arrangement looks like half of the I-toothpick structure. a(n) is also the number of toothpicks after the n-th. - Omar E. Pol, Mar 13 2011

Version "Quarter-circle" (or Q-toothpick): a(n) is also the number of Q-toothpicks after the n-th stage in a Q-toothpick structure in the first quadrant. We start from (0,1) with the first Q-toothpick centered at (1, 1). The structure is asymmetric. For a similar structure but starting from (0, 0) see A187212. See A187210 and A187220 for more information. - Omar E. Pol, Mar 22 2011

Version "Tree": It appears that a(n) is also the number of toothpicks after the n-th stage in a toothpick structure constructed following a special rule: the toothpicks of the new generation have length 4 when they are placed on the infinite square grid (note that every toothpick has four components of length 1), but after every stage, one (or two) of the four components of every toothpick of the new generation is removed, if such component contains an endpoint of the toothpick and if such endpoint is touching the midpoint or the endpoint of another toothpick. The truncated endpoints of the toothpicks remain exposed forever. Note that there are three sizes of toothpicks in the structure: toothpicks of lengths 4, 3 and 2. A159795 gives the total number of components in the structure after the n-th stage. A153006 (the corner sequence of the original version) gives 1/4 of the total of components in the structure after the n-th stage. - Omar E. Pol, Oct 24 2011

From Omar E. Pol, Sep 16 2012: (Start)

It appears that a(n)/A147614(n) converges to 3/4.

It appears that a(n)/A160124(n) converges to 3/2.

It appears that a(n)/A139252(n) converges to 3.

Also:

It appears that A147614(n)/A160124(n) converges to 2.

It appears that A160124(n)/A139252(n) converges to 2.

It appears that A147614(n)/A139252(n) converges to 4.

(End)

It appears that a(n) is also the total number of ON cells after n-th stage in a quadrant of the structure of the cellular automaton described in A169707 plus the total number of ON cells after n+1 stages in a quadrant of the mentioned structure, without its central cell. See the illustration of the NW-NE-SE-SW version in A169707. See also the connection between A160164 and A169707. - Omar E. Pol, Jul 26 2015

On the infinite Cairo pentagonal tiling consider the symmetric figure formed by two non-adjacent pentagons connected by a line segment joining two trivalent nodes. At stage 1 we start with one of these figures turned ON. The rule for the next stages is that the concave part of the figures of the new generation must be adjacent to the complementary convex part of the figures of the old generation. a(n) gives the number of figures that are ON in the structure after n-th stage. A160164(n) gives the number of ON cells in the structure after n-th stage. - Omar E. Pol, Mar 29 2018

From Omar E. Pol, Mar 06 2019: (Start)

The "word" of this sequence is "ab". For further information about the word of cellular automata see A296612.

Version "triangular grid": a(n) is also the total number of toothpicks of length 2 after n-th stage in the toothpick structure on the infinite triangular grid, if we use only two of the three axes. Otherwise, if we use the three axes, so we have the sequence A296510 which has word "abc".

The normal toothpick structure can be considered a superstructure of the Ulam-Warburton celular automaton since A147562(n) equals here the total number of "hidden crosses" after 4*n stages, including the central cross (beginning to count the crosses when their "nuclei" are totally formed with 4 quadrilaterals). Note that every quadrilateral in the structure belongs to a "hidden cross".

Also, the number of "hidden crosses" after n stages equals the total number of "flowers with six petals" after n-th stage in the structure of A323650, which appears to be a "missing link" between this sequence and A147562.

Note that the location of the "nuclei of the hidden crosses" is very similar (essentially the same) to the location of the "flowers with six petals" in the structure of A323650 and to the location of the "ON" cells in the version "one-step bishop" of the Ulam-Warburton cellular automaton of A147562. (End)

From Omar E. Pol, Nov 27 2020: (Start)

The simplest substructures are the arms of the hidden crosses. Each closed region (square or rectangle) of the structure belongs to one of these arms. The narrow arms have regions of area 1, 2, 4, 8, ... The broad arms have regions of area 2, 4, 8, 16 , ... Note that after 2^k stages, with k >= 3, the narrow arms of the main hidden crosses in each quadrant frame the size of the toothpick structure after 2^(k-1) stages.

Another kind of substructure could be called "bar chart" or "bar graph". This substructure is formed by the rectangles and squares of width 2 that are adjacent to any of the four sides of the toothpick structure after 2^k stages, with k >= 2. The height of these successive regions gives the first 2^(k-1) - 1 terms from A006519. For example: if k = 5 the respective heights after 32 stages are [1, 2, 1, 4, 1, 2, 1, 8, 1, 2, 1, 4, 1, 2, 1]. The area of these successive regions gives the first 2^(k-1) - 1 terms of A171977. For example: if k = 5 the respective areas are [2, 4, 2, 8, 2, 4, 2, 16, 2, 4, 2, 8, 2, 4, 2].

