cp's OEIS Frontend

Conjecture: this sequence is the inverse binomial transform of A075272 or, equivalently, the inverse binomial transform of the BinomialMean transform of A075271. - John W. Layman, Sep 12 2002

To win a game, you must flip n+1 heads in a row, where n is the total number of tails flipped so far. Then the probability of winning for the first time after n tails is A005329 / A006125. The probability of having won before n+1 tails is A114604 / A006125. - Joshua Zucker, Dec 14 2005

Number of upper triangular n X n (0,1)-matrices with no zero rows. - Vladeta Jovovic, Mar 10 2008

Equals the q-Fibonacci series for q = (-2), and the series prefaced with a 1: (1, 1, 1, 3, 21, ...) dot (1, -2, 4, -8, ...) if n is even, and (-1, 2, -4, 8, ...) if n is odd. For example, a(3) = 21 = (1, 1, 1, 3) dot (-1, 2, -4, 8) = (-1, 2, -4, 24) and a(4) = 315 = (1, 1, 1, 3, 21) dot (1, -2, 4, -8 16) = (1, -2, 4, -24, 336). - Gary W. Adamson, Apr 17 2009

Number of chambers in an A_n(K) building where K=GF(2) is the field of two elements. This is also the number of maximal flags in an n-dimensional vector space over a field of two elements. - Marcos Spreafico, Mar 22 2012

Given probability p = 1/2^n that an outcome will occur at the n-th stage of an infinite process, then starting at n=1, A114604(n)/A006125(n+2) = 1-a(n)/A006125(n+1) is the probability that the outcome has occurred up to and including the n-th iteration. The limiting ratio is 1-A048651 ~ 0.7112119. These observations are a more formal and generalized statement of Joshua Zucker's Dec 14, 2005 comment. - Bob Selcoe, Mar 02 2016

Also the number of dominating sets in the n-triangular honeycomb rook graph. - Eric W. Weisstein, Jul 14 2017

Empirical: Letting Q denote the Hall-Littlewood Q basis of the symmetric functions over the field of fractions of the univariate polynomial ring in t over the field of rational numbers, and letting h denote the complete homogeneous basis, a(n) is equal to the absolute value of 2^A000292(n) times the coefficient of h_{1^(n*(n+1)/2)} in Q_{(n, n-1, ..., 1)} with t evaluated at 1/2. - John M. Campbell, Apr 30 2018

The series f(x) = Sum_{n>=0} x^(2^n-1)/a(n) satisfies f'(x) = f(x^2), f(0) = 1. - Lucas Larsen, Jan 05 2022

Number of 4-block coverings of an n-set where every element of the set is covered by exactly 3 blocks (if offset is 3), so a(n) = (1/4!)*(4^n-6*2^n+8). - Vladeta Jovovic, Feb 20 2001

Number of non-coprime pairs of polynomials (f,g) with binary coefficients where both f and g have degree n+1 and nonzero constant term. - Luca Mariot and Enrico Formenti, Sep 26 2016

Number of triplets found from the integers 1 to 2^n-1 by converting to binary and performing an XOR operation on the corresponding bits of each pair. Defining addition in this carryless way (0+0=1+1=0, 0+1=1+0=1), each triplet (A,B,C) has the property A+B=C, A+C=B and B+C=A. For example, n=3 gives the 7 triplets (1,2,3), (1,4,5), (1,6,7), (2,4,6), (2,5,7), (3,4,7) and (3,5,6). Each integer appears in the set of triplets 2^(n-1)-1 times, for example 3 for n=3. - Ian Duff, Oct 05 2019

Number of 2-dimensional vector subspaces of (Z_2)^n, so also number of Klein subgroups of the group (C_2)^n. - Robert FERREOL, Jul 28 2021

Note the indexing: the domain starts from 1, while the range includes also zero.

See also the comments at A163511, which is the inverse permutation to this one.

Also triangle read by rows in which T(n,k) is the decimal representation of a binary number whose mirror represents the k-th partition of n according with the list of juxtaposed reverse-lexicographically ordered partitions of the positive integers (A026792).

