cp's OEIS Frontend

First column of the inverse of the number triangle T(n,k)=if(kA001316(n-k-1,k), if(k=n,1,0)). Essentially INVERT transform of A001316.

Note that A001316(n) = Sum_{k=0..n} mod(C(n,k),2).

Therefore, the infinite series Sum_{n>=0} 1/A001316(n) is a special case of Sum_{n>=0} 1/Sum_{k=0..n} (mod(C(n,k),2)*x^k) for x=1 (cf. A001317, A100307, A100308, etc.). For x>1 this series is convergent, while for x=1 it is divergent. It would be interesting to have asymptotic estimates for a(n).

2^0 = 1 is the only odd power of 2.

Number of subsets of an n-set.

There are 2^(n-1) compositions (ordered partitions) of n (see for example Riordan). This is the unlabeled analog of the preferential labelings sequence A000670.

This is also the number of weakly unimodal permutations of 1..n + 1, that is, permutations with exactly one local maximum. E.g., a(4) = 16: 12345, 12354, 12453, 12543, 13452, 13542, 14532 and 15432 and their reversals. - Jon Perry, Jul 27 2003 [Proof: see next line! See also A087783.]

Proof: n must appear somewhere and there are 2^(n-1) possible choices for the subset that precedes it. These must appear in increasing order and the rest must follow n in decreasing order. QED. - N. J. A. Sloane, Oct 26 2003

a(n+1) is the smallest number that is not the sum of any number of (distinct) earlier terms.

Same as Pisot sequences E(1, 2), L(1, 2), P(1, 2), T(1, 2). See A008776 for definitions of Pisot sequences.

With initial 1 omitted, same as Pisot sequences E(2, 4), L(2, 4), P(2, 4), T(2, 4). - David W. Wilson

Not the sum of two or more consecutive numbers. - Lekraj Beedassy, May 14 2004

Least deficient or near-perfect numbers (i.e., n such that sigma(n) = A000203(n) = 2n - 1). - Lekraj Beedassy, Jun 03 2004. [Comment from Max Alekseyev, Jan 26 2005: All the powers of 2 are least deficient numbers but it is not known if there exists a least deficient number that is not a power of 2.]

Almost-perfect numbers referred to as least deficient or slightly defective (Singh 1997) numbers. Does "near-perfect numbers" refer to both almost-perfect numbers (sigma(n) = 2n - 1) and quasi-perfect numbers (sigma(n) = 2n + 1)? There are no known quasi-perfect or least abundant or slightly excessive (Singh 1997) numbers.

The sum of the numbers in the n-th row of Pascal's triangle; the sum of the coefficients of x in the expansion of (x+1)^n.

The Collatz conjecture (the hailstone sequence will eventually reach the number 1, regardless of which positive integer is chosen initially) may be restated as (the hailstone sequence will eventually reach a power of 2, regardless of which positive integer is chosen initially).

The only hailstone sequence which doesn't rebound (except "on the ground"). - Alexandre Wajnberg, Jan 29 2005

With p(n) as the number of integer partitions of n, p(i) is the number of parts of the i-th partition of n, d(i) is the number of different parts of the i-th partition of n, m(i,j) is the multiplicity of the j-th part of the i-th partition of n, one has: a(n) = Sum_{i = 1..p(n)} (p(i)! / (Product_{j=1..d(i)} m(i,j)!)). - Thomas Wieder, May 18 2005

The number of binary relations on an n-element set that are both symmetric and antisymmetric. Also the number of binary relations on an n-element set that are symmetric, antisymmetric and transitive.

The first differences are the sequence itself. - Alexandre Wajnberg and Eric Angelini, Sep 07 2005

a(n) is the largest number with shortest addition chain involving n additions. - David W. Wilson, Apr 23 2006

Beginning with a(1) = 0, numbers not equal to the sum of previous distinct natural numbers. - Giovanni Teofilatto, Aug 06 2006

For n >= 1, a(n) is equal to the number of functions f:{1, 2, ..., n} -> {1, 2} such that for a fixed x in {1, 2, ..., n} and a fixed y in {1, 2} we have f(x) != y. - Aleksandar M. Janjic and Milan Janjic, Mar 27 2007

Let P(A) be the power set of an n-element set A. Then a(n) is the number of pairs of elements {x,y} of P(A) for which x = y. - Ross La Haye, Jan 09 2008

a(n) is the number of permutations on [n+1] such that every initial segment is an interval of integers. Example: a(3) counts 1234, 2134, 2314, 2341, 3214, 3241, 3421, 4321. The map "p -> ascents of p" is a bijection from these permutations to subsets of [n]. An ascent of a permutation p is a position i such that p(i) < p(i+1). The permutations shown map to 123, 23, 13, 12, 3, 2, 1 and the empty set respectively. - David Callan, Jul 25 2008

2^(n-1) is the largest number having n divisors (in the sense of A077569); A005179(n) is the smallest. - T. D. Noe, Sep 02 2008

a(n) appears to match the number of divisors of the modified primorials (excluding 2, 3 and 5). Very limited range examined, PARI example shown. - Bill McEachen, Oct 29 2008

Successive k such that phi(k)/k = 1/2, where phi is Euler's totient function. - Artur Jasinski, Nov 07 2008

A classical transform consists (for general a(n)) in swapping a(2n) and a(2n+1); examples for Jacobsthal A001045 and successive differences: A092808, A094359, A140505. a(n) = A000079 leads to 2, 1, 8, 4, 32, 16, ... = A135520. - Paul Curtz, Jan 05 2009

This is also the (L)-sieve transform of {2, 4, 6, 8, ..., 2n, ...} = A005843. (See A152009 for the definition of the (L)-sieve transform.) - John W. Layman, Jan 23 2009

a(n) = a(n-1)-th even natural number (A005843) for n > 1. - Jaroslav Krizek, Apr 25 2009

For n >= 0, a(n) is the number of leaves in a complete binary tree of height n. For n > 0, a(n) is the number of nodes in an n-cube. - K.V.Iyer, May 04 2009

Permutations of n+1 elements where no element is more than one position right of its original place. For example, there are 4 such permutations of three elements: 123, 132, 213, and 312. The 8 such permutations of four elements are 1234, 1243, 1324, 1423, 2134, 2143, 3124, and 4123. - Joerg Arndt, Jun 24 2009