For a connection to Mersenne primes (A000668) and perfect numbers (A000396) see A153006.

For a representation of the Wagstaff primes (A000979) using the toothpick structure see A194810.

For a connection to stained glass windows and a hidden curve see A336532. (End)

It appears that the graph of a(n) bears a striking resemblance to the cumulative distribution function F(x) for X the random variable taking values in [0,1], where the binary expansion of X is given by a sequence of independent coin tosses with probability 3/4 of being 1 at each bit. It appears that F(n/2^k)*(2^(2k+1)+1)/3 approaches a(n) for k large. - James Coe, Jan 10 2022

For n > 0, a(n) is the degree (n-1) "numbral" power of 5 (see A048888 for the definition of numbral arithmetic). Example: a(3) = 21, since the numbral square of 5 is 5(*)5 = 101(*)101(base 2) = 101 OR 10100 = 10101(base 2) = 21, where the OR is taken bitwise. - John W. Layman, Dec 18 2001

a(n) is composite for all n > 2 and has factors x, (3*x + 2*(-1)^n) where x belongs to A001045. In binary the terms greater than 0 are 1, 101, 10101, 1010101, etc. - John McNamara, Jan 16 2002

Number of n X 2 binary arrays with path of adjacent 1's from upper left corner to right column. - R. H. Hardin, Mar 16 2002

The Collatz-function iteration started at a(n), for n >= 1, will end at 1 after 2*n+1 steps. - Labos Elemer, Sep 30 2002 [corrected by Wolfdieter Lang, Aug 16 2021]

Second binomial transform of A001045. - Paul Barry, Mar 28 2003

All members of sequence are also generalized octagonal numbers (A001082). - Matthew Vandermast, Apr 10 2003

Also sum of squares of divisors of 2^(n-1): a(n) = A001157(A000079(n-1)), for n > 0. - Paul Barry, Apr 11 2003

Binomial transform of A000244 (with leading zero). - Paul Barry, Apr 11 2003

Number of walks of length 2n between two vertices at distance 2 in the cycle graph C_6. For n = 2 we have for example 5 walks of length 4 from vertex A to C: ABABC, ABCBC, ABCDC, AFABC and AFEDC. - Herbert Kociemba, May 31 2004

Also number of walks of length 2n + 1 between two vertices at distance 3 in the cycle graph C_12. - Herbert Kociemba, Jul 05 2004

a(n+1) is the number of steps that are made when generating all n-step random walks that begin in a given point P on a two-dimensional square lattice. To make one step means to mark one vertex on the lattice (compare A080674). - Pawel P. Mazur (Pawel.Mazur(AT)pwr.wroc.pl), Mar 13 2005

a(n+1) is the sum of square divisors of 4^n. - Paul Barry, Oct 13 2005

a(n+1) is the decimal number generated by the binary bits in the n-th generation of the Rule 250 elementary cellular automaton. - Eric W. Weisstein, Apr 08 2006

a(k) = [M^k]2,1, where M is the 3 X 3 matrix defined as follows: M = [1, 1, 1; 1, 3, 1; 1, 1, 1]. - _Simone Severini, Jun 11 2006

a(n-1) / a(n) = percentage of wasted storage if a single image is stored as a pyramid with a each subsequent higher resolution layer containing four times as many pixels as the previous layer. n is the number of layers. - Victor Brodsky (victorbrodsky(AT)gmail.com), Jun 15 2006

k is in the sequence if and only if C(4k + 1, k) (A052203) is odd. - Paul Barry, Mar 26 2007

This sequence also gives the number of distinct 3-colorings of the odd cycle C(2*n - 1). - Keith Briggs, Jun 19 2007

All numbers of the form m*4^m + (4^m-1)/3 have the property that they are sums of two squares and also their indices are the sum of two squares. This follows from the identity m*4^m + (4^m-1)/3 = 4(4(..4(4m + 1) + 1) + 1) + 1 ..) + 1. - Artur Jasinski, Nov 12 2007

For n > 0, terms are the numbers that, in base 4, are repunits: 1_4, 11_4, 111_4, 1111_4, etc. - Artur Jasinski, Sep 30 2008

Let A be the Hessenberg matrix of order n, defined by: A[1, j] = 1, A[i, i] := 5, (i > 1), A[i, i - 1] = -1, and A[i, j] = 0 otherwise. Then, for n >= 1, a(n) = charpoly(A,1). - Milan Janjic, Jan 27 2010

This is the sequence A(0, 1; 3, 4; 2) = A(0, 1; 4, 0; 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

6*a(n) + 1 is every second Mersenne number greater than or equal to M3, hence all Mersenne primes greater than M2 must be a 6*a(n) + 1 of this sequence. - Roderick MacPhee, Nov 01 2010

Smallest number having alternating bit sum n. Cf. A065359.

