cp's OEIS Frontend

Equivalently, integers k such that 2^k - 1 is prime.

It is believed (but unproved) that this sequence is infinite. The data suggest that the number of terms up to exponent N is roughly K log N for some constant K.

Length of prime repunits in base 2.

The associated perfect number N=2^(p-1)*M(p) (=A019279*A000668=A000396), has 2p (=A061645) divisors with harmonic mean p (and geometric mean sqrt(N)). - Lekraj Beedassy, Aug 21 2004

In one of his first publications Euler found the numbers up to 31 but erroneously included 41 and 47.

Equals number of bits in binary expansion of n-th Mersenne prime (A117293). - Artur Jasinski, Feb 09 2007

Number of divisors of n-th even perfect number, divided by 2. Number of divisors of n-th even perfect number that are powers of 2. Number of divisors of n-th even perfect number that are multiples of n-th Mersenne prime A000668(n). - Omar E. Pol, Feb 24 2008

Number of divisors of n-th even superperfect number A061652(n). Numbers of divisors of n-th superperfect number A019279(n), assuming there are no odd superperfect numbers. - Omar E. Pol, Mar 01 2008

Differences between exponents when the even perfect numbers are represented as differences of powers of 2, for example: The 5th even perfect number is 33550336 = 2^25 - 2^12 then a(5)=25-12=13 (see A135655, A133033, A090748). - Omar E. Pol, Mar 01 2008

Number of 1's in binary expansion of n-th even perfect number (see A135650). Number of 1's in binary expansion of divisors of n-th even perfect number that are multiples of n-th Mersenne prime A000668(n) (see A135652, A135653, A135654, A135655). - Omar E. Pol, May 04 2008

Indices of the numbers A006516 that are also even perfect numbers. - Omar E. Pol, Aug 30 2008

Indices of Mersenne numbers A000225 that are also Mersenne primes A000668. - Omar E. Pol, Aug 31 2008

The (prime) number p appears in this sequence if and only if there is no prime q<2^p-1 such that the order of 2 modulo q equals p; a special case is that if p=4k+3 is prime and also q=2p+1 is prime then the order of 2 modulo q is p so p is not a term of this sequence. - Joerg Arndt, Jan 16 2011

Primes p such that sigma(2^p) - sigma(2^p-1) = 2^p-1. - Jaroslav Krizek, Aug 02 2013

Integers k such that every degree k irreducible polynomial over GF(2) is also primitive, i.e., has order 2^k-1. Equivalently, the integers k such that A001037(k) = A011260(k). - Geoffrey Critzer, Dec 08 2019

Conjecture: for k > 1, 2^k-1 is (a Mersenne) prime or k = 2^(2^m)+1 (is a Fermat number) if and only if (k-1)^(2^k-2) == 1 (mod (2^k-1)k^2). - Thomas Ordowski, Oct 05 2023

Conjecture: for p prime, 2^p-1 is (a Mersenne) prime or p = 2^(2^m)+1 (is a Fermat number) if and only if (p-1)^(2^p-2) == 1 (mod 2^p-1). - David Barina, Nov 25 2024

Already as of Dec. 2020, all exponents up to 10^8 had been verified, implying that 74207281, 77232917 and 82589933 are indeed the next three terms. As of today, all exponents up to 130439863 have been tested at least once, see the GIMPS Milestones Report. - M. F. Hasler, Apr 11 2025

On June 23. 2025 all exponents up to 74340751 have been verified, confirming that 74207281 is the exponent of the 49th Mersenne Prime. - Rodolfo Ruiz-Huidobro, Jun 23 2025

For a Mersenne number 2^n - 1 to be prime, the exponent n must itself be prime.

See A000043 for the values of n.

Primes that are repunits in base 2.