Catalan transform of A099087. - R. J. Mathar, Jun 29 2009

a(n) written in base 2: 1,10,100,1000,10000,..., i.e., (n+1) times 1, n times 0 (A011557(n)). - Jaroslav Krizek, Aug 02 2009

Or, phi(n) is equal to the number of perfect partitions of n. - Juri-Stepan Gerasimov, Oct 10 2009

These are the 2-smooth numbers, positive integers with no prime factors greater than 2. - Michael B. Porter, Oct 04 2009

A064614(a(n)) = A000244(n) and A064614(m) < A000244(n) for m < a(n). - Reinhard Zumkeller, Feb 08 2010

a(n) is the largest number m such that the number of steps of iterations of {r - (largest divisor d < r)} needed to reach 1 starting at r = m is equal to n. Example (a(5) = 32): 32 - 16 = 16; 16 - 8 = 8; 8 - 4 = 4; 4 - 2 = 2; 2 - 1 = 1; number 32 has 5 steps and is the largest such number. See A105017, A064097, A175125. - Jaroslav Krizek, Feb 15 2010

a(n) is the smallest proper multiple of a(n-1). - Dominick Cancilla, Aug 09 2010

The powers-of-2 triangle T(n, k), n >= 0 and 0 <= k <= n, begins with: {1}; {2, 4}; {8, 16, 32}; {64, 128, 256, 512}; ... . The first left hand diagonal T(n, 0) = A006125(n + 1), the first right hand diagonal T(n, n) = A036442(n + 1) and the center diagonal T(2*n, n) = A053765(n + 1). Some triangle sums, see A180662, are: Row1(n) = A122743(n), Row2(n) = A181174(n), Fi1(n) = A181175(n), Fi2(2*n) = A181175(2*n) and Fi2(2*n + 1) = 2*A181175(2*n + 1). - Johannes W. Meijer, Oct 10 2010

Records in the number of prime factors. - Juri-Stepan Gerasimov, Mar 12 2011

Row sums of A152538. - Gary W. Adamson, Dec 10 2008

A078719(a(n)) = 1; A006667(a(n)) = 0. - Reinhard Zumkeller, Oct 08 2011

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>=1, a(n) equals the number of 2-colored compositions of n such that no adjacent parts have the same color. - Milan Janjic, Nov 17 2011

Equals A001405 convolved with its right-shifted variant: (1 + 2x + 4x^2 + ...) = (1 + x + 2x^2 + 3x^3 + 6x^4 + 10x^5 + ...) * (1 + x + x^2 + 2x^3 + 3x^4 + 6x^5 + ...). - Gary W. Adamson, Nov 23 2011

The number of odd-sized subsets of an n+1-set. For example, there are 2^3 odd-sized subsets of {1, 2, 3, 4}, namely {1}, {2}, {3}, {4}, {1, 2, 3}, {1, 2, 4}, {1, 3, 4}, and {2, 3, 4}. Also, note that 2^n = Sum_{k=1..floor((n+1)/2)} C(n+1, 2k-1). - Dennis P. Walsh, Dec 15 2011

a(n) is the number of 1's in any row of Pascal's triangle (mod 2) whose row number has exactly n 1's in its binary expansion (see A007318 and A047999). (The result of putting together A001316 and A000120.) - Marcus Jaiclin, Jan 31 2012

A204455(k) = 1 if and only if k is in this sequence. - Wolfdieter Lang, Feb 04 2012

For n>=1 apparently the number of distinct finite languages over a unary alphabet, whose minimum regular expression has alphabetic width n (verified up to n=17), see the Gruber/Lee/Shallit link. - Hermann Gruber, May 09 2012

First differences of A000225. - Omar E. Pol, Feb 19 2013

This is the lexicographically earliest sequence which contains no arithmetic progression of length 3. - Daniel E. Frohardt, Apr 03 2013

a(n-2) is the number of bipartitions of {1..n} (i.e., set partitions into two parts) such that 1 and 2 are not in the same subset. - Jon Perry, May 19 2013

Numbers n such that the n-th cyclotomic polynomial has a root mod 2; numbers n such that the n-th cyclotomic polynomial has an even number of odd coefficients. - Eric M. Schmidt, Jul 31 2013

More is known now about non-power-of-2 "Almost Perfect Numbers" as described in Dagal. - Jonathan Vos Post, Sep 01 2013

Number of symmetric Ferrers diagrams that fit into an n X n box. - Graham H. Hawkes, Oct 18 2013

Numbers n such that sigma(2n) = 2n + sigma(n). - Jahangeer Kholdi, Nov 23 2013

a(1), ..., a(floor(n/2)) are all values of permanent on set of square (0,1)-matrices of order n>=2 with row and column sums 2. - Vladimir Shevelev, Nov 26 2013

Numbers whose base-2 expansion has exactly one bit set to 1, and thus has base-2 sum of digits equal to one. - Stanislav Sykora, Nov 29 2013

A072219(a(n)) = 1. - Reinhard Zumkeller, Feb 20 2014

a(n) is the largest number k such that (k^n-2)/(k-2) is an integer (for n > 1); (k^a(n)+1)/(k+1) is never an integer (for k > 1 and n > 0). - Derek Orr, May 22 2014

If x = A083420(n), y = a(n+1) and z = A087289(n), then x^2 + 2*y^2 = z^2. - Vincenzo Librandi, Jun 09 2014

The mini-sequence b(n) = least number k > 0 such that 2^k ends in n identical digits is given by {1, 18, 39}. The repeating digits are {2, 4, 8} respectively. Note that these are consecutive powers of 2 (2^1, 2^2, 2^3), and these are the only powers of 2 (2^k, k > 0) that are only one digit. Further, this sequence is finite. The number of n-digit endings for a power of 2 with n or more digits id 4*5^(n-1). Thus, for b(4) to exist, one only needs to check exponents up to 4*5^3 = 500. Since b(4) does not exist, it is clear that no other number will exist. - Derek Orr, Jun 14 2014

The least number k > 0 such that 2^k ends in n consecutive decreasing digits is a 3-number sequence given by {1, 5, 25}. The consecutive decreasing digits are {2, 32, 432}. There are 100 different 3-digit endings for 2^k. There are no k-values such that 2^k ends in '987', '876', '765', '654', '543', '321', or '210'. The k-values for which 2^k ends in '432' are given by 25 mod 100. For k = 25 + 100*x, the digit immediately before the run of '432' is {4, 6, 8, 0, 2, 4, 6, 8, 0, 2, ...} for x = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, ...}, respectively. Thus, we see the digit before '432' will never be a 5. So, this sequence is complete. - Derek Orr, Jul 03 2014

a(n) is the number of permutations of length n avoiding both 231 and 321 in the classical sense which are breadth-first search reading words of increasing unary-binary trees. For more details, see the entry for permutations avoiding 231 at A245898. - Manda Riehl, Aug 05 2014