For n = 1, 2, ..., the last digit of a(n) is 1, 5, 1, 5, ... . - Washington Bomfim, Jan 21 2011

Rule 50 elementary cellular automaton generates this sequence. This sequence also appears in the second column of array in A173588. - Paul Muljadi, Jan 27 2011

Sequence found by reading the line from 0, in the direction 0, 5, ... and the line from 1, in the direction 1, 21, ..., in the square spiral whose edges are the Jacobsthal numbers A001045 and whose vertices are the numbers A000975. These parallel lines are two semi-diagonals in the spiral. - Omar E. Pol, Sep 10 2011

a(n), n >= 1, is also the inverse of 3, denoted by 3^(-1), Modd(2^(2*n - 1)). For Modd n see a comment on A203571. E.g., a(2) = 5, 3 * 5 = 15 == 1 (Modd 8), because floor(15/8) = 1 is odd and -15 == 1 (mod 8). For n = 1 note that 3 * 1 = 3 == 1 (Modd 2) because floor(3/2) = 1 and -3 == 1 (mod 2). The inverse of 3 taken Modd 2^(2*n) coincides with 3^(-1) (mod 2^(2*n)) given in A007583(n), n >= 1. - Wolfdieter Lang, Mar 12 2012

If an AVL tree has a leaf at depth n, then the tree can contain no more than a(n+1) nodes total. - Mike Rosulek, Nov 20 2012

Also, this is the Lucas sequence V(5, 4). - Bruno Berselli, Jan 10 2013

Also, for n > 0, a(n) is an odd number whose Collatz trajectory contains no odd number other than n and 1. - Jayanta Basu, Mar 24 2013

Sum_{n >= 1} 1/a(n) converges to (3*(log(4/3) - QPolyGamma[0, 1, 1/4]))/log(4) = 1.263293058100271... = A321873. - K. G. Stier, Jun 23 2014

Consider n spheres in R^n: the i-th one (i=1, ..., n) has radius r(i) = 2^(1-i) and the coordinates of its center are (0, 0, ..., 0, r(i), 0, ..., 0) where r(i) is in position i. The coordinates of the intersection point in the positive orthant of these spheres are (2/a(n), 4/a(n), 8/a(n), 16/a(n), ...). For example in R^2, circles centered at (1, 0) and (0, 1/2), and with radii 1 and 1/2, meet at (2/5, 4/5). - Jean M. Morales, May 19 2015

From Peter Bala, Oct 11 2015: (Start)

a(n) gives the values of m such that binomial(4*m + 1,m) is odd. Cf. A003714, A048716, A263132.

2*a(n) = A020988(n) gives the values of m such that binomial(4*m + 2, m) is odd.

4*a(n) = A080674(n) gives the values of m such that binomial(4*m + 4, m) is odd. (End)

Collatz Conjecture Corollary: Except for powers of 2, the Collatz iteration of any positive integer must eventually reach a(n) and hence terminate at 1. - Gregory L. Simay, May 09 2016

Number of active (ON, black) cells at stage 2^n - 1 of the two-dimensional cellular automaton defined by "Rule 598", based on the 5-celled von Neumann neighborhood. - Robert Price, May 16 2016

From Luca Mariot and Enrico Formenti, Sep 26 2016: (Start)

a(n) is also the number of coprime pairs of polynomials (f, g) over GF(2) where both f and g have degree n + 1 and nonzero constant term.

a(n) is also the number of pairs of one-dimensional binary cellular automata with linear and bipermutive local rule of neighborhood size n+1 giving rise to orthogonal Latin squares of order 2^m, where m is a multiple of n. (End)

Except for 0, 1 and 5, all terms are Brazilian repunits numbers in base 4, and so belong to A125134. For n >= 3, all these terms are composite because a(n) = {(2^n-1) * (2^n + 1)}/3 and either (2^n - 1) or (2^n + 1) is a multiple of 3. - Bernard Schott, Apr 29 2017

Given the 3 X 3 matrix A = [2, 1, 1; 1, 2, 1; 1, 1, 2] and the 3 X 3 unit matrix I_3, A^n = a(n)(A - I_3) + I_3. - Nicolas Patrois, Jul 05 2017

The binary expansion of a(n) (n >= 1) consists of n 1's alternating with n - 1 0's. Example: a(4) = 85 = 1010101_2. - Emeric Deutsch, Aug 30 2017

a(n) (n >= 1) is the viabin number of the integer partition [n, n - 1, n - 2, ..., 2, 1] (for the definition of viabin number see comment in A290253). Example: a(4) = 85 = 1010101_2; consequently, the southeast border of the Ferrers board of the corresponding integer partition is ENENENEN, where E = (1, 0), N = (0, 1); this leads to the integer partition [4, 3, 2, 1]. - Emeric Deutsch, Aug 30 2017