In order to construct this sequence as a triangle we use the following rules:

- In the list of A026792 we replace each part of size j of the k-th partition of n by concatenation of j - 1 zeros and only one 1.

- Then replace this new set of parts by the concatenation of its parts.

- Then replace this string by its mirror version which is a binary number.

T(n,k) is the decimal value of this binary number, which represents the k-th partition of n (see example).

The partitions of n are represented by a subsequence with A000041(n) integers starting with 2^(n-1) and ending with 2^n - 1, n >= 1. The odd numbers of the sequence are in A000225.

First differs from A065609 at a(23).

Conjecture: this sequence is a sorted version of b(n) where b(2^k) = 2^k for k >= 0, b(n) = A080100(n)*(2*b(A053645(n)) + 1) otherwise. - Mikhail Kurkov, Oct 21 2023

Number of palindromic structures using exactly three different symbols; Mobius transform: A056279. - Marks R. Nester

Number of ways of placing n labeled balls into k=3 indistinguishable boxes. - Thomas Wieder, Nov 30 2004

With two leading zeros, this is the second binomial transform of cosh(x)-1 and the binomial transform of A000225 (with extra leading zero). - Paul Barry, May 13 2003

Let [m] denote the first m positive integers. Then a(n) is the number of functions f from [n] to [n+1] that satisfy (i) f(x) > x for all x, (ii) f(x) = n+1 for exactly 3 elements and (iii) f(f(x)) = n+1 for the remaining n-3 elements of [n]. For example, a(4)=6 since there are exactly 6 functions from {1,2,3,4} to {1,2,3,4,5} such that f(x) > x, f(x) = 5 for 3 elements and f(f(x)) = 5 for the remaining element. The functions are f1 = {(1,5), (2,5), (3,4), (4,5)}, f2 = {(1,5), (2,3), (3,5), (4,5)}, f3 = {(1,5), (2,4), (3,5), (4,5)}, f4 = {(1,2), (2,5), (3,5), (4,5)}, f5 = {(1,3), (2,5), (3,5), (4,5)}, f6 = {(1,4), (2,5), (3,5), (4,5)}. - Dennis P. Walsh, Feb 20 2007

Conjecture. Let S(1)={1} and, for n > 1, let S(n) be the smallest set containing x, 2x and 3x for each element x in S(n-1). Then a(n+2) is the sum of the elements in S(n). (It is easy to prove that the number of elements in S(n) is the n-th triangular number given by A001952.) See A122554 for a sequence defined in this way. - John W. Layman, Nov 21 2007; corrected (a(n) to a(n+2) due to offset change) by Fred Daniel Kline, Oct 02 2014

Let P(A) be the power set of an n-element set A. Then a(n+1) = the number of pairs of elements {x,y} of P(A) for which x and y are disjoint and for which x is not a subset of y and y is not a subset of x. Wieder calls these "disjoint strict 2-combinations". - Ross La Haye, Jan 11 2008; corrected by Ross La Haye, Oct 29 2008

Also, let P(A) be the power set of an n-element set A. Then a(n+2) = the number of pairs of elements {x,y} of P(A) for which either 0) x and y are disjoint and for which either x is a subset of y or y is a subset of x, or 1) x and y are disjoint and for which x is not a subset of y and y is not a subset of x, or 2) x and y are intersecting and for which either x is a proper subset of y or y is a proper subset of x. - Ross La Haye, Jan 11 2008

3 * a(n+1) = p(n+1) where p(x) is the unique degree-n polynomial such that p(k) = a(k+1) for k = 0, 1, ..., n. - Michael Somos, Apr 29 2012

John W. Layman's conjecture that a(n+2) is the sum of elements in S(n) follows from the identification of S(n) with the first n rows of A036561, whose row sums are A001047. - Fred Daniel Kline, Oct 02 2014

From M. Sinan Kul, Sep 08 2016: (Start)

Let m be equal to the product of n-1 distinct primes. Then a(n) is equal to the number of distinct fractions >=1 that may be created by dividing a divisor of m by another divisor. For example for m = 2*3*5 = 30, we would have the following 6 fractions: 6/5, 3/2, 5/3, 5/2, 10/3, 15/2.