Numbers n such that sigma(n) = sigma(2n) - phi(4n). - Farideh Firoozbakht, Aug 14 2014

This is a B_2 sequence: for i < j, differences a(j) - a(i) are all distinct. Here 2*a(n) < a(n+1) + 1, so a(n) - a(0) < a(n+1) - a(n). - Thomas Ordowski, Sep 23 2014

a(n) counts n-walks (closed) on the graph G(1-vertex; 1-loop, 1-loop). - David Neil McGrath, Dec 11 2014

a(n-1) counts walks (closed) on the graph G(1-vertex; 1-loop, 2-loop, 3-loop, 4-loop, ...). - David Neil McGrath, Jan 01 2015

b(0) = 4; b(n+1) is the smallest number not in the sequence such that b(n+1) - Prod_{i=0..n} b(i) divides b(n+1) - Sum_{i=0..n} b(i). Then b(n) = a(n) for n > 2. - Derek Orr, Jan 15 2015

a(n) counts the permutations of length n+2 whose first element is 2 such that the permutation has exactly one descent. - Ran Pan, Apr 17 2015

a(0)-a(30) appear, with a(26)-a(30) in error, in tablet M 08613 (see CDLI link) from the Old Babylonian period (c. 1900-1600 BC). - Charles R Greathouse IV, Sep 03 2015

Subsequence of A028982 (the squares or twice squares sequence). - Timothy L. Tiffin, Jul 18 2016

A000120(a(n)) = 1. A000265(a(n)) = 1. A000593(a(n)) = 1. - Juri-Stepan Gerasimov, Aug 16 2016

Number of monotone maps f : [0..n] -> [0..n] which are order-increasing (i <= f(i)) and idempotent (f(f(i)) = f(i)). In other words, monads on the n-th ordinal (seen as a posetal category). Any monad f determines a subset of [0..n] that contains n, by considering its set of monad algebras = fixed points { i | f(i) = i }. Conversely, any subset S of [0..n] containing n determines a monad on [0..n], by the function i |-> min { j | i <= j, j in S }. - Noam Zeilberger, Dec 11 2016

Consider n points lying on a circle. Then for n>=2 a(n-2) gives the number of ways to connect two adjacent points with nonintersecting chords. - Anton Zakharov, Dec 31 2016

Satisfies Benford's law [Diaconis, 1977; Berger-Hill, 2017] - N. J. A. Sloane, Feb 07 2017

Also the number of independent vertex sets and vertex covers in the n-empty graph. - Eric W. Weisstein, Sep 21 2017

Also the number of maximum cliques in the n-halved cube graph for n > 4. - Eric W. Weisstein, Dec 04 2017

Number of pairs of compositions of n corresponding to a seaweed algebra of index n-1. - Nick Mayers, Jun 25 2018

The multiplicative group of integers modulo a(n) is cyclic if and only if n = 0, 1, 2. For n >= 3, it is a product of two cyclic groups. - Jianing Song, Jun 27 2018

k^n is the determinant of n X n matrix M_(i, j) = binomial(k + i + j - 2, j) - binomial(i+j-2, j), in this case k=2. - Tony Foster III, May 12 2019

Solutions to the equation Phi(2n + 2*Phi(2n)) = 2n. - M. Farrokhi D. G., Jan 03 2020

a(n-1) is the number of subsets of {1,2,...,n} which have an element that is the size of the set. For example, for n = 4, a(3) = 8 and the subsets are {1}, {1,2}, {2,3}, {2,4}, {1,2,3}, {1,3,4}, {2,3,4}, {1,2,3,4}. - Enrique Navarrete, Nov 21 2020

a(n) is the number of self-inverse (n+1)-order permutations with 231-avoiding. E.g., a(3) = 8: [1234, 1243, 1324, 1432, 2134, 2143, 3214, 4321]. - Yuchun Ji, Feb 26 2021

For any fixed k > 0, a(n) is the number of ways to tile a strip of length n+1 with tiles of length 1, 2, ... k, where the tile of length k can be black or white, with the restriction that the first tile cannot be black. - Greg Dresden and Bora Bursalı, Aug 31 2023

The binary weight of n is also called Hamming weight of n. [The term "Hamming weight" was named after the American mathematician Richard Wesley Hamming (1915-1998). - Amiram Eldar, Jun 16 2021]

a(n) is also the largest integer such that 2^a(n) divides binomial(2n, n) = A000984(n). - Benoit Cloitre, Mar 27 2002

To construct the sequence, start with 0 and use the rule: If k >= 0 and a(0), a(1), ..., a(2^k-1) are the first 2^k terms, then the next 2^k terms are a(0) + 1, a(1) + 1, ..., a(2^k-1) + 1. - Benoit Cloitre, Jan 30 2003

An example of a fractal sequence. That is, if you omit every other number in the sequence, you get the original sequence. And of course this can be repeated. So if you form the sequence a(0 * 2^n), a(1 * 2^n), a(2 * 2^n), a(3 * 2^n), ... (for any integer n > 0), you get the original sequence. - Christopher.Hills(AT)sepura.co.uk, May 14 2003

The n-th row of Pascal's triangle has 2^k distinct odd binomial coefficients where k = a(n) - 1. - Lekraj Beedassy, May 15 2003

Fixed point of the morphism 0 -> 01, 1 -> 12, 2 -> 23, 3 -> 34, 4 -> 45, etc., starting from a(0) = 0. - Robert G. Wilson v, Jan 24 2006

a(n) is the number of times n appears among the mystery calculator sequences: A005408, A042964, A047566, A115419, A115420, A115421. - Jeremy Gardiner, Jan 25 2006

a(n) is the number of solutions of the Diophantine equation 2^m*k + 2^(m-1) + i = n, where m >= 1, k >= 0, 0 <= i < 2^(m-1); a(5) = 2 because only (m, k, i) = (1, 2, 0) [2^1*2 + 2^0 + 0 = 5] and (m, k, i) = (3, 0, 1) [2^3*0 + 2^2 + 1 = 5] are solutions. - Hieronymus Fischer, Jan 31 2006

