cp's OEIS Frontend

Or, n >= 0 appears 2^n times. - Jon Perry, Sep 21 2002

a(n) + 1 = number of bits in binary expansion of n.

Largest power of 2 dividing lcm(1..n): A007814(A003418(n)).

log_2(0) = -infinity.

Also Max_{k=1..n} Omega(k), where Omega(n) = A001222(n), number of prime factors with repetition; see A080613. - Reinhard Zumkeller, Feb 25 2003

From Paul Weisenhorn, Sep 29 2010, updated Aug 11 2020: (Start)

Arithmetic mean: m(1,(c+1)/c) = (2*c+1)/(2*c); harmonic mean: h(1,(c+1)/c) = 2*(c+1)/(2*c+1);

a(n) is the number of means to reach (n+1)/n from 2/1; with m for 0 and h for 1, the inverse binary expansion of n, without the leading 1, gives the sequence of means.

For example, n=20; inverse binary expansion without the leading 1: 0010 ---> m m h m or m(1, m(1, h(1, m(1, 2)))) = 21/20.

The 4 twofold means for n from 4 to 7:

m(1,m(1,2)) = m(1,3/2) = 5/4,

h(1,m(1,2)) = h(1,3/2) = 6/5,

m(1,h(1,2)) = m(1,4/3) = 7/6,

h(1,h(1,2)) = h(1,4/3) = 8/7. (End) [Edited by Petros Hadjicostas, Jul 23 2020]

As function of the absolute value, defines the minimal Euclidean function v on Z\{0}. A ring R is Euclidean if for some function v : R\{0}->N a division by nonzero b can be defined with remainder r satisfying either r=0 or v(r) < v(b). For the integers taking v(n)=|n| works, but v(n) = floor(log_2(|n|)) works as well; moreover it is the possibility with smallest possible values. For division by b>0 one can always choose |r| <= floor(b/2); this sequence satisfies a(1) = 0 and recursively a(n) = 1 + max(a(1), ..., a(floor(n/2))) for n > 1. - Marc A. A. van Leeuwen, Feb 16 2011

Maximum number of guesses required to find any k in a range of 1..n, with 'higher', 'lower' and 'correct' as answers. - Jon Perry, Nov 02 2013

Number of powers of 2 <= n. - Ralph-Joseph Tatt, Apr 23 2018

a(n) + 1 is the minimum number of pairwise disjoint subsets of an n-element set such that for each k from 1 to n there is a set with cardinality k which is the union of some of those subsets. - Wojciech Raszka, Apr 15 2019

Minimum height of an n-node binary tree. - Yuchun Ji, Mar 22 2021

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

Coordination sequence for infinite tree with valency 3.

Number of Hamiltonian cycles in K_3 X P_n.

Number of ternary words of length n avoiding aa, bb, cc.

For n > 0, row sums of A029635. - Paul Barry, Jan 30 2005

Binomial transform is {1, 4, 13, 40, 121, 364, ...}, see A003462. - Philippe Deléham, Jul 23 2005

Convolved with the Jacobsthal sequence A001045 = A001786: (1, 4, 12, 32, 80, ...). - Gary W. Adamson, May 23 2009

Equals (n+1)-th row sums of triangle A161175. - Gary W. Adamson, Jun 05 2009

a(n) written in base 2: a(0) = 1, a(n) for n >= 1: 11, 110, 11000, 110000, ..., i.e.: 2 times 1, (n-1) times 0 (see A003953(n)). - Jaroslav Krizek, Aug 17 2009

INVERTi transform of A003688. - Gary W. Adamson, Aug 05 2010

An elephant sequence, see A175655. For the central square four A[5] vectors, with decimal values 42, 138, 162 and 168, lead to this sequence. For the corner squares these vectors lead to the companion sequence A083329. - Johannes W. Meijer, Aug 15 2010

A216022(a(n)) != 2 and A216059(a(n)) != 3. - Reinhard Zumkeller, Sep 01 2012

Number of length-n strings of 3 letters with no two adjacent letters identical. The general case (strings of r letters) is the sequence with g.f. (1+x)/(1-(r-1)*x). - Joerg Arndt, Oct 11 2012

Sums of pairs of rows of Pascal's triangle A007318, T(2n,k)+T(2n+1,k); Sum_{n>=1} A000290(n)/a(n) = 4. - John Molokach, Sep 26 2013

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

From Peter Bala, Oct 29 2013: (Start)

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

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

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

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

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

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

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

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

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

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

(End)