Here finding the number of fractions would be equivalent to distributing n-1 balls (distinct primes) to two bins (numerator and denominator) with no empty bins which can be found by Stirling numbers of the second kind. So another definition for a(n) is a(n) = Sum_{i=2..n-1} Stirling2(i,2)*binomial(n-1,i).

Also for n > 0, a(n) = (d(m^2)+1)/2 - d(m) where m is equal to the product of n-1 distinct primes. Example for a(5): m = 2*3*5*7 = 210 (product of four distinct primes) so a(5) = (d(210^2)+1)/2 - d(210) = 41 - 16 = 25. (End)

6*a(n) is the number of ternary strings of length n that contain at least one of each of the 3 symbols on which they are defined. For example, for n=4, the strings are the 12 permutations of 0012, the 12 permutations of 0112, and the 12 permutations of 0122. - Enrique Navarrete, Aug 23 2021

A simpler form of La Haye's first comment is: a(n+1) is the number of ways we can form disjoint unions of two nonempty subsets of [n] (see example below). Cf. A001047 for the requirement that the union contains n. - Enrique Navarrete, Aug 24 2021

As partial sums of the Nicomachus triangle's rows and the differences of the powers of 3 and 2 (A001047), each iteration corresponds to two figurate variations of the Sierpinski triangle (3^n) with cross-correlation to the Nicomachus triangle, see illustrations in links. The Sierpinski half-hexagons of (A001047) stack and conform to the footprint of 2^n - 1 triangular numbers. The 3^n Sierpinski triangle minus its 2^n bottom row, also correlates to the Nicomachus triangle according to each Sierpinski triangular sub-row. - John Elias, Oct 04 2021

For n>0: largest m<=n such that no carry occurs when adding m to n in binary arithmetic: A003817(n+1) = a(n) + n = a(n) XOR n. - Reinhard Zumkeller, Nov 14 2009

a(0) could be considered to be 0 (it was set so from 2004 to 2008) if the binary representation of zero was chosen to be the empty string. - Jason Kimberley, Sep 19 2011

For n > 0: A240857(n,a(n)) = 0. - Reinhard Zumkeller, Apr 14 2014

This is a base-2 analog of A048379. Another variant, without converting back to decimal, is given in A256078. - M. F. Hasler, Mar 22 2015

For n >= 2, a(n) is the least nonnegative k that must be added to n+1 to make a power of 2. Hence in a single-elimination tennis tournament with n entrants, a(n-1) is the number of players given a bye in round one, so that the number of players remaining at the start of round two is a power of 2. For example, if 39 players register, a(38)=25 players receive a round-one bye leaving 14 to play, so that round two will have 25+(14/2)=32 players. - Mathew Englander, Jan 20 2024

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

Number of walks of length 2n+1 between two adjacent vertices in the cycle graph C_8. Example: a(1)=3 because in the cycle ABCDEFGH we have three walks of length 3 between A and B: ABAB, ABCB and AHAB. - Emeric Deutsch, Apr 01 2004

Smallest number containing in its binary representation two equal non-overlapping subwords of length n: A097295(a(n))=n and A097295(m)Reinhard Zumkeller, Aug 04 2004

a(n)^2 + (A006516(n))^2 = a(2n). E.g., a(3) = 36, A006516(3) = 28, a(6) = 2080. 36^2 + 28^2 = 2080. - Gary W. Adamson, Jun 17 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 x equals y or x does not equal y. - Ross La Haye, Jan 02 2008

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). This is just a simpler statement of my previous comment for this sequence. - Ross La Haye, Jan 10 2008