Define f(k) = 2k+1; begin with k = 2, a(n+1) = least prime of the form f(f(f(...(a(n))))). - Amarnath Murthy, Dec 26 2003

Mersenne primes other than the first are of the form 6n+1. - Lekraj Beedassy, Aug 27 2004. Mersenne primes other than the first are of the form 24n+7; see also A124477. - Artur Jasinski, Nov 25 2007

A034876(a(n)) = 0 and A034876(a(n)+1) = 1. - Jonathan Sondow, Dec 19 2004

Mersenne primes are solutions to sigma(n+1)-sigma(n) = n as perfect numbers (A000396(n)) are solutions to sigma(n) = 2n. In fact, appears to give all n such that sigma(n+1)-sigma(n) = n. - Benoit Cloitre, Aug 27 2002

If n is in the sequence then sigma(sigma(n)) = 2n+1. Is it true that this sequence gives all numbers n such that sigma(sigma(n)) = 2n+1? - Farideh Firoozbakht, Aug 19 2005

It is easily proved that if n is a Mersenne prime then sigma(sigma(n)) - sigma(n) = n. Is it true that Mersenne primes are all the solutions of the equation sigma(sigma(x)) - sigma(x) = x? - Farideh Firoozbakht, Feb 12 2008

Sum of divisors of n-th even superperfect number A061652(n). Sum of divisors of n-th superperfect number A019279(n), if there are no odd superperfect numbers. - Omar E. Pol, Mar 11 2008

Indices of both triangular numbers and generalized hexagonal numbers (A000217) that are also even perfect numbers. - Omar E. Pol, May 10 2008, Sep 22 2013

Number of positive integers (1, 2, 3, ...) whose sum is the n-th perfect number A000396(n). - Omar E. Pol, May 10 2008

Vertex number where the n-th perfect number A000396(n) is located in the square spiral whose vertices are the positive triangular numbers A000217. - Omar E. Pol, May 10 2008

Mersenne numbers A000225 whose indices are the prime numbers A000043. - Omar E. Pol, Aug 31 2008

The digital roots are 1 if p == 1 (mod 6) and 4 if p == 5 (mod 6). [T. Koshy, Math Gaz. 89 (2005) p. 465]

Primes p such that for all primes q < p, p XOR q = p - q. - Brad Clardy, Oct 26 2011

All these primes, except 3, are Brazilian primes, so they are also in A085104 and A023195. - Bernard Schott, Dec 26 2012

All prime numbers p can be classified by k = (p mod 12) into four classes: k=1, 5, 7, 11. The Mersennne prime numbers 2^p-1, p > 2 are in the class k=7 with p=12*(n-1)+7, n=1,2,.... As all 2^p (p odd) are in class k=8 it follows that all 2^p-1, p > 2 are in class k=7. - Freimut Marschner, Jul 27 2013

From "The Guinness Book of Primes": "During the reign of Queen Elizabeth I, the largest known prime number was the number of grains of rice on the chessboard up to and including the nineteenth square: 524,287 [= 2^19 - 1]. By the time Lord Nelson was fighting the Battle of Trafalgar, the record for the largest prime had gone up to the thirty-first square of the chessboard: 2,147,483,647 [= 2^31 - 1]. This ten-digits number was proved to be prime in 1772 by the Swiss mathematician Leonard Euler, and it held the record until 1867." [du Sautoy] - Robert G. Wilson v, Nov 26 2013

If n is in the sequence then A024816(n) = antisigma(n) = antisigma(n+1) - 1. Is it true that this sequence gives all numbers n such that antisigma(n) = antisigma(n+1) - 1? Are there composite numbers with this property? - Jaroslav Krizek, Jan 24 2014

If n is in the sequence then phi(n) + sigma(sigma(n)) = 3n. Is it true that Mersenne primes are all the solutions of the equation phi(x) + sigma(sigma(x)) = 3x? - Farideh Firoozbakht, Sep 03 2014

a(5) = A229381(2) = 8191 is the "Simpsons' Mersenne prime". - Jonathan Sondow, Jan 02 2015

Equivalently, prime powers of the form 2^n - 1, see Theorem 2 in Lemos & Cambraia Junior. - Charles R Greathouse IV, Jul 07 2016

Primes whose sum of divisors is a power of 2. Primes p such that p + 1 is a power of 2. Primes in A046528. - Omar E. Pol, Jul 09 2016

From Jaroslav Krizek, Jan 19 2017: (Start)

Primes p such that sigma(p+1) = 2p+1.