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

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

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

a(n) is also the number of different lines determined by pair of vertices in an n-dimensional hypercube. The number of these lines modulo being parallel is in A003462. - Ola Veshta (olaveshta(AT)my-deja.com), Feb 15 2001

Let G_n be the elementary Abelian group G_n = (C_2)^n for n >= 1: A006516 is the number of times the number -1 appears in the character table of G_n and A007582 is the number of times the number 1. Together the two sequences cover all the values in the table, i.e., A006516(n) + A007582(n) = 2^(2n). - Ahmed Fares (ahmedfares(AT)my-deja.com), Jun 01 2001

a(n) is the number of n-letter words formed using four distinct letters, one of which appears an odd number of times. - Lekraj Beedassy, Jul 22 2003 [See, e.g., the Balakrishnan reference, problems 2.67 and 2.68, p. 69. - Wolfdieter Lang, Jul 16 2017]

Number of 0's making up the central triangle in a Pascal's triangle mod 2 gasket. - Lekraj Beedassy, May 14 2004

m-th triangular number, where m is the n-th Mersenne number, i.e., a(n)=A000217(A000225(n)). - Lekraj Beedassy, May 25 2004

Number of walks of length 2n+1 between two nodes at distance 3 in the cycle graph C_8. - Herbert Kociemba, Jul 02 2004

The sequence of fractions a(n+1)/(n+1) is the 3rd binomial transform of (1, 0, 1/3, 0, 1/5, 0, 1/7, ...). - Paul Barry, Aug 05 2005

Number of monic irreducible polynomials of degree 2 in GF(2^n)[x]. - Max Alekseyev, Jan 23 2006

(A007582(n))^2 + a(n)^2 = A007582(2n). E.g., A007582(3) = 36, a(3) = 28; A007582(6) = 2080. 36^2 + 28^2 = 2080. - Gary W. Adamson, Jun 17 2006

The sequence 6*a(n), n>=1, gives the number of edges of the Hanoi graph H_4^{n} with 4 pegs and n>=1 discs. - Daniele Parisse, Jul 28 2006

8*a(n) is the total border length of the 4*n masks used when making an order n regular DNA chip, using the bidimensional Gray code suggested by Pevzner in the book "Computational Molecular Biology." - Bruno Petazzoni (bruno(AT)enix.org), Apr 05 2007

If we start with 1 in binary and at each step we prepend 1 and append 0, we construct this sequence: 1 110 11100 1111000 etc.; see A109241(n-1). - Artur Jasinski, Nov 26 2007

Let P(A) be the power set of an n-element set A. Then a(n) = the number of pairs of elements {x,y} of P(A) for which x does not equal y. - Ross La Haye, Jan 02 2008

Wieder calls these "conjoint usual 2-combinations." The set of "conjoint strict k-combinations" is the subset of conjoint usual k-combinations where the empty set and the set itself are excluded from possible selection. These numbers C(2^n - 2,k), which for k = 2 (i.e., {x,y} of the power set of a set) give {1, 0, 1, 15, 91, 435, 1891, 7875, 32131, 129795, 521731, ...}. - Ross La Haye, Jan 15 2008

If n is a member of A000043 then a(n) is also a perfect number (A000396). - Omar E. Pol, Aug 30 2008

a(n) is also the number whose binary representation is A109241(n-1), for n>0. - Omar E. Pol, Aug 31 2008

From Daniel Forgues, Nov 10 2009: (Start)

If we define a spoof-perfect number as:

A spoof-perfect number is a number that would be perfect if some (one or more) of its odd composite factors were wrongly assumed to be prime, i.e., taken as a spoof prime.

And if we define a "strong" spoof-perfect number as:

A "strong" spoof-perfect number is a spoof-perfect number where sigma(n) does not reveal the compositeness of the odd composite factors of n which are wrongly assumed to be prime, i.e., taken as a spoof prime.

The odd composite factors of n which are wrongly assumed to be prime then have to be obtained additively in sigma(n) and not multiplicatively.

Then:

If 2^n-1 is odd composite but taken as a spoof prime then 2^(n-1)*(2^n - 1) is an even spoof perfect number (and moreover "strong" spoof-perfect).