For n>0: A000120(a(n))=2, A023414(a(n))=2*(n-1), A087117(a(n))=n-1. - Reinhard Zumkeller, Jun 23 2009

a(n+1) written in base 2: 11, 1010, 100100, 10001000, 1000010000, ..., i.e., number 1, n times 0, number 1, n times 0 (A163449(n)). - Jaroslav Krizek, Jul 27 2009

a(n) for n >= 1 is a bisection of A001445(n+1). - Jaroslav Krizek, Aug 14 2009

Related to A102573: letting T(q,r) be the coefficient of n^(r+1) in the polynomial 2^(q-n)/n times sum_{k=0..n} binomial(n,k)*k^q, then A007582(x)= sum_{k=0..x-1} T(x,k)*2^k. - John M. Campbell, Nov 16 2011

a(n) gives the number of pairs (r, s) such that 0 <= r <= s <= (2^n)-1 that satisfy AND(r, s, XOR(r, s)) = 0. - Ramasamy Chandramouli, Aug 30 2012

a(n) = A000217(2^n) = 2^(2n-1) + 2^(n-1) is the nearest triangular number above 2^(2n-1); cf. A006516, A233327. - Antti Karttunen, Feb 26 2014

Consider the quantum spin-1/2 chain with even number of sites L (physics, condensed matter theory). The spectrum of the Hamiltonian can be classified according to symmetries. If the only symmetry of the spin Hamiltonian is Parity, i.e., reflection with respect to the middle of the chain (see e.g. the transverse-field Ising model with open boundary conditions), then the dimension of the p=+1 parity sector is given by a(n) with n=L/2. - Marin Bukov, Mar 11 2016

a(n) is also the total number of words of length n, over an alphabet of four letters, of which one of them appears an even number of times. See the Lekraj Beedassy, Jul 22 2003, comment on A006516 (4-letter odd case), and the Balakrishnan reference there. For the 1- to 11-letter cases, see the crossrefs. - Wolfdieter Lang, Jul 17 2017

a(n) is the number of nonisomorphic spanning trees of the cyclic snake formed with n+1 copies of the cycle on 4 vertices. A cyclic snake is a connected graph whose block-cutpoint is a path and all its n blocks are isomorphic to the cycle C_m. - Christian Barrientos, Sep 05 2024

Also, with offset 1, the cogrowth sequence of the dihedral group with 16 elements, D8 = . - Sean A. Irvine, Nov 06 2024

n XOR n-1, i.e., nim-sum of a pair of consecutive numbers.

Denominator of quotient sigma(2*n)/sigma(n). - Labos Elemer, Nov 04 2003

a(n) = the Towers of Hanoi disc moved at the n-th move, using standard moves with discs labeled (1, 3, 7, 15, ...) starting from top (smallest = 1). - Gary W. Adamson, Oct 26 2009

Equals row sums of triangle A168312. - Gary W. Adamson, Nov 22 2009

In the binary expansion of n, delete everything left of the rightmost 1 bit, and set all bits to the right of it. - Ralf Stephan, Aug 22 2013

Every finite sequence of positive integers summing to n may be termwise dominated by a subsequence of the first n values in this sequence [see Bannister et al., 2013]. - David Eppstein, Aug 31 2013

Sum of powers of 2 dividing n. - Omar E. Pol, Aug 18 2019

Given the binary expansion of (n-1) as {b[k-1], b[k-2], ..., b[2], b[1], b[0]}, then the binary expansion of a(n) is {bitand(b[k-1], b[k-2], ..., b[2], b[1], b[0]), bitand(b[k-2], ..., b[2], b[1], b[0]), ..., bitand(b[2], b[1], b[0]), bitand(b[1], b[0]), b[0], 1}. Recursively stated - 0th bit (L.S.B) of a(n), a(n)[0] = 1, a(n)[i] = bitand(a(n)[i-1], (n-1)[i-1]), where n[i] = i-th bit in the binary expansion of n. - Chinmaya Dash, Jun 27 2020

Joe Keane (jgk(AT)jgk.org) observes that this sequence (beginning at 4) is "size of raises in pot-limit poker, one blind, maximum raising."