Primes p such that A051027(p) = sigma(sigma(p)) = 2^k-1 for some k > 1.

Primes p of the form sigma(2^prime(n)-1)-1 for some n. Corresponding values of numbers n are in A016027.

Primes p of the form sigma(2^(n-1)) for some n > 1. Corresponding values of numbers n are in A000043 (Mersenne exponents).

Primes of the form sigma(2^(n+1)) for some n > 1. Corresponding values of numbers n are in A153798 (Mersenne exponents-2).

Primes p of the form sigma(n) where n is even; subsequence of A023195. Primes p of the form sigma(n) for some n. Conjecture: 31 is the only prime p such that p = sigma(x) = sigma(y) for distinct numbers x and y; 31 = sigma(16) = sigma(25).

Conjecture: numbers n such that n = sigma(sigma(n+1)-n-1)-1, i.e., A072868(n)-1.

Conjecture: primes of the form sigma(4*(n-1)) for some n. Corresponding values of numbers n are in A281312. (End)

[Conjecture] For n > 2, the Mersenne number M(n) = 2^n - 1 is a prime if and only if 3^M(n-1) == -1 (mod M(n)). - Thomas Ordowski, Aug 12 2018 [This needs proof! - Joerg Arndt, Mar 31 2019]

Named "Mersenne's numbers" by W. W. Rouse Ball (1892, 1912) after Marin Mersenne (1588-1648). - Amiram Eldar, Feb 20 2021

Theorem. Let b = 2^p - 1 (where p is a prime). Then b is a Mersenne prime iff (c = 2^p - 2 is totient or a term of A002202). Otherwise, if c is (nontotient or a term of A005277) then b is composite. Proof. Trivial, since, while b = v^g - 1 where v is even, v > 2, g is an integer, g > 1, b is always composite, and c = v^g - 2 is nontotient (or a term of A005277), and so is for any composite b = 2^g - 1 (in the last case, c = v^g - 2 is also nontotient, or a term of A005277). - Sergey Pavlov, Aug 30 2021 [Disclaimer: This proof has not been checked. - N. J. A. Sloane, Oct 01 2021]

Also total number of parts in all partitions of n into equal parts that do not contain 1 as a part. - Omar E. Pol, Jan 16 2013

Related concepts: If a(n) < n, n is said to be deficient, if a(n) > n, n is abundant, and if a(n) = n, n is perfect. If there is a cycle of length 2, so that a(n) = b and a(b) = n, b and n are said to be amicable. If there is a longer cycle, the numbers in the cycle are said to be sociable. See examples. - Juhani Heino, Jul 17 2017

Sum of the smallest parts in the partitions of n into two parts such that the smallest part divides the largest. - Wesley Ivan Hurt, Dec 22 2017

a(n) is also the total number of parts congruent to 0 mod k in the partitions of k*n into equal parts that do not contain k as a part (the comment dated Jan 16 2013 is the case for k = 1). - Omar E. Pol, Nov 23 2019

Fixed points are in A000396. - Alois P. Heinz, Mar 10 2024

Row n is a palindromic composition of 2*n.

T(n,k) is also the length of the k-th segment in a Dyck path on the first quadrant of the square grid, connecting the x-axis with the y-axis, from (n, 0) to (0, n), starting with a segment in vertical direction, see example.

Conjecture 1: the area under the n-th Dyck path equals A024916(n), the sum of all divisors of all positive integers <= n.

If the conjecture is true then the n-th Dyck path represents the boundary segments after the alternating sum of the elements of the n-th row of A236104.