For example:

a(8) = 2^(8-1)*(2^8 - 1) = 128*255 = 32640 (where 255 (with factors 3*5*17) is taken as a spoof prime);

sigma(a(8)) = (2^8 - 1)*(255 + 1) = 255*256 = 2*(128*255) = 2*32640 = 2n is spoof-perfect (and also "strong" spoof-perfect since 255 is obtained additively);

a(11) = 2^(11-1)*(2^11 - 1) = 1024*2047 = 2096128 (where 2047 (with factors 23*89) is taken as a spoof prime);

sigma(a(11)) = (2^11 - 1)*(2047 + 1) = 2047*2048 = 2*(1024*2047) = 2*2096128 = 2n is spoof-perfect (and also "strong" spoof-perfect since 2047 is obtained additively).

I did a Google search and didn't find anything about the distinction between "strong" versus "weak" spoof-perfect numbers. Maybe some other terminology is used.

An example of an even "weak" spoof-perfect number would be:

n = 90 = 2*5*9 (where 9 (with factors 3^2) is taken as a spoof prime);

sigma(n) = (1+2)*(1+5)*(1+9) = 3*(2*3)*(2*5) = 2*(2*5*(3^2)) = 2*90 = 2n is spoof-perfect (but is not "strong" spoof-perfect since 9 is obtained multiplicatively as 3^2 and is thus revealed composite).

Euler proved:

If 2^k - 1 is a prime number, then 2^(k-1)*(2^k - 1) is a perfect number and every even perfect number has this form.

The following seems to be true (is there a proof?):

If 2^k - 1 is an odd composite number taken as a spoof prime, then 2^(k-1)*(2^k - 1) is a "strong" spoof-perfect number and every even "strong" spoof-perfect number has this form?

There is only one known odd spoof-perfect number (found by Rene Descartes) but it is a "weak" spoof-perfect number (cf. 'Descartes numbers' and 'Unsolved problems in number theory' links below). (End)

a(n+1) = A173787(2*n+1,n); cf. A020522, A059153. - Reinhard Zumkeller, Feb 28 2010

Also, row sums of triangle A139251. - Omar E. Pol, May 25 2010

Starting with "1" = (1, 1, 2, 4, 8, ...) convolved with A002450: (1, 5, 21, 85, 341, ...); and (1, 3, 7, 15, 31, ...) convolved with A002001: (1, 3, 12, 48, 192, ...). - Gary W. Adamson, Oct 26 2010

a(n) is also the number of toothpicks in the corner toothpick structure of A153006 after 2^n - 1 stages. - Omar E. Pol, Nov 20 2010

The number of n-dimensional odd theta functions of half-integral characteristic. (Gunning, p.22) - Michael Somos, Jan 03 2014

a(n) = A000217((2^n)-1) = 2^(2n-1) - 2^(n-1) is the nearest triangular number below 2^(2n-1); cf. A007582, A233327. - Antti Karttunen, Feb 26 2014

a(n) is the sum of all the remainders when all the odd numbers < 2^n are divided by each of the powers 2,4,8,...,2^n. - J. M. Bergot, May 07 2014

Let b(m,k) = number of ways to form a sequence of m selections, without replacement, from a circular array of m labeled cells, such that the first selection of a cell whose adjacent cells have already been selected (a "first connect") occurs on the k-th selection. b(m,k) is defined for m >=3, and for 3 <= k <= m. Then b(m,k)/2m ignores rotations and reflection. Let m=n+2, then a(n) = b(m,m-1)/2m. Reiterated, a(n) is the (m-1)th column of the triangle b(m,k)/2m, whose initial rows are (1), (1 2), (2 6 4), (6 18 28 8), (24 72 128 120 16), (120 360 672 840 496 32), (720 2160 4128 5760 5312 2016 64); see A249796. Note also that b(m,3)/2m = n!, and b(m,m)/2m = 2^n. Proofs are easy. - Tony Bartoletti, Oct 30 2014

Beginning at a(1) = 1, this sequence is the sum of the first 2^(n-1) numbers of the form 4*k + 1 = A016813(k). For example, a(4) = 120 = 1 + 5 + 9 + 13 + 17 + 21 + 25 + 29. - J. M. Bergot, Dec 07 2014

a(n) is the number of edges in the (2^n - 1)-dimensional simplex. - Dimitri Boscainos, Oct 05 2015

a(n) is the number of linear elements in a complete plane graph in 2^n points. - Dimitri Boscainos, Oct 05 2015

a(n) is the number of linear elements in a complete parallelotope graph in n dimensions. - Dimitri Boscainos, Oct 05 2015

a(n) is the number of lattices L in Z^n such that the quotient group Z^n / L is C_4. - Álvar Ibeas, Nov 26 2015

a(n) gives the quadratic coefficient of the polynomial ((x + 1)^(2^n) + (x - 1)^(2^n))/2, cf. A201461. - Martin Renner, Jan 14 2017