Numbers whose binary and Gray-code representations are both palindromes (i.e., intersection of A006995 and A281379). - Amiram Eldar, May 17 2021

Starting with n = 1 the sequence satisfies {a(n) mod 6} = repeat{1, 5, 3}. - Wolfdieter Lang, Jan 14 2022

Terms >= 5 are those q for which the multiplicative order of 2 mod q is floor(log_2(q)) + 2 (and which is 1 more than the smallest possible order for any q). - Tim Seuré, Mar 09 2024

The order of 2 modulo a(n) is 2*n for n >= 2. - Joerg Arndt, Mar 09 2024

Number of ordered trees with n edges and height at most 4.

Number of palindromic structures using a maximum of three different symbols. - Marks R. Nester

Number of compositions of all even natural numbers into n parts <= 2 (0 is counted as a part), see example. - Adi Dani, May 14 2011

Consider the mapping f(a/b) = (a + 2*b)/(2*a + b). Taking a = 1, b = 2 to start with, and carrying out this mapping repeatedly on each new (reduced) rational number gives the sequence 1/2, 4/5, 13/14, 40/41, ... converging to 1. The sequence contains the denominators = (3^n+1)/2. The same mapping for N, i.e., f(a/b) = (a + N*b)/(a+b) gives fractions converging to N^(1/2). - Amarnath Murthy, Mar 22 2003

Second binomial transform of the expansion of cosh(x). - Paul Barry, Apr 05 2003

The sequence (1, 1, 2, 5, ...) = 3^n/6 + 1/2 + 0^n/3 has binomial transform A007581. - Paul Barry, Jul 20 2003

Number of (s(0), s(1), ..., s(2n+2)) such that 0 < s(i) < 6 and |s(i) - s(i-1)| = 1 for i = 1, 2, ..., 2n+2, s(0) = 1, s(2n+2) = 1. - Herbert Kociemba, Jun 10 2004

Density of regular language L over {1,2,3}^* (i.e., number of strings of length n in L) described by regular expression 11*+11*2(1+2)*+11*2(1+2)*3(1+2+3)*. - Nelma Moreira, Oct 10 2004

Sums of rows of the triangle in A119258. - Reinhard Zumkeller, May 11 2006

Number of n-words from the alphabet A = {a,b,c} which contain an even number of a's. - Fung Cheok Yin (cheokyin_restart(AT)yahoo.com.hk), Aug 30 2006

Let P(A) be the power set of an n-element set A. Then a(n) = the number of pairs of elements {x,y} of P(A) for which either 0) x and y are disjoint and for which x is not a subset of y and y is not a subset of x, or 1) x = y. - Ross La Haye, Jan 10 2008

a(n+1) gives the number of primitive periodic multiplex juggling sequences of length n with base state <2>. - Steve Butler, Jan 21 2008

a(n) is also the number of idempotent order-preserving and order-decreasing partial transformations (of an n-chain). - Abdullahi Umar, Oct 02 2008

Equals row sums of triangle A147292. - Gary W. Adamson, Nov 05 2008

Equals leftmost column of A071919^3. - Gary W. Adamson, Apr 13 2009

A010888(a(n))=5 for n >= 2, that is, the digital root of the terms >= 5 equals 5. - Parthasarathy Nambi, Jun 03 2009

Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=5, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n-1)=(-1)^n*charpoly(A,2). - Milan Janjic, Jan 27 2010

Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=6, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n)=(-1)^(n-1)*charpoly(A,3). - Milan Janjic, Feb 21 2010

It appears that if s(n) is a rational sequence of the form s(1)=2, s(n)= (2*s(n-1)+1)/(s(n-1)+2), n>1 then s(n)=a(n)/(a(n-1)-1).

Form an array with m(1,n)=1 and m(i,j) = Sum_{k=1..i-1} m(k,j) + Sum_{k=1..j-1} m(i,k), which is the sum of the terms to the left of m(i,j) plus the sum above m(i,j). The sum of the terms in antidiagonal(n-1) = a(n). - J. M. Bergot, Jul 16 2013

From Peter Bala, Oct 29 2013: (Start)

An Engel expansion of 3 to the base b := 3/2 as defined in A181565, with the associated series expansion 3 = b + b^2/2 + b^3/(2*5) + b^4/(2*5*14) + .... Cf. A034472.

More generally, for a positive integer n >= 3, the sequence [1, n - 1, n^2 - n - 1, ..., ( (n - 2)*n^k + 1 )/(n - 1), ...] is an Engel expansion of n/(n - 2) to the base n/(n - 1). Cases include A007583 (n = 4), A083065 (n = 5) and A083066 (n = 6). (End)

Diagonal elements (and one more than antidiagonal elements) of the matrix A^n where A=(2,1;1,2). - David Neil McGrath, Aug 17 2014

From M. Sinan Kul, Sep 07 2016: (Start)

a(n) is equal to the number of integer solutions to the following equation when x is equal to the product of n distinct primes: 1/x = 1/y + 1/z where 0 < x < y <= z.