Conjecture 2: two adjacent Dyck paths never cross (checked by hand up to n = 128), hence the total area between the n-th Dyck path and the (n-1)-st Dyck path is equal to sigma(n) = A000203(n), the sum of divisors of n.

The connection between A196020 and A237271 is as follows: A196020 --> A236104 --> A235791 --> A237591 --> this sequence --> A239660 --> A237270 --> A237271.

PARI scripts area(n) and chkcross(n) have been written to check the 2 properties and have been run up to n=10000. - Michel Marcus, Mar 27 2014

Mathematica functions have been written that verified the 2 properties through n=30000. - Hartmut F. W. Hoft, Apr 07 2014

Comments from Franklin T. Adams-Watters on sequences related to the "symmetric representation of sigma" in A235791 and related sequences, Mar 31 2014: (Start)

The place to start is with A235791, which is very simple. Then go to A237591, also very simple, and A237593, still very simple.

You then need to interpret the rows of A237593 as Dyck paths. This interpretation is in terms of run lengths, so 2,1,1,2 means up twice, down once, up once, and down twice. Because the rows of A237593 are symmetric and of even length, this path will always be symmetric.

Now the surprising fact is that the areas enclosed by the Dyck path for n (laid on its side) always includes the area enclosed for n-1; and the number of squares added is sigma(n).

Finally, look at the connected areas enclosed by n but not by n-1; the size of these areas is the symmetric representation of sigma. (End)

The symmetric representation of sigma, so defined, is row n of A237270. - Peter Munn, Jan 06 2025

It appears that, for the n-th set, the number of cells lying on the first diagonal is equal to A067742(n), the number of middle divisors of n. - Michel Marcus, Jun 21 2014

Checked Michel Marcus's conjecture with two Mathematica functions up to n=100000, for more information see A240542. - Hartmut F. W. Hoft, Jul 17 2014

A003056(n) is also the number of peaks of the Dyck path related to the n-th row of triangle. - Omar E. Pol, Nov 03 2015

The number of peaks of the Dyck path associated to the row A000396(n) of this triangle equals the n-th Mersenne prime A000668(n), hence Mersenne primes are visible in two ways at the pyramid described in A245092. - Omar E. Pol, Dec 19 2016

The limit as n approaches infinity (area under the Dyck path described in the n-th row of triangle divided by n^2) equals Pi^2/12 = zeta(2)/2. (Cf. A072691.) - Omar E. Pol, Dec 18 2021

The connection between the isosceles triangle and the stepped pyramid is due to the fact that this object can also be interpreted as a pop-up card. - Omar E. Pol, Nov 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

Number of edges in the join of two complete graphs, each of order n, K_n * K_n. - Roberto E. Martinez II, Jan 07 2002

The power series expansion of the entropy function H(x) = (1+x)log(1+x) + (1-x)log(1-x) has 1/a_i as the coefficient of x^(2i) (the odd terms being zero). - Tommaso Toffoli (tt(AT)bu.edu), May 06 2002

Partial sums of A016813 (4n+1). Also with offset = 0, a(n) = (2n+1)(n+1) = A005408 * A000027 = 2n^2 + 3n + 1, i.e., a(0) = 1. - Jeremy Gardiner, Sep 29 2002

Sequence also gives the greatest semiperimeter of primitive Pythagorean triangles having inradius n-1. Such a triangle has consecutive longer sides, with short leg 2n-1, hypotenuse a(n) - (n-1) = A001844(n), and area (n-1)*a(n) = 6*A000330(n-1). - Lekraj Beedassy, Apr 23 2003

Number of divisors of 12^(n-1), i.e., A000005(A001021(n-1)). - Henry Bottomley, Oct 22 2001

More generally, if p1 and p2 are two arbitrarily chosen distinct primes then a(n) is the number of divisors of (p1^2*p2)^(n-1) or equivalently of any member of A054753^(n-1). - Ant King, Aug 29 2011

Number of standard tableaux of shape (2n-1,1,1) (n>=1). - Emeric Deutsch, May 30 2004

It is well known that for n>0, A014105(n) [0,3,10,21,...] is the first of 2n+1 consecutive integers such that the sum of the squares of the first n+1 such integers is equal to the sum of the squares of the last n; e.g., 10^2 + 11^2 + 12^2 = 13^2 + 14^2.

