cp's OEIS Frontend

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

This is a bisection of the Fibonacci sequence A000045. a(n) = F(2*n-1), with F(n) = A000045(n) and F(-1) = 1.

Number of ordered trees with n+1 edges and height at most 3 (height=number of edges on a maximal path starting at the root). Number of directed column-convex polyominoes of area n+1. Number of nondecreasing Dyck paths of length 2n+2. - Emeric Deutsch, Jul 11 2001

Terms are the solutions x to: 5x^2-4 is a square, with 5x^2-4 in A081071 and sqrt(5x^2-4) in A002878. - Benoit Cloitre, Apr 07 2002

a(0) = a(1) = 1, a(n+1) is the smallest Fibonacci number greater than the n-th partial sum. - Amarnath Murthy, Oct 21 2002

The fractional part of tau*a(n) decreases monotonically to zero. - Benoit Cloitre, Feb 01 2003

Numbers k such that floor(phi^2*k^2) - floor(phi*k)^2 = 1 where phi=(1+sqrt(5))/2. - Benoit Cloitre, Mar 16 2003

Number of leftist horizontally convex polyominoes with area n+1.

Number of 31-avoiding words of length n on alphabet {1,2,3} which do not end in 3. (E.g., at n=3, we have 111, 112, 121, 122, 132, 211, 212, 221, 222, 232, 321, 322 and 332.) See A028859. - Jon Perry, Aug 04 2003

Appears to give all solutions > 1 to the equation: x^2 = ceiling(x*r*floor(x/r)) where r=phi=(1+sqrt(5))/2. - Benoit Cloitre, Feb 24 2004

a(1) = 1, a(2) = 2, then the least number such that the square of any term is just less than the geometric mean of its neighbors. a(n+1)*a(n-1) > a(n)^2. - Amarnath Murthy, Apr 06 2004

All positive integer solutions of Pell equation b(n)^2 - 5*a(n+1)^2 = -4 together with b(n)=A002878(n), n >= 0. - Wolfdieter Lang, Aug 31 2004

Essentially same as Pisot sequence E(2,5).

Number of permutations of [n+1] avoiding 321 and 3412. E.g., a(3) = 13 because the permutations of [4] avoiding 321 and 3412 are 1234, 2134, 1324, 1243, 3124, 2314, 2143, 1423, 1342, 4123, 3142, 2413, 2341. - Bridget Tenner, Aug 15 2005

Number of 1324-avoiding circular permutations on [n+1].

A subset of the Markoff numbers (A002559). - Robert G. Wilson v, Oct 05 2005

(x,y) = (a(n), a(n+1)) are the solutions of x/(yz) + y/(xz) + z/(xy) = 3 with z=1. - Floor van Lamoen, Nov 29 2001

Number of (s(0), s(1), ..., s(2n)) such that 0 < s(i) < 5 and |s(i) - s(i-1)| = 1 for i = 1,2,...,2n, s(0) = 1, s(2n) = 1. - Herbert Kociemba, Jun 10 2004

With interpolated zeros, counts closed walks of length n at the start or end node of P_4. a(n) counts closed walks of length 2n at the start or end node of P_4. The sequence 0,1,0,2,0,5,... counts walks of length n between the start and second node of P_4. - Paul Barry, Jan 26 2005

a(n) is the number of ordered trees on n edges containing exactly one non-leaf vertex all of whose children are leaves (every ordered tree must contain at least one such vertex). For example, a(0) = 1 because the root of the tree with no edges is not considered to be a leaf and the condition "all children are leaves" is vacuously satisfied by the root and a(4) = 13 counts all 14 ordered trees on 4 edges (A000108) except (ignore dots)

|..|

.\/.

which has two such vertices. - David Callan, Mar 02 2005

Number of directed column-convex polyominoes of area n. Example: a(2)=2 because we have the 1 X 2 and the 2 X 1 rectangles. - Emeric Deutsch, Jul 31 2006

Same as the number of Kekulé structures in polyphenanthrene in terms of the number of hexagons in extended (1,1)-nanotubes. See Table 1 on page 411 of I. Lukovits and D. Janezic. - Parthasarathy Nambi, Aug 22 2006

Number of free generators of degree n of symmetric polynomials in 3-noncommuting variables. - Mike Zabrocki, Oct 24 2006