If z = k*y where k is a fraction >= 1 then the solutions can be given as: y = ((k+1)/k)*x and z = (k+1)*x.

Here k can be equal to any divisor of x or to the ratio of two divisors.

For example for x = 2*3*5 = 30 (product of three distinct primes), k would have the following 14 values: 1, 6/5, 3/2, 5/3, 2, 5/2, 3, 10/3, 5, 6, 15/2, 10, 15, 30.

As an example for k = 10/3, we would have y=39, z=130 and 1/39 + 1/130 = 1/30.

Here finding the number of fractions would be equivalent to distributing n balls (distinct primes) to two bins (numerator and denominator) with no empty bins which can be found using Stirling numbers of the second kind. So another definition for a(n) is: a(n) = 2^n + Sum_{i=2..n} Stirling2(i,2)*binomial(n,i).

(End)

a(n+1) is the smallest i for which the Catalan number C(i) (see A000108) is divisible by 3^n for n > 0. This follows from the rule given by Franklin T. Adams-Watters for determining the multiplicity with which a prime divides C(n). We need to find the smallest number in base 3 to achieve a given count. Applied to prime 3, 1 is the smallest digit that counts but requires to be followed by 2 which cannot be at the end to count. Therefore the number in base 3 of the form 1{n-1 times}20 = (3^(n+1) + 1)/2 + 1 = a(n+1)+1 is the smallest number to achieve count n which implies the claim. - Peter Schorn, Mar 06 2020

Let A be a Toeplitz matrix of order n, defined by: A[i,j]=1, if ij; A[i,i]=2. Then, for n>=1, a(n) = det A. - Dmitry Efimov, Oct 28 2021

a(n) is the least number k such that A065363(k) = -(n-1), for n > 0. - Amiram Eldar, Sep 03 2022

Binomial transform of A046717. Second binomial transform of A000302 (with interpolated zeros). Partial sums are A007583.

Counts closed walks of length 2n at a vertex of the cyclic graph on 4 nodes C_4. With interpolated zeros, counts closed walks of length n at a vertex of the cyclic graph on 4 nodes C_4. - Paul Barry, Mar 10 2004

In general, Sum_{k=0..n} Sum_{j=0..n} C(2*(n-k), j)*C(2*k, j)*r^j has expansion (1 - (r+1)*x)/(1 - (r+3)*x - (r-1)*(r+3)*x^2 + (r-1)^3*x^3). - Paul Barry, Jun 04 2005 [corrected by Jason Yuen, Jan 20 2025]

a(n) is the number of binary strings of length 2n with an even number of 0's (and hence an even number of 1's). - Toby Gottfried, Mar 22 2010

Number of compositions of n where there are 2 sorts of part 1, 4 sorts of part 2, 8 sorts of part 3, ..., 2^k sorts of part k. - Joerg Arndt, Aug 04 2014

a(n) is also the number of permutations simultaneously avoiding 231 and 321 in the classical sense which can be realized as labels on an increasing strict binary tree with 2n-1 nodes. See A245904 for more information on increasing strict binary trees. - Manda Riehl Aug 07 2014

INVERT transform of powers of 2 (A000079). - Alois P. Heinz, Feb 11 2021

a(n) is the number of elements in an n-interval of the binomial poset of even-sized subsets of positive integers, cf. Stanley reference and second formula by Paul Barry. Each multichain 0 = x_0 <= x_1 <= x_2 = 1 in such an n-interval corresponds to a closed walk described above by Paul Barry. More generally, each multichain 0 = x_0 <= x_1 <= ... <= x_k = 1 corresponds to a closed walk of length 2n on the k-dimensional hypercube, cf. A054879, A092812, A121822. - Geoffrey Critzer, Apr 21 2023

Numbers whose binary representation is 10, n times (see A163662(n) for n >= 1). - Alexandre Wajnberg, May 31 2005

Numbers whose base-4 representation consists entirely of 2's; twice base-4 repunits. - Franklin T. Adams-Watters, Mar 29 2006

Expected time to finish a random Tower of Hanoi problem with 2n disks using optimal moves, so (since 2n is even and A010684(2n) = 1) a(n) = A060590(2n). - Henry Bottomley, Apr 05 2001

a(n) is the number of derangements of [2n + 3] with runs consisting of consecutive integers. E.g., a(1) = 10 because the derangements of {1, 2, 3, 4, 5} with runs consisting of consecutive integers are 5|1234, 45|123, 345|12, 2345|1, 5|4|123, 5|34|12, 45|23|1, 345|2|1, 5|4|23|1, 5|34|2|1 (the bars delimit the runs). - Emeric Deutsch, May 26 2003