Less well known is that for n>1, a(n) [0,1,6,15,28,...] is the first of 2n consecutive integers such that sum of the squares of the first n such integers is equal to the sum of the squares of the last n-1 plus n^2; e.g., 15^2 + 16^2 + 17^2 = 19^2 + 20^2 + 3^2. - Charlie Marion, Dec 16 2006

a(n) is also a perfect number A000396 when n is an even superperfect number A061652. - Omar E. Pol, Sep 05 2008

Sequence found by reading the line from 0, in the direction 0, 6, ... and the line from 1, in the direction 1, 15, ..., in the square spiral whose vertices are the generalized hexagonal numbers A000217. - Omar E. Pol, Jan 09 2009

For n>=1, 1/a(n) = Sum_{k=0..2*n-1} ((-1)^(k+1)*binomial(2*n-1,k)*binomial(2*n-1+k,k)*H(k)/(k+1)) with H(k) harmonic number of order k.

The number of possible distinct colorings of any 2 colors chosen from n colors of a square divided into quadrants. - Paul Cleary, Dec 21 2010

Central terms of the triangle in A051173. - Reinhard Zumkeller, Apr 23 2011

For n>0, a(n-1) is the number of triples (w,x,y) with all terms in {0,...,n} and max(|w-x|,|x-y|) = |w-y|. - Clark Kimberling, Jun 12 2012

a(n) is the number of positions of one domino in an even pyramidal board with base 2n. - César Eliud Lozada, Sep 26 2012

Partial sums give A002412. - Omar E. Pol, Jan 12 2013

Let a triangle have T(0,0) = 0 and T(r,c) = |r^2 - c^2|. The sum of the differences of the terms in row(n) and row(n-1) is a(n). - J. M. Bergot, Jun 17 2013

With T_(i+1,i)=a(i+1) and all other elements of the lower triangular matrix T zero, T is the infinitesimal generator for A176230, analogous to A132440 for the Pascal matrix. - Tom Copeland, Dec 11 2013

a(n) is the number of length 2n binary sequences that have exactly two 1's. a(2) = 6 because we have: {0,0,1,1}, {0,1,0,1}, {0,1,1,0}, {1,0,0,1}, {1,0,1,0}, {1,1,0,0}. The ordinary generating function with interpolated zeros is: (x^2 + 3*x^4)/(1-x^2)^3. - Geoffrey Critzer, Jan 02 2014

For n > 0, a(n) is the largest integer k such that k^2 + n^2 is a multiple of k + n. More generally, for m > 0 and n > 0, the largest integer k such that k^(2*m) + n^(2*m) is a multiple of k + n is given by k = 2*n^(2*m) - n. - Derek Orr, Sep 04 2014

Binomial transform of (0, 1, 4, 0, 0, 0, ...) and second partial sum of (0, 1, 4, 4, 4, ...). - Gary W. Adamson, Oct 05 2015

a(n) also gives the dimension of the simple Lie algebras D_n, for n >= 4. - Wolfdieter Lang, Oct 21 2015

For n > 0, a(n) equals the number of compositions of n+11 into n parts avoiding parts 2, 3, 4. - Milan Janjic, Jan 07 2016

Also the number of minimum dominating sets and maximal irredundant sets in the n-cocktail party graph. - Eric W. Weisstein, Jun 29 and Aug 17 2017