Let f(x)=x+2*sqrt(x) and g(x)=x-2*sqrt(x). Then f(4^n*x)=b(n)*f(x)+a(n)*g(x) and g(4^n*x)=a(n)*f(x)+b(n)*g(x), where b is A007582. - Luc Rousseau, Dec 06 2018

For n>=1, a(n) is the covering radius of the first order Reed-Muller code RM(1,2n). - Christof Beierle, Dec 22 2021

a(n) =

Companion numbers to A003462.

a(n) = A024101(n)/A024023(n). - Reinhard Zumkeller, Feb 14 2009

Mahler exhibits this sequence with n>=2 as a proof that there exists an infinite number of x coprime to 3, such that x belongs to A005836 and x^2 belong to A125293. - Michel Marcus, Nov 12 2012

a(n-1) is the number of n-digit base 3 numbers that have an even number of digits 0. - Yifan Xie, Jul 13 2024

5^a(n) is the highest power of 5 dividing (5^n)!. - Benoit Cloitre, Feb 04 2002

n such that A002294(n) is not divisible by 5. - Benoit Cloitre, Jan 14 2003

Without leading zero, i.e., sequence {a(n+1) = (5*5^n-1)/4}, this is the binomial transform of A003947. - Paul Barry, May 19 2003 [Edited by M. F. Hasler, Oct 31 2014]

Numbers n such that a(n) is prime are listed in A004061(n) = {3, 7, 11, 13, 47, 127, 149, 181, 619, 929, ...}. Corresponding primes a(n) are listed in A086122(n) = {31, 19531, 12207031, 305175781, 177635683940025046467781066894531, ...}. 3^(m+1) divides a(2*3^m*k). 31 divides a(3k). 13 divides a(4k). 11 divides a(5k). 71 divides a(5k). 7 divides a(6k). 19531 divides a(7k). 313 divides a(8k). 19 divides a(9k). 829 divides a(9k). 71 divides a(10k). 521 divides a(10k). 17 divides a(16k). p divides a(p-1) for all prime p except p = {2,5}. p^(m+1) divides a(p^m*(p-1)) for all prime p except p = {2,5}. p divides a((p-1)/2) for prime p = {11, 19, 29, 31, 41, 59, 61, 71, 79, 89, 101, 109, ...} = A045468, Primes congruent to {1, 4} mod 5. p divides a((p-1)/3) for prime p = {13, 67, 127, 163, 181, 199, 211, 241, 313, 337, 367, 379, 409, 457, ...}. p divides a((p-1)/4) for prime p = {101, 109, 149, 181, 269, 389, 401, 409, 449, 461, 521, 541, ...} = A107219, Primes of the form x^2+100y^2. p divides a((p-1)/5) for prime p = {31, 191, 251, 271, 601, 641, 761, 1091, 1861, ...}. p divides a((p-1)/6) for prime p = {181, 199, 211, 241, 379, 409, 631, 691, 739, 769, 1039, ...}. - Alexander Adamchuk, Jan 23 2007

Starting with 1 = convolution square of A026375: (1, 3, 11, 45, 195, 873, ...). - Gary W. Adamson, May 17 2009

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

This is the sequence A(0,1;4,5;2) = A(0,1;6,-5;0) of the family of sequences [a,b:c,d:k] considered by Gary Detlefs, and treated as A(a,b;c,d;k) in the W. Lang link given below. - Wolfdieter Lang, Oct 18 2010

It is the Lucas sequence U(6,5). - Felix P. Muga II, Mar 21 2014

a(2*n+1) is the sum of the numerators and denominators of the reduced fractions 0 < b/5^n < 1 plus 1, with b < 5^n. - J. M. Bergot, Jul 24 2015

The sequence multiplied by 10 (0, 10, 60, 310, 1560, ...) is the maximum number of coins which can be decided by n weighings on 2 balances in the counterfeit coin problem with undecided under/overweight. [Halbeisen and Hungerbuhler, Disc. Math. 147 (1995) 139 Theorem 1]. - R. J. Mathar, Sep 10 2015

Order of the rank-n projective geometry PG(n-1,5) over the finite field GF(5). - Anthony Hernandez, Oct 05 2016

Number of zeros in the substitution system {0 -> 11100, 1 -> 11110} at step n from initial string "1" (1 -> 11110 -> 1111011110111101111011100 -> ...). - Ilya Gutkovskiy, Apr 10 2017

a(n) is the numerator of Sum_{k=1..n} 1/5^k, which approaches a limit of 1/4. The denominators are 5^n. In general, Sum_{k=1..n} 1/x^k approaches a limit of 1/(x-1). It is of interest to note that as x increases, so does the rate of convergence. See Crossrefs for numerators for other values of x which have the general form (x^n-1)/(x-1). - Gary Detlefs, Aug 31 2021

From David W. Wilson: Numbers triangular, differences square.