For n > 0, also smallest numbers having in binary representation exactly n + 1 maximal groups of consecutive zeros: A087120(n) = a(n-1), see A087116. - Reinhard Zumkeller, Aug 14 2003

Number of walks of length 2n + 3 between any two diametrically opposite vertices of the cycle graph C_6. Example: a(0) = 2 because in the cycle ABCDEF we have two walks of length 3 between A and D: ABCD and AFED. - Emeric Deutsch, Apr 01 2004

From Paul Barry, May 18 2003: (Start)

Row sums of triangle using cumulative sums of odd-indexed rows of Pascal's triangle (start with zeros for completeness):

0 0

1 1

1 4 4 1

1 6 14 14 6 1

1 8 27 49 49 27 8 1 (End)

a(n) gives the position of the n-th zero in A173732, i.e., A173732(a(n)) = 0 for all n and this gives all the zeros in A173732. - Howard A. Landman, Mar 14 2010

Smallest number having alternating bit sum -n. Cf. A065359. For n = 0, 1, ..., the last digit of a(n) is 0, 2, 0, 2, ... . - Washington Bomfim, Jan 22 2011

Number of toothpicks minus 1 in the toothpick structure of A139250 after 2^n stages. - Omar E. Pol, Mar 15 2012

For n > 0 also partial sums of the odd powers of 2 (A004171). - K. G. Stier, Nov 04 2013

Values of m such that binomial(4*m + 2, m) is odd. Cf. A002450. - Peter Bala, Oct 06 2015

For a(n) > 2, values of m such that m is two steps away from a power of 2 under the Collatz iteration. - Roderick MacPhee, Nov 10 2016

a(n) is the position of the first occurrence of 2^(n+1)-1 in A020986. See the Brillhart and Morton link, pp. 856-857. - John Keith, Jan 12 2021

a(n) is the number of monotone paths in the n-dimensional cross-polytope for a generic linear orientation. See the Black and De Loera link. - Alexander E. Black, Feb 15 2023

Might be called the "Arima sequence" after Yoriyuki Arima who in 1769 constructed this sequence as the number of moves of the outer ring in the optimal solution for the Chinese Rings puzzle (baguenaudier). - Andreas M. Hinz, Feb 15 2017

Let u(k), v(k), w(k) be the 3 sequences defined by u(1)=1, v(1)=0, w(1)=0 and u(k+1) = u(k) + v(k), v(k+1) = u(k) + w(k), w(k+1) = v(k) + w(k); let M(k) = Max(u(k), v(k), w(k)); then a(n) = M(n). - Benoit Cloitre, Mar 25 2002

Unimodal analog of Fibonacci numbers: a(n+1) = Sum_{k=0..n/2} A071922(n-k, n-2*k). Based on the observation that F_{n+1} = Sum_{k} binomial (n-k, k). - Michele Dondi (bik.mido(AT)tiscalinet.it), Jun 30 2002

Numbers n at which the length of the symmetric signed digit expansion of n with q=2 (i.e., the length of the representation of n in the (-1,0,1)2 number system) increases. - _Ralf Stephan, Jun 30 2003

Row sums of Riordan array (1/(1-x), x/(1-2*x^2)). - Paul Barry, Apr 24 2005

For n > 0, record-values of A107910: a(n) = A107910(A023548(n)). - Reinhard Zumkeller, May 28 2005

2^(n+1) = 2*a(n) + 2*A001045(n) + A000975(n-1); e.g., 2^6 = 64 = 2*a(5) + 2*A001045(5) + 2*A000975(4) = 2*11 + 2*11 + 2*10. Let a(n), A001045(n) and A000975(n-1) = the legs of a triangle (a, b, c). Then a(n-1), A001045(n-1) and A000975(n-2) = (S-c), (S-b), (S-a), where S = the triangle semiperimeter. Example: a(5), A001045(5) and A000975(4) = triangle (a, b, c) = (11, 11, 10). Then a(4), A001045(4), A000975(3) = (S-c), (S-b), (S-a) = (6, 5, 5). - Gary W. Adamson, Dec 24 2007

a(n) is the number of length-n binary representations of a nonnegative integer that is divisible by 3. The initial digits are allowed to be 0's. a(4) = 6 because we have 0000, 0011, 0110, 1001, 1100, 1111. - Geoffrey Critzer, Jan 13 2014

a(n) is the top left entry of the n-th power of the 3 X 3 matrix [1, 0, 1; 0, 1, 1; 1, 1, 0] or of the 3 X 3 matrix [1, 1, 0; 1, 0, 1; 0, 1, 1]. - R. J. Mathar, Feb 04 2014