The first appearance of k, k >= 0, is at a(2^k-1). - Robert G. Wilson v, Jul 27 2006

Sequence is given by T^(infinity)(0) where T is the operator transforming any word w = w(1)w(2)...w(m) into T(w) = w(1)(w(1)+1)w(2)(w(2)+1)...w(m)(w(m)+1). I.e., T(0) = 01, T(01) = 0112, T(0112) = 01121223. - Benoit Cloitre, Mar 04 2009

For n >= 2, the minimal k for which a(k(2^n-1)) is not multiple of n is 2^n + 3. - Vladimir Shevelev, Jun 05 2009

Triangle inequality: a(k+m) <= a(k) + a(m). Equality holds if and only if C(k+m, m) is odd. - Vladimir Shevelev, Jul 19 2009

a(k*m) <= a(k) * a(m). - Robert Israel, Sep 03 2023

The number of occurrences of value k in the first 2^n terms of the sequence is equal to binomial(n, k), and also equal to the sum of the first n - k + 1 terms of column k in the array A071919. Example with k = 2, n = 7: there are 21 = binomial(7,2) = 1 + 2 + 3 + 4 + 5 + 6 2's in a(0) to a(2^7-1). - Brent Spillner (spillner(AT)acm.org), Sep 01 2010, simplified by R. J. Mathar, Jan 13 2017

Let m be the number of parts in the listing of the compositions of n as lists of parts in lexicographic order, a(k) = n - length(composition(k)) for all k < 2^n and all n (see example); A007895 gives the equivalent for compositions into odd parts. - Joerg Arndt, Nov 09 2012

From Daniel Forgues, Mar 13 2015: (Start)

Just tally up row k (binary weight equal k) from 0 to 2^n - 1 to get the binomial coefficient C(n,k). (See A007318.)

0 1 3 7 15

0: O | . | . . | . . . . | . . . . . . . . |

1: | O | O . | O . . . | O . . . . . . . |

2: | | O | O O . | O O . O . . . |

3: | | | O | O O O . |

4: | | | | O |

Due to its fractal nature, the sequence is quite interesting to listen to.

(End)

The binary weight of n is a particular case of the digit sum (base b) of n. - Daniel Forgues, Mar 13 2015

The mean of the first n terms is 1 less than the mean of [a(n+1),...,a(2n)], which is also the mean of [a(n+2),...,a(2n+1)]. - Christian Perfect, Apr 02 2015

a(n) is also the largest part of the integer partition having viabin number n. The viabin number of an integer partition is defined in the following way. Consider the southeast border of the Ferrers board of the integer partition and consider the binary number obtained by replacing each east step with 1 and each north step, except the last one, with 0. The corresponding decimal form is, by definition, the viabin number of the given integer partition. "Viabin" is coined from "via binary". For example, consider the integer partition [2, 2, 2, 1]. The southeast border of its Ferrers board yields 10100, leading to the viabin number 20. - Emeric Deutsch, Jul 20 2017

a(n) is also known as the population count of the binary representation of n. - Chai Wah Wu, May 19 2020

Devadoss refers to these numbers as type B Catalan numbers (cf. A000108).

Equal to the binomial coefficient sum Sum_{k=0..n} binomial(n,k)^2.

Number of possible interleavings of a program with n atomic instructions when executed by two processes. - Manuel Carro (mcarro(AT)fi.upm.es), Sep 22 2001

Convolving a(n) with itself yields A000302, the powers of 4. - T. D. Noe, Jun 11 2002

Number of ordered trees with 2n+1 edges, having root of odd degree and nonroot nodes of outdegree 0 or 2. - Emeric Deutsch, Aug 02 2002

Also number of directed, convex polyominoes having semiperimeter n+2.

Also number of diagonally symmetric, directed, convex polyominoes having semiperimeter 2n+2. - Emeric Deutsch, Aug 03 2002

The second inverse binomial transform of this sequence is this sequence with interpolated zeros. Its g.f. is (1 - 4*x^2)^(-1/2), with n-th term C(n,n/2)(1+(-1)^n)/2. - Paul Barry, Jul 01 2003

Number of possible values of a 2n-bit binary number for which half the bits are on and half are off. - Gavin Scott (gavin(AT)allegro.com), Aug 09 2003

Ordered partitions of n with zeros to n+1, e.g., for n=4 we consider the ordered partitions of 11110 (5), 11200 (30), 13000 (20), 40000 (5) and 22000 (10), total 70 and a(4)=70. See A001700 (esp. Mambetov Bektur's comment). - Jon Perry, Aug 10 2003

Number of nondecreasing sequences of n integers from 0 to n: a(n) = Sum_{i_1=0..n} Sum_{i_2=i_1..n}...Sum_{i_n=i_{n-1}..n}(1). - J. N. Bearden (jnb(AT)eller.arizona.edu), Sep 16 2003

Number of peaks at odd level in all Dyck paths of semilength n+1. Example: a(2)=6 because we have U*DU*DU*D, U*DUUDD, UUDDU*D, UUDUDD, UUU*DDD, where U=(1,1), D=(1,-1) and * indicates a peak at odd level. Number of ascents of length 1 in all Dyck paths of semilength n+1 (an ascent in a Dyck path is a maximal string of up steps). Example: a(2)=6 because we have uDuDuD, uDUUDD, UUDDuD, UUDuDD, UUUDDD, where an ascent of length 1 is indicated by a lower case letter. - Emeric Deutsch, Dec 05 2003

a(n-1) = number of subsets of 2n-1 distinct elements taken n at a time that contain a given element. E.g., n=4 -> a(3)=20 and if we consider the subsets of 7 taken 4 at a time with a 1 we get (1234, 1235, 1236, 1237, 1245, 1246, 1247, 1256, 1257, 1267, 1345, 1346, 1347, 1356, 1357, 1367, 1456, 1457, 1467, 1567) and there are 20 of them. - Jon Perry, Jan 20 2004

The dimension of a particular (necessarily existent) absolutely universal embedding of the unitary dual polar space DSU(2n,q^2) where q>2. - J. Taylor (jt_cpp(AT)yahoo.com), Apr 02 2004.