Inverse: With phi = (sqrt(5) + 1)/2, log_phi((sqrt(5)*a(n) + sqrt(5*a(n)^2 - 4))/2) = n for n >= 1. - David W. Cantrell (DWCantrell(AT)sigmaxi.net), Feb 19 2007

Consider a teacher who teaches one student, then he finds he can teach two students while the original student learns to teach a student. And so on with every generation an individual can teach one more student then he could before. a(n) starting at a(2) gives the total number of new students/teachers (see program). - Ben Paul Thurston, Apr 11 2007

The Diophantine equation a(n)=m has a solution (for m >= 1) iff ceiling(arcsinh(sqrt(5)*m/2)/log(phi)) != ceiling(arccosh(sqrt(5)*m/2)/log(phi)) where phi is the golden ratio. An equivalent condition is A130255(m)=A130256(m). - Hieronymus Fischer, May 24 2007

a(n+1) = B^(n)(1), n >= 0, 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., 2=`0`, 5=`00`, 13=`000`, ..., in Wythoff code.

Bisection of the Fibonacci sequence into odd-indexed nonzero terms (1, 2, 5, 13, ...) and even-indexed terms (1, 3, 8, 21, ...) may be represented as row sums of companion triangles A140068 and A140069. - Gary W. Adamson, May 04 2008

a(n) is the number of partitions pi of [n] (in standard increasing form) such that Flatten[pi] is a (2-1-3)-avoiding permutation. Example: a(4)=13 counts all 15 partitions of [4] except 13/24 and 13/2/4. Here "standard increasing form" means the entries are increasing in each block and the blocks are arranged in increasing order of their first entries. Also number that avoid 3-1-2. - David Callan, Jul 22 2008

Let P be the partial sum operator, A000012: (1; 1,1; 1,1,1; ...) and A153463 = M, the partial sum & shift operator. It appears that beginning with any randomly taken sequence S(n), iterates of the operations M * S(n), -> M * ANS, -> P * ANS, etc. (or starting with P) will rapidly converge upon a two-sequence limit cycle of (1, 2, 5, 13, 34, ...) and (1, 1, 3, 8, 21, ...). - Gary W. Adamson, Dec 27 2008

Number of musical compositions of Rhythm-music over a time period of n-1 units. Example: a(4)=13; indeed, denoting by R a rest over a time period of 1 unit and by N[j] a note over a period of j units, we have (writing N for N[1]): NNN, NNR, NRN, RNN, NRR, RNR, RRN, RRR, N[2]R, RN[2], NN[2], N[2]N, N[3] (see the J. Groh reference, pp. 43-48). - Juergen K. Groh (juergen.groh(AT)lhsystems.com), Jan 17 2010

Given an infinite lower triangular matrix M with (1, 2, 3, ...) in every column but the leftmost column shifted upwards one row. Then (1, 2, 5, ...) = lim_{n->infinity} M^n. (Cf. A144257.) - Gary W. Adamson, Feb 18 2010

As a fraction: 8/71 = 0.112676 or 98/9701 = 0.010102051334... (fraction 9/71 or 99/9701 for sequence without initial term). 19/71 or 199/9701 for sequence in reverse. - Mark Dols, May 18 2010

For n >= 1, a(n) is the number of compositions (ordered integer partitions) of 2n-1 into an odd number of odd parts. O.g.f.: (x-x^3)/(1-3x^2+x^4) = A(A(x)) where A(x) = 1/(1-x)-1/(1-x^2).

For n > 0, determinant of the n X n tridiagonal matrix with 1's in the super and subdiagonals, (1,3,3,3,...) in the main diagonal, and the rest zeros. - Gary W. Adamson, Jun 27 2011

The Gi3 sums, see A180662, of the triangles A108299 and A065941 equal the terms of this sequence without a(0). - Johannes W. Meijer, Aug 14 2011

The number of permutations for which length equals reflection length. - Bridget Tenner, Feb 22 2012

Number of nonisomorphic graded posets with 0 and 1 and uniform Hasse graph of rank n+1, with exactly 2 elements of each rank between 0 and 1. (Uniform used in the sense of Retakh, Serconek and Wilson. Graded used in R. Stanley's sense that all maximal chains have the same length.)