As Beedassy's formula shows, this Hexagonal number sequence is the odd bisection of the Triangle number sequence. Both of these sequences are figurative number sequences. For A000384, a(n) can be found by multiplying its triangle number by its hexagonal number. For example let's use the number 153. 153 is said to be the 17th triangle number but is also said to be the 9th hexagonal number. Triangle(17) Hexagonal(9). 17*9=153. Because the Hexagonal number sequence is a subset of the Triangle number sequence, the Hexagonal number sequence will always have both a triangle number and a hexagonal number. n* (2*n-1) because (2*n-1) renders the triangle number. - Bruce J. Nicholson, Nov 05 2017

Also numbers k with the property that in the symmetric representation of sigma(k) the smallest Dyck path has a central valley and the largest Dyck path has a central peak, n >= 1. Thus all hexagonal numbers > 0 have middle divisors. (Cf. A237593.) - Omar E. Pol, Aug 28 2018

k^a(n-1) mod n = 1 for prime n and k=2..n-1. - Joseph M. Shunia, Feb 10 2019

Consider all Pythagorean triples (X, Y, Z=Y+1) ordered by increasing Z: a(n+1) gives the semiperimeter of related triangles; A005408, A046092 and A001844 give the X, Y and Z values. - Ralf Steiner, Feb 25 2020

See A002939(n) = 2*a(n) for the corresponding perimeters. - M. F. Hasler, Mar 09 2020

It appears that these are the numbers k with the property that the smallest subpart in the symmetric representation of sigma(k) is 1. - Omar E. Pol, Aug 28 2021

The above conjecture is true. See A280851 for a proof. - Hartmut F. W. Hoft, Feb 02 2022

The n-th hexagonal number equals the sum of the n consecutive integers with the same parity starting at n; for example, 1, 2+4, 3+5+7, 4+6+8+10, etc. In general, the n-th 2k-gonal number is the sum of the n consecutive integers with the same parity starting at (k-2)*n - (k-3). When k = 1 and 2, this result generates the positive integers, A000027, and the squares, A000290, respectively. - Charlie Marion, Mar 02 2022

Conjecture: For n>0, min{k such that there exist subsets A,B of {0,1,2,...,a(n)} such that |A|=|B|=k and A+B={0,1,2,...,2*a(n)}} = 2*n. - Michael Chu, Mar 09 2022

A number m is abundant if sigma(m) > 2m (this sequence), perfect if sigma(m) = 2m (cf. A000396), or deficient if sigma(m) < 2m (cf. A005100), where sigma(m) is the sum of the divisors of m (A000203).

While the first even abundant number is 12 = 2^2*3, the first odd abundant is 945 = 3^3*5*7, the 232nd abundant number!

It appears that for m abundant and > 23, 2*A001055(m) - A101113(m) is NOT 0. - Eric Desbiaux, Jun 01 2009

If m is a term so is every positive multiple of m. "Primitive" terms are in A091191.

If m=6k (k>=2), then sigma(m) >= 1 + k + 2*k + 3*k + 6*k > 12*k = 2*m. Thus all such m are in the sequence.

According to Deléglise (1998), the abundant numbers have natural density 0.2474 < A(2) < 0.2480. Thus the n-th abundant number is asymptotic to 4.0322*n < n/A(2) < 4.0421*n. - Daniel Forgues, Oct 11 2015

From Bob Selcoe, Mar 28 2017 (prompted by correspondence with Peter Seymour): (Start)

Applying similar logic as the proof that all multiples of 6 >= 12 appear in the sequence, for all odd primes p:

i) all numbers of the form j*p*2^k (j >= 1) appear in the sequence when p < 2^(k+1) - 1;

ii) no numbers appear when p > 2^(k+1) - 1 (i.e., are deficient and are in A005100);

iii) when p = 2^(k+1) - 1 (i.e., perfect numbers, A000396), j*p*2^k (j >= 2) appear.

Note that redundancies are eliminated when evaluating p only in the interval [2^k, 2^(k+1)].

The first few even terms not of the forms i or iii are {70, 350, 490, 550, 572, 650, 770, ...}. (End)

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

Number of toothpicks added to the toothpick structure at the n-th step (see A139250).