Number of standard tableaux of shape (n+1, 1^n). - Emeric Deutsch, May 13 2004

Erdős, Graham et al. conjectured that a(n) is never squarefree for sufficiently large n (cf. Graham, Knuth, Patashnik, Concrete Math., 2nd ed., Exercise 112). Sárközy showed that if s(n) is the square part of a(n), then s(n) is asymptotically (sqrt(2)-2) * (sqrt(n)) * zeta(1/2). Granville and Ramare proved that the only squarefree values are a(1)=2, a(2)=6 and a(4)=70. - Jonathan Vos Post, Dec 04 2004 [For more about this conjecture, see A261009. - N. J. A. Sloane, Oct 25 2015]

The MathOverflow link contains the following comment (slightly edited): The Erdős squarefree conjecture (that a(n) is never squarefree for n>4) was proved in 1980 by Sárközy, A. (On divisors of binomial coefficients. I. J. Number Theory 20 (1985), no. 1, 70-80.) who showed that the conjecture holds for all sufficiently large values of n, and by A. Granville and O. Ramaré (Explicit bounds on exponential sums and the scarcity of squarefree binomial coefficients. Mathematika 43 (1996), no. 1, 73-107) who showed that it holds for all n>4. - Fedor Petrov, Nov 13 2010. [From N. J. A. Sloane, Oct 29 2015]

p divides a((p-1)/2)-1=A030662(n) for prime p=5, 13, 17, 29, 37, 41, 53, 61, 73, 89, 97, ... = A002144(n) Pythagorean primes: primes of form 4n+1. - Alexander Adamchuk, Jul 04 2006

The number of direct routes from my home to Granny's when Granny lives n blocks south and n blocks east of my home in Grid City. To obtain a direct route, from the 2n blocks, choose n blocks on which one travels south. For example, a(2)=6 because there are 6 direct routes: SSEE, SESE, SEES, EESS, ESES and ESSE. - Dennis P. Walsh, Oct 27 2006

Inverse: With q = -log(log(16)/(pi a(n)^2)), ceiling((q + log(q))/log(16)) = n. - David W. Cantrell (DWCantrell(AT)sigmaxi.net), Feb 26 2007

Number of partitions with Ferrers diagrams that fit in an n X n box (including the empty partition of 0). Example: a(2) = 6 because we have: empty, 1, 2, 11, 21 and 22. - Emeric Deutsch, Oct 02 2007

So this is the 2-dimensional analog of A008793. - William Entriken, Aug 06 2013

The number of walks of length 2n on an infinite linear lattice that begins and ends at the origin. - Stefan Hollos (stefan(AT)exstrom.com), Dec 10 2007

The number of lattice paths from (0,0) to (n,n) using steps (1,0) and (0,1). - Joerg Arndt, Jul 01 2011

Integral representation: C(2n,n)=1/Pi Integral [(2x)^(2n)/sqrt(1 - x^2),{x,-1, 1}], i.e., C(2n,n)/4^n is the moment of order 2n of the arcsin distribution on the interval (-1,1). - N-E. Fahssi, Jan 02 2008

Also the Catalan transform of A000079. - R. J. Mathar, Nov 06 2008

Straub, Amdeberhan and Moll: "... it is conjectured that there are only finitely many indices n such that C_n is not divisible by any of 3, 5, 7 and 11." - Jonathan Vos Post, Nov 14 2008

Equals INVERT transform of A081696: (1, 1, 3, 9, 29, 97, 333, ...). - Gary W. Adamson, May 15 2009

Also, in sports, the number of ordered ways for a "Best of 2n-1 Series" to progress. For example, a(2) = 6 means there are six ordered ways for a "best of 3" series to progress. If we write A for a win by "team A" and B for a win by "team B" and if we list the played games chronologically from left to right then the six ways are AA, ABA, BAA, BB, BAB, and ABB. (Proof: To generate the a(n) ordered ways: Write down all a(n) ways to designate n of 2n games as won by team A. Remove the maximal suffix of identical letters from each of these.) - Lee A. Newberg, Jun 02 2009

Number of n X n binary arrays with rows, considered as binary numbers, in nondecreasing order, and columns, considered as binary numbers, in nonincreasing order. - R. H. Hardin, Jun 27 2009

Hankel transform is 2^n. - Paul Barry, Aug 05 2009

It appears that a(n) is also the number of quivers in the mutation class of twisted type BC_n for n>=2.

Central terms of Pascal's triangle: a(n) = A007318(2*n,n). - Reinhard Zumkeller, Nov 09 2011

Number of words on {a,b} of length 2n such that no prefix of the word contains more b's than a's. - Jonathan Nilsson, Apr 18 2012

From Pascal's triangle take row(n) with terms in order a1,a2,..a(n) and row(n+1) with terms b1,b2,..b(n), then 2*(a1*b1 + a2*b2 + ... + a(n)*b(n)) to get the terms in this sequence. - J. M. Bergot, Oct 07 2012. For example using rows 4 and 5: 2*(1*(1) + 4*(5) + 6*(10) + 4*(10) + 1*(5)) = 252, the sixth term in this sequence.

Take from Pascal's triangle row(n) with terms b1, b2, ..., b(n+1) and row(n+2) with terms c1, c2, ..., c(n+3) and find the sum b1*c2 + b2*c3 + ... + b(n+1)*c(n+2) to get A000984(n+1). Example using row(3) and row(5) gives sum 1*(5)+3*(10)+3*(10)+1*(5) = 70 = A000984(4). - J. M. Bergot, Oct 31 2012

a(n) == 2 mod n^3 iff n is a prime > 3. (See Mestrovic link, p. 4.) - Gary Detlefs, Feb 16 2013

Conjecture: For any positive integer n, the polynomial sum_{k=0}^n a(k)x^k is irreducible over the field of rational numbers. In general, for any integer m>1 and n>0, the polynomial f_{m,n}(x) = Sum_{k=0..n} (m*k)!/(k!)^m*x^k is irreducible over the field of rational numbers. - Zhi-Wei Sun, Mar 23 2013

This comment generalizes the comment dated Oct 31 2012 and the second of the sequence's original comments. For j = 1 to n, a(n) = Sum_{k=0..j} C(j,k)* C(2n-j, n-k) = 2*Sum_{k=0..j-1} C(j-1,k)*C(2n-j, n-k). - Charlie Marion, Jun 07 2013

The differences between consecutive terms of the sequence of the quotients between consecutive terms of this sequence form a sequence containing the reciprocals of the triangular numbers. In other words, a(n+1)/a(n)-a(n)/a(n-1) = 2/(n*(n+1)). - Christian Schulz, Jun 08 2013