HANKEL transform of sequence and the sequence omitting a(0) is the sequence A019590(n). This is the unique sequence with that property. - Michael Somos, May 03 2012

The number of Dyck paths of length 2n and height at most 3. - Ira M. Gessel, Aug 06 2012

Pisano period lengths: 1, 3, 4, 3, 10, 12, 8, 6, 12, 30, 5, 12, 14, 24, 20, 12, 18, 12, 9, 30, ... - R. J. Mathar, Aug 10 2012

Primes in the sequence are 2, 5, 13, 89, 233, 1597, 28657, ... (apparently A005478 without the 3). - R. J. Mathar, May 09 2013

a(n+1) is the sum of rising diagonal of the Pascal triangle written as a square - cf. comments in A085812. E.g., 13 = 1+5+6+1. - John Molokach, Sep 26 2013

a(n) is the top left entry of the n-th power of any of the 3 X 3 matrices [1, 1, 1; 1, 1, 1; 0, 1, 1] or [1, 1, 1; 0, 1, 1; 1, 1, 1] or [1, 1, 0; 1, 1, 1; 1, 1, 1] or [1, 0, 1; 1, 1, 1; 1, 1, 1]. - R. J. Mathar, Feb 03 2014

Except for the initial term, positive values of x (or y) satisfying x^2 - 3xy + y^2 + 1 = 0. - Colin Barker, Feb 04 2014

Except for the initial term, positive values of x (or y) satisfying x^2 - 18xy + y^2 + 64 = 0. - Colin Barker, Feb 16 2014

Positive values of x such that there is a y satisfying x^2 - xy - y^2 - 1 = 0. - Ralf Stephan, Jun 30 2014

a(n) is also the number of permutations simultaneously avoiding 231, 312 and 321 in the classical sense which can be realized as labels on an increasing strict binary tree with 2n-1 nodes. See A245904 for more information on increasing strict binary trees. - Manda Riehl, Aug 07 2014

(1, a(n), a(n+1)), n >= 0, are Markoff triples (see A002559 and Robert G. Wilson v's Oct 05 2005 comment). In the Markoff tree they give one of the outer branches. Proof: a(n)*a(n+1) - 1 = A001906(2*n)^2 = (a(n+1) - a(n))^2 = a(n)^2 + a(n+1)^2 - 2*a(n)*a(n+1), thus 1^2 + a(n)^2 + a(n+1)^2 = 3*a(n)*a(n+1). - Wolfdieter Lang, Jan 30 2015

For n > 0, a(n) is the smallest positive integer not already in the sequence such that a(1) + a(2) + ... + a(n) is a Fibonacci number. - Derek Orr, Jun 01 2015

Number of vertices of degree n-2 (n >= 3) in all Fibonacci cubes, see Klavzar, Mollard, & Petkovsek. - Emeric Deutsch, Jun 22 2015

Except for the first term, this sequence can be generated by Corollary 1 (ii) of Azarian's paper in the references for this sequence. - Mohammad K. Azarian, Jul 02 2015

Precisely the numbers F(n)^k + F(n+1)^k that are also Fibonacci numbers with k > 1, see Luca & Oyono. - Charles R Greathouse IV, Aug 06 2015

a(n) = MA(n) - 2*(-1)^n where MA(n) is exactly the maximum area of a quadrilateral with lengths of sides in order L(n-2), L(n-2), F(n+1), F(n+1) for n > 1 and L(n)=A000032(n). - J. M. Bergot, Jan 28 2016

a(n) is the number of bargraphs of semiperimeter n+1 having no valleys (i.e., convex bargraphs). Equivalently, number of bargraphs of semiperimeter n+1 having exactly 1 peak. Example: a(5) = 34 because among the 35 (=A082582(6)) bargraphs of semiperimeter 6 only the one corresponding to the composition [2,1,2] has a valley. - Emeric Deutsch, Aug 12 2016

Integers k such that the fractional part of k*phi is less than 1/k. See Byszewski link p. 2. - Michel Marcus, Dec 10 2016

Number of words of length n-1 over {0,1,2,3} in which binary subwords appear in the form 10...0. - Milan Janjic, Jan 25 2017

With a(0) = 0 this is the Riordan transform with the Riordan matrix A097805 (of the associated type) of the Fibonacci sequence A000045. See a Feb 17 2017 comment on A097805. - Wolfdieter Lang, Feb 17 2017

Number of sequences (e(1), ..., e(n)), 0 <= e(i) < i, such that there is no triple i < j < k with e(i) < e(j) < e(k). [Martinez and Savage, 2.12] - Eric M. Schmidt, Jul 17 2017