With 0 prefixed, this sequence is an autosequence of the first kind because the sequence of first differences A001045 is. Its companion is A052950. - Paul Curtz, Dec 18 2018, edited by M. F. Hasler, Dec 21 2018

Apparently, the sequence gives the distinct values taken by A129761, the first differences of fibbinary numbers. - Rémy Sigrist, Oct 26 2019

The sequence with offset 1 can be generated in three steps starting with A158780. First, put in alternate signs (1, -1, 1, -2, 2, -4, ...) and take the inverse; getting (1, 1, 0, 1, 1, 2, 3, 5, 8, 13, 21, ...). Take the invert transform of the latter, resulting in the sequence. It follows from the inverti transform being 1, 1, 0, 1, 1, 2, 3, ... that (for example), a(9) = 171 = (1, 1, 0, 1, 1, 2, 3, 5, 8) dot (86, 43, 0, 11, 6, 6, 6, 5, 8) = (86 + 43 + 0 + 11 + 6 + 6 + 6 + 5 + 8). A similar procedure is shown in the Aug 08 2019 comment of A006356. - Gary W. Adamson, Feb 04 2022

Let u(k), v(k), w(k) be defined by u(1)=1, v(1)=0, w(1)=0 and u(k+1)=u(k)+v(k)+w(k), v(k+1)=u(k)+v(k), w(k+1)=u(k); then {u(n)} = 1,1,3,6,14,31,... (A006356 with an extra initial 1), {v(n)} = 0,1,2,5,11,25,... (this sequence with its initial 0 deleted) and {w(n)} = {u(n)} prefixed by an extra 0 = A077998 with an extra initial 0. - Benoit Cloitre, Apr 05 2002. Also u(k)^2+v(k)^2+w(k)^2 = u(2k). - Gary W. Adamson, Dec 23 2003

Form the graph with matrix A=[1, 1, 1; 1, 0, 0; 1, 0, 1]. Then A006054 counts walks of length n between the vertex of degree 1 and the vertex of degree 3. - Paul Barry, Oct 02 2004

Form the digraph with matrix [1,1,0; 1,0,1; 1,1,1]. A006054(n) counts walks of length n between the vertices with loops. - Paul Barry, Oct 15 2004

Nonzero terms = INVERT transform of (1, 1, 2, 2, 3, 3, ...). Example: 56 = (1, 1, 2, 5, 11, 25) dot (3, 3, 2, 2, 1, 1) = (3 + 3 + 4 + 10 + 11 + 25). - Gary W. Adamson, Apr 20 2009

-a(n+1) appears in the formula for the nonpositive powers of rho:= 2*cos(Pi/7), the ratio of the smaller diagonal in the heptagon to the side length s=2*sin(Pi/7), when expressed in the basis <1,rho,sigma>, with sigma:=rho^2-1, the ratio of the larger heptagon diagonal to the side length, as follows. rho^(-n) = C(n)*1 + C(n-1)*rho - a(n+1)*sigma, n >= 0, with C(n)=A077998(n), C(-1):=0. See the Steinbach reference, and a comment under A052547.

If, with the above notations, the power basis of the field Q(rho) is taken one has for nonpositive powers of rho, rho^(-n) = a(n+2)*1 + A077998(n-1)*rho - a(n+1)*rho^2. For nonnegative powers see A006053. See also the Steinbach reference. - Wolfdieter Lang, May 06 2011

a(n) appears also in the nonnegative powers of sigma,(defined in the above comment, where also the basis is given). See a comment in A106803.

The sequence b(n):=(-1)^(n+1)*a(n) forms the negative part (i.e., with nonpositive indices) of the sequence (-1)^n*A006053(n+1). In this way we obtain what we shall call the Ramanujan-type sequence number 2a for the argument 2*Pi/7 (see the comment to Witula's formula in A006053). We have b(n) = -2*b(n-1) + b(n-2) + b(n-3) and b(n) * 49^(1/3) = (c(1)/c(4))^(1/3) * (c(1))^(-n) + (c(2)/c(1))^(1/3) * (c(2))^(-n) + (c(4)/c(2))^(1/3) * (c(4))^(-n) = (c(2)/c(1))^(1/3) * (c(1))^(-n+1) + (c(4)/c(2))^(1/3) * (c(2))^(-n+1) + (c(1)/c(4))^(1/3) * (c(4))^(-n+1), where c(j) := 2*cos(2*Pi*j/7) (for the proof, see the comments to A215112). - Roman Witula, Aug 06 2012

(1, 1, 2, 5, 11, 25, 56, ...) * (1, 0, 1, 0, 1, ...) = the variant of A006356: (1, 1, 3, 6, 14, 31, ...). - Gary W. Adamson, May 15 2013