Number of distinct strings of length 2n using n letters A and n letters B. - Hans Havermann, May 07 2014

From Fung Lam, May 19 2014: (Start)

Expansion of G.f. A(x) = 1/(1+q*x*c(x)), where parameter q is positive or negative (except q=-1), and c(x) is the g.f. of A000108 for Catalan numbers. The case of q=-1 recovers the g.f. of A000108 as xA^2-A+1=0. The present sequence A000984 refers to q=-2. Recurrence: (1+q)*(n+2)*a(n+2) + ((q*q-4*q-4)*n + 2*(q*q-q-1))*a(n+1) - 2*q*q*(2*n+1)*a(n) = 0, a(0)=1, a(1)=-q. Asymptotics: a(n) ~ ((q+2)/(q+1))*(q^2/(-q-1))^n, q<=-3, a(n) ~ (-1)^n*((q+2)/(q+1))*(q^2/(q+1))^n, q>=5, and a(n) ~ -Kq*2^(2*n)/sqrt(Pi*n^3), where the multiplicative constant Kq is given by K1=1/9 (q=1), K2=1/8 (q=2), K3=3/25 (q=3), K4=1/9 (q=4). These formulas apply to existing sequences A126983 (q=1), A126984 (q=2), A126982 (q=3), A126986 (q=4), A126987 (q=5), A127017 (q=6), A127016 (q=7), A126985 (q=8), A127053 (q=9), and to A007854 (q=-3), A076035 (q=-4), A076036 (q=-5), A127628 (q=-6), A126694 (q=-7), A115970 (q=-8). (End)

a(n)*(2^n)^(j-2) equals S(n), where S(n) is the n-th number in the self-convolved sequence which yields the powers of 2^j for all integers j, n>=0. For example, when n=5 and j=4, a(5)=252; 252*(2^5)^(4-2) = 252*1024 = 258048. The self-convolved sequence which yields powers of 16 is {1, 8, 96, 1280, 17920, 258048, ...}; i.e., S(5) = 258048. Note that the convolved sequences will be composed of numbers decreasing from 1 to 0, when j<2 (exception being j=1, where the first two numbers in the sequence are 1 and all others decreasing). - Bob Selcoe, Jul 16 2014

The variance of the n-th difference of a sequence of pairwise uncorrelated random variables each with variance 1. - Liam Patrick Roche, Jun 04 2015

Number of ordered trees with n edges where vertices at level 1 can be of 2 colors. Indeed, the standard decomposition of ordered trees leading to the equation C = 1 + zC^2 (C is the Catalan function), yields this time G = 1 + 2zCG, from where G = 1/sqrt(1-4z). - Emeric Deutsch, Jun 17 2015

Number of monomials of degree at most n in n variables. - Ran Pan, Sep 26 2015

Let V(n, r) denote the volume of an n-dimensional sphere with radius r, then V(n, 2^n) / Pi = V(n-1, 2^n) * a(n/2) for all even n. - Peter Luschny, Oct 12 2015

a(n) is the number of sets {i1,...,in} of length n such that n >= i1 >= i2 >= ... >= in >= 0. For instance, a(2) = 6 as there are only 6 such sets: (2,2) (2,1) (2,0) (1,1) (1,0) (0,0). - Anton Zakharov, Jul 04 2016

From Ralf Steiner, Apr 07 2017: (Start)

By analytic continuation to the entire complex plane there exist regularized values for divergent sums such as:

Sum_{k>=0} a(k)/(-2)^k = 1/sqrt(3).

Sum_{k>=0} a(k)/(-1)^k = 1/sqrt(5).

Sum_{k>=0} a(k)/(-1/2)^k = 1/3.

Sum_{k>=0} a(k)/(1/2)^k = -1/sqrt(7)i.

Sum_{k>=0} a(k)/(1)^k = -1/sqrt(3)i.

Sum_{k>=0} a(k)/2^k = -i. (End)

Number of sequences (e(1), ..., e(n+1)), 0 <= e(i) < i, such that there is no triple i < j < k with e(i) > e(j). [Martinez and Savage, 2.18] - Eric M. Schmidt, Jul 17 2017

The o.g.f. for the sequence equals the diagonal of any of the following the rational functions: 1/(1 - (x + y)), 1/(1 - (x + y*z)), 1/(1 - (x + x*y + y*z)) or 1/(1 - (x + y + y*z)). - Peter Bala, Jan 30 2018

From Colin Defant, Sep 16 2018: (Start)

Let s denote West's stack-sorting map. a(n) is the number of permutations pi of [n+1] such that s(pi) avoids the patterns 132, 231, and 321. a(n) is also the number of permutations pi of [n+1] such that s(pi) avoids the patterns 132, 312, and 321.

a(n) is the number of permutations of [n+1] that avoid the patterns 1342, 3142, 3412, and 3421. (End)

All binary self-dual codes of length 4n, for n>0, must contain at least a(n) codewords of weight 2n. More to the point, there will always be at least one, perhaps unique, binary self-dual code of length 4n that will contain exactly a(n) codewords that have a hamming weight equal to half the length of the code (2n). This code can be constructed by direct summing the unique binary self-dual code of length 2 (up to permutation equivalence) to itself an even number of times. A permutation equivalent code can be constructed by augmenting two identity matrices of length 2n together. - Nathan J. Russell, Nov 25 2018

From Isaac Saffold, Dec 28 2018: (Start)

Let [b/p] denote the Legendre symbol and 1/b denote the inverse of b mod p. Then, for m and n, where n is not divisible by p,

[(m+n)/p] == [n/p]*Sum_{k=0..(p-1)/2} (-m/(4*n))^k * a(k) (mod p).

Evaluating this identity for m = -1 and n = 1 demonstrates that, for all odd primes p, Sum_{k=0..(p-1)/2} (1/4)^k * a(k) is divisible by p. (End)

Number of vertices of the subgraph of the (2n-1)-dimensional hypercube induced by all bitstrings with n-1 or n many 1s. The middle levels conjecture asserts that this graph has a Hamilton cycle. - Torsten Muetze, Feb 11 2019

a(n) is the number of walks of length 2n from the origin with steps (1,1) and (1,-1) that stay on or above the x-axis. Equivalently, a(n) is the number of walks of length 2n from the origin with steps (1,0) and (0,1) that stay in the first octant. - Alexander Burstein, Dec 24 2019

Number of permutations of length n>0 avoiding the partially ordered pattern (POP) {3>1, 1>2} of length 4. That is, number of length n permutations having no subsequences of length 4 in which the first element is larger than the second element but smaller than the third elements. - Sergey Kitaev, Dec 08 2020

From Gus Wiseman, Jul 21 2021: (Start)

Also the number of integer compositions of 2n+1 with alternating sum 1, where the alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i. For example, the a(0) = 1 through a(2) = 6 compositions are:

(1) (2,1) (3,2)

(1,1,1) (1,2,2)

(2,2,1)

(1,1,2,1)

(2,1,1,1)

(1,1,1,1,1)

The following relate to these compositions:

- The unordered version is A000070.