Number of permutations of [n] that avoid the patterns 321 and 2341. - Colin Defant, May 11 2018

The sequence solves the following problem: find all the pairs (i,j) such that i divides 1+j^2 and j divides 1+i^2. In fact, the pairs (a(n), a(n+1)), n > 0, are all the solutions. - Tomohiro Yamada, Dec 23 2018

Number of permutations in S_n whose principal order ideals in the Bruhat order are lattices (equivalently, modular, distributive, Boolean lattices). - Bridget Tenner, Jan 16 2020

From Wolfdieter Lang, Mar 30 2020: (Start)

a(n) is the upper left entry of the n-th power of the 2 X 2 tridiagonal matrix M_2 = Matrix([1,1], [1,2]) from A322602: a(n) = ((M_2)^n)[1,1].

Proof: (M_2)^2 = 3*M + 1_2 (with the 2 X 2 unit matrix 1_2) from the characteristic polynomial of M_2 (see a comment in A322602) and the Cayley-Hamilton theorem. The recurrence M^n = M*M^(n-1) leads to (M_n)^n = S(n, 3)*1_2 + S(n-a, 3)*(M - 3*1_2), for n >= 0, with S(n, 3) = F(2(n+1)) = A001906(n+1). Hence ((M_2)^n)[1,1] = S(n, 3) - 2*S(n-1, 3) = a(n) = F(2*n-1) = (1/(2*r+1))*r^(2*n-1)*(1 + (1/r^2)^(2*n-1)), with r = rho(5) = A001622 (golden ratio) (see the first Aug 31 2004 formula, using the recurrence of S(n, 3), and the Michael Somos Oct 28 2002 formula). This proves a conjecture of Gary W. Adamson in A322602.

The ratio a(n)/a(n-1) converges to r^2 = rho(5)^2 = A104457 for n -> infinity (see the a(n) formula in terms of r), which is one of the statements by Gary W. Adamson in A322602. (End)

a(n) is the number of ways to stack coins with a bottom row of n coins such that any coin not on the bottom row touches exactly two coins in the row below, and all the coins on any row are contiguous [Wilf, 2.12]. - Greg Dresden, Jun 29 2020

a(n) is the upper left entry of the (2*n)-th power of the 4 X 4 Jacobi matrix L with L(i,j)=1 if |i-j| = 1 and L(i,j)=0 otherwise. - Michael Shmoish, Aug 29 2020

All positive solutions of the indefinite binary quadratic F(1, -3, 1) := x^2 - 3*x*y + y^2, of discriminant 5, representing -1 (special Markov triples (1, y=x, z=y) if y <= z) are [x(n), y(n)] = [abs(F(2*n+1)), abs(F(2*n-1))], for n = -infinity..+infinity. (F(-n) = (-1)^(n+1)*F(n)). There is only this single family of proper solutions, and there are no improper solutions. [See also the Floor van Lamoen Nov 29 2001 comment, which uses this negative n, and my Jan 30 2015 comment.] - Wolfdieter Lang, Sep 23 2020

These are the denominators of the lower convergents to the golden ratio, tau; they are also the numerators of the upper convergents (viz. 1/1 < 3/2 < 8/5 < 21/13 < ... < tau < ... 13/8 < 5/3 < 2/1). - Clark Kimberling, Jan 02 2022

a(n+1) is the number of subgraphs of the path graph on n vertices. - Leen Droogendijk, Jun 17 2023

For n > 4, a(n+2) is the number of ways to tile this 3 x n "double-box" shape with squares and dominos (reflections or rotations are counted as distinct tilings). The double-box shape is made up of two horizontal strips of length n, connected by three vertical columns of length 3, and the center column can be located anywhere not touching the two outside columns.

_ _ _ _

|||_|||_|||_|||_|||

|| _ |_| _ _ ||

|||_|||_|||_|||_|||. - Greg Dresden and Ruishan Wu, Aug 25 2024

a(n+1) is the number of integer sequences a_1, ..., a_n such that for any number 1 <= k <= n, (a_1 + ... + a_k)^2 = a_1^3 + ... + a_k^3. - Yifan Xie, Dec 07 2024

Number of partitions of n in which greatest part is odd.

Number of partitions of n+1 into an even number of parts, the least being 1. Example: a(5)=4 because we have [5,1], [3,1,1,1], [2,1,1] and [1,1,1,1,1,1].