The limit of a(n+1)/a(n) for n -> infinity is, for all generic sequences with this recurrence of signature (2,1,-1), sigma = rho^2-1, approximately 2.246979603, the length ratio (largest diagonal)/side in the regular heptagon (7-gon). For rho = 2*cos(Pi/7) and sigma see a comment above, and the P. Steinbach reference. Proof: a(n+1)/a(n) = 2 + 1/(a(n)/a(n-1)) - 1/((a(n)/a(n-1))*(a(n-1)/a(n-2))), leading in the limit to sigma^3 -2*sigma^2 - sigma + 1, which is solved by sigma = rho^2-1, due to C(7, rho) = 0 , with the minimal polynomial C(7, x) = x^3 - x^2 - 2*x + 1 of rho (see A187360). - Wolfdieter Lang, Nov 07 2013

Numbers of straight-chain aliphatic amino acids involving single, double or triple bonds (allowing adjacent double bonds) when cis/trans isomerism is neglected. - Stefan Schuster, Apr 19 2018

Let A(r,n) be the total number of ordered arrangements of an n+r tiling of r red squares and white tiles of total length n, where the individual tile lengths can range from 1 to n. A(r,0) corresponds to a tiling of r red squares only, and so A(r,0) = 1. Also, A(r,n)=0 for n<0. Let A_1(r,n) = Sum_{j=0..n} A(r,j). Then the expansion of 1/(1 - 2*x - x^2 + x^3) is A_1(0,n) + A_1(1,n-2) + A_1(n-4) + ... = a(n) without the initial two 0's. In general, the expansion of 1/(1 - 2*x -x^k + x^(k+1)) is equal to Sum_{j>=0} A_1(j, n-j*k). - Gregory L. Simay, May 25 2018

For n>1, a(n) is the number of ways to tile a strip of length n-1 with one color of squares and dominos, two colors of trominos and quadrominos, 3 colors of 5-minos and 6-minos, and so on. - Greg Dresden and Zhiyu Zhang, Jun 26 2025

T(n,k) is the number of lattice paths from (0,0) to (n,k) with steps (1,0) and two kinds of steps (1,1). The number of paths with steps (1,0) and s kinds of steps (1,1) corresponds to the expansion of (1+s*x)^n. - Joerg Arndt, Jul 01 2011

Also sum of rows in A046816. - Lior Manor, Apr 24 2004

Also square array of unsigned coefficients of Chebyshev polynomials of second kind. - Philippe Deléham, Aug 12 2005

The rows give the number of k-simplices in the n-cube. For example, 1, 6, 12, 8 shows that the 3-cube has 1 volume, 6 faces, 12 edges and 8 vertices. - Joshua Zucker, Jun 05 2006

Triangle whose (i, j)-th entry is binomial(i, j)*2^j.

With offset [1,1] the triangle with doubled numbers, 2*a(n,m), enumerates sequences of length m with nonzero integer entries n_i satisfying sum(|n_i|) <= n. Example n=4, m=2: [1,3], [3,1], [2,2] each in 2^2=4 signed versions: 2*a(4,2) = 2*6 = 12. The Sum over m (row sums of 2*a(n,m)) gives 2*3^(n-1), n >= 1. See the W. Lang comment and a K. A. Meissner reference under A024023. - Wolfdieter Lang, Jan 21 2008

n-th row of the triangle = leftmost column of nonzero terms of X^n, where X = an infinite bidiagonal matrix with (1,1,1,...) in the main diagonal and (2,2,2,...) in the subdiagonal. - Gary W. Adamson, Jul 19 2008

Numerators of a matrix square-root of Pascal's triangle A007318, where the denominators for the n-th row are set to 2^n. - Gerald McGarvey, Aug 20 2009

From Johannes W. Meijer, Sep 22 2010: (Start)

The triangle sums (see A180662 for their definitions) link the Pell-Jacobsthal triangle, whose mirror image is A038207, with twenty-four different sequences; see the crossrefs.

This triangle may very well be called the Pell-Jacobsthal triangle in view of the fact that A000129 (Kn21) are the Pell numbers and A001045 (Kn11) the Jacobsthal numbers.

(End)

T(n,k) equals the number of n-length words on {0,1,2} having n-k zeros. - Milan Janjic, Jul 24 2015

T(n-1,k-1) is the number of 2-compositions of n with zeros having k positive parts; see Hopkins & Ouvry reference. - Brian Hopkins, Aug 16 2020

T(n,k) is the number of chains 0=x_0Geoffrey Critzer, Oct 01 2022

Excluding the initial 1, T(n,k) is the number of k-faces of a regular n-cross polytope. See A038207 for n-cube and A135278 for n-simplex. - Mohammed Yaseen, Jan 14 2023

The numbers written in base -2.

a(A007583(n)) are the only terms with all 1s digits; the number of digits = 2n + 1. - Bob Selcoe, Aug 21 2016