- The alternating sum -1 version is counted by A001791, ranked by A345910/A345912.

- The alternating sum 0 version is counted by A088218, ranked by A344619.

- Including even indices gives A126869.

- The complement is counted by A202736.

- Ranked by A345909 (reverse: A345911).

Equivalently, a(n) counts binary numbers with 2n+1 digits and one more 1 than 0's. For example, the a(2) = 6 binary numbers are: 10011, 10101, 10110, 11001, 11010, 11100.

(End)

From Michael Wallner, Jan 25 2022: (Start)

a(n) is the number of nx2 Young tableaux with a single horizontal wall between the first and second column. If there is a wall between two cells, the entries may be decreasing; see [Banderier, Wallner 2021].

Example for a(2)=6:

3 4 2 4 3 4 3|4 4|3 2|4

1|2, 1|3, 2|1, 1 2, 1 2, 1 3

a(n) is also the number of nx2 Young tableaux with n "walls" between the first and second column.

Example for a(2)=6:

3|4 2|4 4|3 3|4 4|3 4|2

1|2, 1|3, 1|2, 2|1, 2|1, 3|1 (End)

From Shel Kaphan, Jan 12 2023: (Start)

a(n)/4^n is the probability that a fair coin tossed 2n times will come up heads exactly n times and tails exactly n times, or that a random walk with steps of +-1 will return to the starting point after 2n steps (not necessarily for the first time). As n becomes large, this number asymptotically approaches 1/sqrt(n*Pi), using Stirling's approximation for n!.

a(n)/(4^n*(2n-1)) is the probability that a random walk with steps of +-1 will return to the starting point for the first time after 2n steps. The absolute value of the n-th term of A144704 is denominator of this fraction.

Considering all possible random walks of exactly 2n steps with steps of +-1, a(n)/(2n-1) is the number of such walks that return to the starting point for the first time after 2n steps. See the absolute values of A002420 or A284016 for these numbers. For comparison, as mentioned by Stefan Hollos, Dec 10 2007, a(n) is the number of such walks that return to the starting point after 2n steps, but not necessarily for the first time. (End)

p divides a((p-1)/2) + 1 for primes p of the form 4*k+3 (A002145). - Jules Beauchamp, Feb 11 2023

Also the size of the shuffle product of two words of length n, such that the union of the two words consist of 2n distinct elements. - Robert C. Lyons, Mar 15 2023

a(n) is the number of vertices of the n-dimensional cyclohedron W_{n+1}. - Jose Bastidas, Mar 25 2025

Consider a stack of pancakes of height n, where the only allowed operation is reversing the top portion of the stack. First, perform a series of reversals of increasing sizes, followed by a series of reversals of decreasing sizes. The number of distinct permutations of the initial stack that can be reached through these operations is a(n). - Thomas Baruchel, May 12 2025

Named after Axel Thue, whose name is pronounced as if it were spelled "Tü" where the ü sound is roughly as in the German word üben. (It is incorrect to say "Too-ee" or "Too-eh".) - N. J. A. Sloane, Jun 12 2018

Also called the Thue-Morse infinite word, or the Morse-Hedlund sequence, or the parity sequence.

Fixed point of the morphism 0 --> 01, 1 --> 10, see example. - Joerg Arndt, Mar 12 2013

The sequence is cubefree (does not contain three consecutive identical blocks) [see Offner for a direct proof] and is overlap-free (does not contain XYXYX where X is 0 or 1 and Y is any string of 0's and 1's).

a(n) = "parity sequence" = parity of number of 1's in binary representation of n.

To construct the sequence: alternate blocks of 0's and 1's of successive lengths A003159(k) - A003159(k-1), k = 1, 2, 3, ... (A003159(0) = 0). Example: since the first seven differences of A003159 are 1, 2, 1, 1, 2, 2, 2, the sequence starts with 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0. - Emeric Deutsch, Jan 10 2003

Characteristic function of A000069 (odious numbers). - Ralf Stephan, Jun 20 2003

a(n) = S2(n) mod 2, where S2(n) = sum of digits of n, n in base-2 notation. There is a class of generalized Thue-Morse sequences: Let Sk(n) = sum of digits of n; n in base-k notation. Let F(t) be some arithmetic function. Then a(n)= F(Sk(n)) mod m is a generalized Thue-Morse sequence. The classical Thue-Morse sequence is the case k=2, m=2, F(t)= 1*t. - Ctibor O. Zizka, Feb 12 2008 (with correction from Daniel Hug, May 19 2017)

More generally, the partial sums of the generalized Thue-Morse sequences a(n) = F(Sk(n)) mod m are fractal, where Sk(n) is sum of digits of n, n in base k; F(t) is an arithmetic function; m integer. - Ctibor O. Zizka, Feb 25 2008

Starting with offset 1, = running sums mod 2 of the kneading sequence (A035263, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, ...); also parity of A005187: (1, 3, 4, 7, 8, 10, 11, 15, 16, 18, 19, ...). - Gary W. Adamson, Jun 15 2008

Generalized Thue-Morse sequences mod n (n>1) = the array shown in A141803. As n -> infinity the sequences -> (1, 2, 3, ...). - Gary W. Adamson, Jul 10 2008

The Thue-Morse sequence for N = 3 = A053838, (sum of digits of n in base 3, mod 3): (0, 1, 2, 1, 2, 0, 2, 0, 1, 1, 2, ...) = A004128 mod 3. - Gary W. Adamson, Aug 24 2008

For all positive integers k, the subsequence a(0) to a(2^k-1) is identical to the subsequence a(2^k+2^(k-1)) to a(2^(k+1)+2^(k-1)-1). That is to say, the first half of A_k is identical to the second half of B_k, and the second half of A_k is identical to the first quarter of B_{k+1}, which consists of the k/2 terms immediately following B_k.