Also number of partitions of n+1 such that the largest part is even and occurs only once. Example: a(5)=4 because we have [6], [4,2], [4,1,1] and [2,1,1,1,1]. - Emeric Deutsch, Apr 05 2006

Also the number of partitions of n such that the number of odd parts and the number of even parts have opposite parities. Example: a(8)=10 is a count of these partitions: 8, 611, 521, 431, 422, 41111, 332, 32111, 22211, 2111111. - Clark Kimberling, Feb 01 2014, corrected Jan 06 2021

In Chaves 2011 see page 38 equation (3.20). - Michael Somos, Dec 28 2014

Suppose that c(0) = 1, that c(1), c(2), ... are indeterminates, that d(0) = 1, and that d(n) = -c(n) - c(n-1)*d(1) - ... - c(0)*d(n-1). When d(n) is expanded as a polynomial in c(1), c(2),..,c(n), the terms are of the form H*c(i_1)*c(i_2)*...*c(i_k). Let P(n) = [c(i_1), c(i_2), ..., c(i_k)], a partition of n. Then H is negative if P has an odd number of parts, and H is positive if P has an even number of parts. That is, d(n) has A027193(n) negative coefficients, A027187(n) positive coefficients, and A000041 terms. The maximal coefficient in d(n), in absolute value, is A102462(n). - Clark Kimberling, Dec 15 2016

Ramanujan theta functions: phi(q) := Sum_{k=-oo..oo} q^(k^2) (A000122), chi(q) := Prod_{k>=0} (1+q^(2k+1)) (A000700).

For all k >= 1, the k-generalized Fibonacci number F(k,n) satisfies the recurrence obtained by adding more terms to the recurrence of the Fibonacci numbers.

The number of tilings of an 1 X n rectangle with tiles of size 1 X 1, 1 X 2, ..., 1 X k is F(k,n).

T(k,n) is the number of 0-balanced ordered trees with n edges and height k (height is the number of edges from root to a leaf). - Emeric Deutsch, Jan 19 2007

Brlek et al. (2006) call this table "number of psp-polyominoes with flat bottom". - N. J. A. Sloane, Oct 30 2018

a(n) is the number of permutation matrices with a negative contribution to the determinant that is the Möbius function. See A174725 for how the determinant is defined. - Mats Granvik, May 26 2017

From Gus Wiseman, Jan 04 2021: (Start)

Also the number of ordered factorizations of n into an odd number of factors > 1. The unordered case is A339890. For example, the a(n) factorizations for n = 8, 12, 24, 30, 32, 36 are:

(8) (12) (24) (30) (32) (36)

(2*2*2) (2*2*3) (2*2*6) (2*3*5) (2*2*8) (2*2*9)

(2*3*2) (2*3*4) (2*5*3) (2*4*4) (2*3*6)

(3*2*2) (2*4*3) (3*2*5) (2*8*2) (2*6*3)

(2*6*2) (3*5*2) (4*2*4) (2*9*2)

(3*2*4) (5*2*3) (4*4*2) (3*2*6)

(3*4*2) (5*3*2) (8*2*2) (3*3*4)

(4*2*3) (2*2*2*2*2) (3*4*3)

(4*3*2) (3*6*2)

(6*2*2) (4*3*3)

(6*2*3)

(6*3*2)

(9*2*2)

(End)

The sequence of column k satisfies a linear recurrence with constant coefficients of order k*(k+1)/2 = A000217(k).

Stirling transform of A005212(n)=[1,0,6,0,120,0,5040,...] is a(n)=[1,1,7,37,271,...]. - Michael Somos, Mar 04 2004

Occurs also as first column of a matrix-inversion occurring in a sum-of-like-powers problem. Consider the problem for any fixed natural number m>2 of finding solutions to sum(k=1,n,k^m) = (k+1)^m. Erdos conjectured that there are no solutions for n,m>2. Let D be the matrix of differences of D[m,n] := sum(k=1,n,k^m) - (k+1)^m. Then the generating functions for the rows of this matrix D constitute a set of polynomials in n (for varying n along columns) and the m-th polynomial defining the m-th row. Let GF_D be the matrix of the coefficients of this set of polynomials. Then the present sequence is the (unsigned) second column of GF_D^-1. - Gottfried Helms, Apr 01 2007