It appears that this sequence is the BinomialMean transform of A002315 - see A075271. - John W. Layman, Oct 02 2002

Number of (s(0), s(1), ..., s(2n+3)) such that 0 < s(i) < 8 and |s(i) - s(i-1)| = 1 for i = 1,2,...,2n+3, s(0) = 1, s(2n+3) = 4. - Herbert Kociemba, Jun 11 2004

a(n) = number of distinct matrix products in (A+B+C+D)^n where commutators [A,B]=[C,D]=0 but neither A nor B commutes with C or D. - Paul D. Hanna and Joshua Zucker, Feb 01 2006

The n-th term of the sequence is the entry (1,2) in the n-th power of the matrix M=[1,-1;-1,3]. - Simone Severini, Feb 15 2006

Hankel transform of this sequence is [1,-2,0,0,0,0,0,0,0,0,0,...]. - Philippe Deléham, Nov 21 2007

A204089 convolved with A000225, e.g., a(4) = 164 = (1*31 + 1*15 + 4*7 + 14*3 + 48*1) = (31 + 15 + 28 + 42 + 48). - Gary W. Adamson, Dec 23 2008

Equals INVERT transform of A000225: (1, 3, 7, 15, 31, ...). - Gary W. Adamson, May 03 2009

For n>=1, a(n-1) is the number of generalized compositions of n when there are 2^i-1 different types of the part i, (i=1,2,...). - Milan Janjic, Sep 24 2010

Binomial transform of A078057. - R. J. Mathar, Mar 28 2011

Pisano period lengths: 1, 1, 8, 1, 24, 8, 6, 1, 24, 24, 120, 8, 168, 6, 24, 1, 8, 24, 360, 24, ... . - R. J. Mathar, Aug 10 2012

a(n) is the diagonal of array A228405. - Richard R. Forberg, Sep 02 2013

From Wolfdieter Lang, Oct 01 2013: (Start)

a(n) appears together with A106731, both interspersed with zeros, in the representation of nonnegative powers of the algebraic number rho(8) = 2*cos(Pi/8) = A179260 of degree 4, which is the length ratio of the smallest diagonal and the side in the regular octagon.

The minimal polynomial for rho(8) is C(8,x) = x^4 - 4*x^2 + 2, hence rho(8)^n = A(n+1)*1 + A(n)*rho(8) + B(n+1)*rho(8)^2 + B(n)*rho(8)^3, n >= 0, with A(2*k) = 0, k >= 0, A(1) = 1, A(2*k+1) = A106731(k-1), k >= 1, and B(2*k) = 0, k >= 0, B(1) = 0, B(2*k+1) = a(k-1), k >= 1. See also the P. Steinbach reference given under A049310. (End)

The ratio a(n)/A006012(n) converges to 1+sqrt(2). - Karl V. Keller, Jr., May 16 2015

From Tom Copeland, Dec 04 2015: (Start)

An aerated version of this sequence is given by the o.g.f. = 1 / (1 - 4 x^2 + 2 x^4) = 1 / [x^4 a_4(1/x)] = 1 / determinant(I - x M) = exp[-log(1 -4 x + 2 x^4)], where M is the adjacency matrix for the simple Lie algebra B_4 given in A265185 with the characteristic polynomial a_4(x) = x^4 - 4 x^2 + 2 = 2 T_4(x/2) = A127672(4,x), where T denotes a Chebyshev polynomial of the first kind.

A133314 relates a(n) to the reciprocal of the e.g.f. 1 - 4 x + 4 x^2/2!. (End)

a(n) is the number of vertices of the Minkowski sum of n simplices with vertices e_(2*i+1), e_(2*i+2), e_(2*i+3), e_(2*i+4) for i=0,...,n-1, where e_i is a standard basis vector. - Alejandro H. Morales, Oct 03 2022

Also T(2n+1,n+1), T given by A027935. Also first row of Inverse Stolarsky array.