Proof: The subsequence a(2^k+2^(k-1)) to a(2^(k+1)-1), the second half of B_k, is by definition formed from the subsequence a(2^(k-1)) to a(2^k-1), the second half of A_k, by interchanging its 0's and 1's. In turn, the subsequence a(2^(k-1)) to a(2^k-1), the second half of A_k, which is by definition also B_{k-1}, is by definition formed from the subsequence a(0) to a(2^(k-1)-1), the first half of A_k, which is by definition also A_{k-1}, by interchanging its 0's and 1's. Interchanging the 0's and 1's of a subsequence twice leaves it unchanged, so the subsequence a(2^k+2^(k-1)) to a(2^(k+1)-1), the second half of B_k, must be identical to the subsequence a(0) to a(2^(k-1)-1), the first half of A_k.

Also, the subsequence a(2^(k+1)) to a(2^(k+1)+2^(k-1)-1), the first quarter of B_{k+1}, is by definition formed from the subsequence a(0) to a(2^(k-1)-1), the first quarter of A_{k+1}, by interchanging its 0's and 1's. As noted above, the subsequence a(2^(k-1)) to a(2^k-1), the second half of A_k, which is by definition also B_{k-1}, is by definition formed from the subsequence a(0) to a(2^(k-1)-1), which is by definition A_{k-1}, by interchanging its 0's and 1's, as well. If two subsequences are formed from the same subsequence by interchanging its 0's and 1's then they must be identical, so the subsequence a(2^(k+1)) to a(2^(k+1)+2^(k-1)-1), the first quarter of B_{k+1}, must be identical to the subsequence a(2^(k-1)) to a(2^k-1), the second half of A_k.

Therefore the subsequence a(0), ..., a(2^(k-1)-1), a(2^(k-1)), ..., a(2^k-1) is identical to the subsequence a(2^k+2^(k-1)), ..., a(2^(k+1)-1), a(2^(k+1)), ..., a(2^(k+1)+2^(k-1)-1), QED.

According to the German chess rules of 1929 a game of chess was drawn if the same sequence of moves was repeated three times consecutively. Euwe, see the references, proved that this rule could lead to infinite games. For his proof he reinvented the Thue-Morse sequence. - Johannes W. Meijer, Feb 04 2010

"Thue-Morse 0->01 & 1->10, at each stage append the previous with its complement. Start with 0, 1, 2, 3 and write them in binary. Next calculate the sum of the digits (mod 2) - that is divide the sum by 2 and use the remainder." Pickover, The Math Book.

Let s_2(n) be the sum of the base-2 digits of n and epsilon(n) = (-1)^s_2(n), the Thue-Morse sequence, then prod(n >= 0, ((2*n+1)/(2*n+2))^epsilon(n) ) = 1/sqrt(2). - Jonathan Vos Post, Jun 06 2012

Dekking shows that the constant obtained by interpreting this sequence as a binary expansion is transcendental; see also "The Ubiquitous Prouhet-Thue-Morse Sequence". - Charles R Greathouse IV, Jul 23 2013

Drmota, Mauduit, and Rivat proved that the subsequence a(n^2) is normal--see A228039. - Jonathan Sondow, Sep 03 2013

Although the probability of a 0 or 1 is equal, guesses predicated on the latest bit seen produce a correct match 2 out of 3 times. - Bill McEachen, Mar 13 2015

From a(0) to a(2n+1), there are n+1 terms equal to 0 and n+1 terms equal to 1 (see Hassan Tarfaoui link, Concours Général 1990). - Bernard Schott, Jan 21 2022

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

The primes become the powers of 2 (2 -> 1, 3 -> 2, 5 -> 4, 7 -> 8); the composite numbers are formed by taking the values for the factors in the increasing order, multiplying them by the consecutive powers of 2, and summing. See the Example section.

From Antti Karttunen, Jun 27 2014: (Start)

The odd bisection (containing even terms) halved gives A244153.

The even bisection (containing odd terms), when one is subtracted from each and halved, gives this sequence back.

(End)

Question: Are there any other solutions that would satisfy the recurrence r(1) = 0; and for n > 1, r(n) = Sum_{d|n, d>1} 2^A033265(r(d)), apart from simple variants 2^k * A156552(n)? See also A297112, A297113. - Antti Karttunen, Dec 30 2017

3 does not divide binomial(2s, s) if and only if s is a member of this sequence, where binomial(2s, s) = A000984(s) are the central binomial coefficients.

This is the lexicographically earliest increasing sequence of nonnegative numbers that contains no arithmetic progression of length 3. - Robert Craigen (craigenr(AT)cc.umanitoba.ca), Jan 29 2001

In the notation of A185256 this is the Stanley Sequence S(0,1). - N. J. A. Sloane, Mar 19 2010

Complement of A074940. - Reinhard Zumkeller, Mar 23 2003

Sums of distinct powers of 3. - Ralf Stephan, Apr 27 2003

Numbers n such that central trinomial coefficient A002426(n) == 1 (mod 3). - Emeric Deutsch and Bruce E. Sagan, Dec 04 2003

A039966(a(n)+1) = 1; A104406(n) = number of terms <= n.

Subsequence of A125292; A125291(a(n)) = 1 for n>1. - Reinhard Zumkeller, Nov 26 2006

Also final value of n - 1 written in base 2 and then read in base 3 and with finally the result translated in base 10. - Philippe LALLOUET (philip.lallouet(AT)wanadoo.fr), Jun 23 2007

a(n) modulo 2 is the Thue-Morse sequence A010060. - Dennis Tseng, Jul 16 2009

Also numbers such that the balanced ternary representation is the same as the base 3 representation. - Alonso del Arte, Feb 25 2011

Fixed point of the morphism: 0 -> 01; 1 -> 34; 2 -> 67; ...; n -> (3n)(3n+1), starting from a(1) = 0. - Philippe Deléham, Oct 22 2011

It appears that this sequence lists the values of n which satisfy the condition sum(binomial(n, k)^(2*j), k = 0..n) mod 3 <> 0, for any j, with offset 0. See Maple code. - Gary Detlefs, Nov 28 2011

Also, it follows from the above comment by Philippe Lallouet that the sequence must be generated by the rules: a(1) = 0, and if m is in the sequence then so are 3*m and 3*m + 1. - L. Edson Jeffery, Nov 20 2015

Add 1 to each term and we get A003278. - N. J. A. Sloane, Dec 01 2019