To be precise, the differences are the squares of the powers of three with positive indices. Hence a(n+1) - a(n) = (A000244(n+1))^2 = A001019(n+1). [Added by Ant King, Jan 05 2011]

Partial sums of A001019. This is m-th triangular number, where m is partial sums of A000244. a(n) = A000217(A003462(n)). - Lekraj Beedassy, May 25 2004

With offset 0, binomial transform of A003951. - Philippe Deléham, Jul 22 2005

Numbers in base 9: 1, 11, 111, 1111, 11111, 111111, 1111111, etc. - Zerinvary Lajos, Apr 26 2009

Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=9, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n)=det(A). - Milan Janjic, Feb 21 2010

Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=10, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n >= 2, a(n-1) = (-1)^n*charpoly(A,1). - Milan Janjic, Feb 21 2010

From Hieronymus Fischer, Jan 30 2013: (Start)

Least index k such that A052382(k) >= 10^(n-1), for n > 0.

Also index k such that A052382(k) = (10^n-1)/9, n > 0.

A052382(a(n)) is the least zerofree number with n digits, for n > 0.

For n > 1: A052382(a(n)-1) is the greatest zerofree number with n-1 digits. (End)

For n > 0, 4*a(n) is the total number of holes in a certain triangle fractal (start with 9 triangles, 4 holes) after n iterations. See illustration in links. - Kival Ngaokrajang, Feb 21 2015

For n > 0, a(n) is the sum of the numerators and denominators of the reduced fractions 0 < (b/3^(n-1)) < 1 plus 1. Example for n=3 gives fractions 1/9, 2/9, 1/3, 4/9, 5/9, 2/3, 7/9, and 8/9 plus 1 has sum of numerators and denominators +1 = a(3) = 91. - J. M. Bergot, Jul 11 2015

Except for 0 and 1, all terms are Brazilian repunits numbers in base 9, so belong to A125134. All these terms are composite because a(n) is the ((3^n - 1)/2)-th triangular number. - Bernard Schott, Apr 23 2017

These are also the second steps after the junctions of the Collatz trajectories of 2^(2k-1)-1 and 2^2k-1. - David Rabahy, Nov 01 2017

Number of different directions along lines and hyper-diagonals in an n-dimensional cubic lattice for the attacking queens problem (A036464 in n=2, A068940 in n=3 and A068941 in n=4). The n-dimensional direction vectors have the a(n)+1 Cartesian coordinates (i,j,k,l,...) where i,j,k,l,... = -1, 0, or +1, excluding the zero-vector i=j=k=l=...=0. The corresponding hyper-line count is A003462. - R. J. Mathar, May 01 2006

Total number of sequences of length m=1,...,n with nonzero integer elements satisfying the condition Sum_{k=1..m} |n_k| <= n. See the K. A. Meissner link p. 6 (with a typo: it should be 3^([2a]-1)-1). - Wolfdieter Lang, Jan 21 2008

Let P(A) be the power set of an n-element set A and R be a relation on P(A) such that for all x, y of P(A), xRy if x and y are disjoint and either 0) x is a proper subset of y or y is a proper subset of x, or 1) x is not a subset of y and y is not a subset of x. Then a(n) = |R|. - Ross La Haye, Mar 19 2009

Number of neighbors in Moore's neighborhood in n dimensions. - Dmitry Zaitsev, Nov 30 2015

Number of terms in conjunctive normal form of Boolean expression with n variables. E.g., a(2) = 8: [~x, ~y, x, y, ~x|~y, ~x|y, x|~y, x|y]. - Yuchun Ji, May 12 2023

Number of rays of the Coxeter arrangement of type B_n. Equivalently, number of facets of the n-dimensional type B permutahedron. - Jose Bastidas, Sep 12 2023

Glaisher calls this zeta(n) or zeta_1(n). - N. J. A. Sloane, Nov 24 2018

Coefficients in expansion of Sum_{n >= 1} x^n/(1+x^n)^2 = Sum_{n >= 1} (-1)^(n-1)*n*x^n/(1-x^n).

Unsigned sequence is A113184. - Peter Bala, Dec 14 2020