It appears that if n is equal to 1 plus a power of 2 with positive exponent then a(n) = 4. (For proof see the second Applegate link.)

It appears that there is a relation between this sequence, even superperfect numbers, Mersenne primes and even perfect numbers. Conjecture: The sum of the toothpicks added to the toothpick structure between the stage A061652(k) and the stage A000668(k) is equal to the k-th even perfect number, for k >= 1. For example: A000396(1) = 2+4 = 6. A000396(2) = 4+4+8+12 = 28. A000396(3) = 16+4+8+12+12+16+28+32+20+16+28+36+40+60+88+80 = 496. - Omar E. Pol, May 04 2009

Concerning this conjecture, see David Applegate's comments on the conjectures in A153006. - N. J. A. Sloane, May 14 2009

In the triangle (See example lines), the sum of row k is equal to A006516(k), for k >= 1. - Omar E. Pol, May 15 2009

Equals (1, 2, 2, 2, ...) convolved with A160762: (1, 0, 2, -2, 2, 2, 2, -6, ...). - Gary W. Adamson, May 25 2009

Convolved with the Jacobsthal sequence A001045 = A160704: (1, 3, 9, 19, 41, ...). - Gary W. Adamson, May 24 2009

It appears that the sums of two successive terms of A160552 give the positive terms of this sequence. - Omar E. Pol, Feb 19 2015

From Omar E. Pol, Feb 28 2019: (Start)

The study of the toothpick automaton on triangular grid (A296510), and other C.A. of the same family, reveals that some cellular automata that have recurrent periods can be represented in general by irregular triangles (of first differences) whose row lengths are the terms of A011782 multiplied by k, where k >= 1, is the length of an internal cycle. This internal cycle is called "word" of a cellular automaton. For example: A160121 has word "a", so k = 1. This sequence has word "ab", so k = 2. A296511 has word "abc", so k = 3. A299477 has word "abcb" so k = 4. A299479 has word "abcbc", so k = 5.

The structure of this triangle (with word "ab" and k = 2) for the nonzero terms is as follows:

a,b;

a,b;

a,b,a,b;

a,b,a,b,a,b,a,b;

a,b,a,b,a,b,a,b,a,b,a,b,a,b,a,b;

...

The row lengths are the terms of A011782 multiplied by 2, equaling the column 2 of the square array A296612: 2, 2, 4, 8, 16, ...

This arrangement has the property that the odd-indexed columns (a) contain numbers of the toothpicks that are parallel to initial toothpick, and the even-indexed columns (b) contain numbers of the toothpicks that are orthogonal to the initial toothpick (see the third triangle in the Example section).

An associated sound to the animation could be (tick, tock), (tick, tock), ..., the same as the ticking clock sound.

For further information about the "word" of a cellular automaton see A296612. (End)

A number k is abundant if sigma(k) > 2k (cf. A005101), perfect if sigma(k) = 2k (cf. A000396), or deficient if sigma(k) < 2k (this sequence), where sigma(k) is the sum of the divisors of k (A000203).

Also, numbers k such that A033630(k) = 1. - Reinhard Zumkeller, Mar 02 2007

According to Deléglise (1998), the abundant numbers have natural density 0.2474 < A(2) < 0.2480. Since the perfect numbers have density 0, the deficient numbers have density 0.7520 < 1 - A(2) < 0.7526. Thus the n-th deficient number is asymptotic to 1.3287*n < n/(1 - A(2)) < 1.3298*n. - Daniel Forgues, Oct 10 2015

The data begins with 3 runs of 5 consecutive terms, from 1 to 5, 7 to 11 and 13 to 17. The maximal length of a run of consecutive terms is 5 because 6 is a perfect number and its proper multiples are abundant numbers. - Bernard Schott, May 19 2019

If p and q are primes such that phi(p*q) > p+1, then p*q^n is a term in the sequence for all n >= 1 where phi is the Euler totient function. - Amrit Awasthi, Sep 10 2024