Third diagonal of array defined by T(i, 1)=T(1, j)=1, T(i, j)=Max(T(i-1, j)+T(i-1, j-1); T(i-1, j-1)+T(i, j-1)). - Benoit Cloitre, Aug 05 2003

Number of Schroeder paths of length 2(n+1) having exactly one up step starting at an even height (a Schroeder path is a lattice path starting from (0,0), ending at a point on the x-axis, consisting only of steps U=(1,1) (up steps), D=(1,-1) (down steps) and H=(2,0) (level steps) and never going below the x-axis). Schroeder paths are counted by the large Schroeder numbers (A006318). Example: a(1)=4 because among the six Schroeder paths of length 4 only the paths (U)HD, (U)UDD, H(U)D, (U)DH have exactly one U step that starts at an even height (shown between parentheses). - Emeric Deutsch, Dec 19 2004

Also: smallest number not writeable as the sum of fewer than n positive Fibonacci numbers. E.g., a(5)=88 because it is the smallest number that needs at least 5 Fibonacci numbers: 88 = 55 + 21 + 8 + 3 + 1. - Johan Claes, Apr 19 2005 [corrected for offset and clarification by Mike Speciner, Sep 19 2023] In general, a(n) is the sum of n positive Fibonacci numbers as a(n) = Sum_{i=1..n} A000045(2*i). See A001076 when negative Fibonacci numbers can be included in the sum. - Mike Speciner, Sep 24 2023

Except for first term, numbers a(n) that set a new record in the number of Fibonacci numbers needed to sum up to n. Position of records in sequence A007895. - Ralf Stephan, May 15 2005

Successive extremal petal bends beta(n) = a(n-2). See the Ring Lemma of Rodin and Sullivan in K. Stephenson, Introduction to Circle Packing (Cambridge U. P., 2005), pp. 73-74 and 318-321. - David W. Cantrell (DWCantrell(AT)sigmaxi.net)

a(n+1)= AAB^(n)(1), n>=1, with compositions of Wythoff's complementary A(n):=A000201(n) and B(n)=A001950(n) sequences. See the W. Lang link under A135817 for the Wythoff representation of numbers (with A as 1 and B as 0 and the argument 1 omitted). E.g., 4=`110`, 12=`1100`, 33=`11000`, 88=`110000`, ..., in Wythoff code. AA(1)=1=a(1) but for uniqueness reason 1=A(1) in Wythoff code. - N. J. A. Sloane, Jun 29 2008

Start with n. Each n generates a sublist {n-1,n-1,n-2,..,1}. Each element of each sublist also generates a sublist. Add numbers in all terms. For example, 3->{2,2,1} and both 2->{1,1}, so a(3) = 3 + 2 + 2 + 1 + 1 + 1 + 1 + 1 = 12. - Jon Perry, Sep 01 2012

For n>0: smallest number such that the inner product of Zeckendorf binary representation and its reverse equals n: A216176(a(n)) = n, see also A189920. - Reinhard Zumkeller, Mar 10 2013

Also, numbers m such that 5*m*(m+2)+1 is a square. - Bruno Berselli, May 19 2014

Also, number of nonempty submultisets of multisets of weight n that span an initial interval of integers (see 2nd example). - Gus Wiseman, Feb 10 2015

From Robert K. Moniot, Oct 04 2020: (Start)

Including a(-1):=0, consecutive terms (a(n-1),a(n))=(u,v) or (v,u) give all points on the hyperbola u^2-u+v^2-v-4*u*v=0 with both coordinates nonnegative integers. Note that this follows from identifying (1,u+1,v+1) with the Markov triple (1,Fibonacci(2n-1),Fibonacci(2n+1)). See A001519 (comments by Robert G. Wilson, Oct 05 2005, and Wolfdieter Lang, Jan 30 2015).

Let T(n) denote the n-th triangular number. If i, j are any two successive elements of the above sequence then (T(i-1) + T(j-1))/T(i+j-1) = 3/5. (End)