cp's OEIS Frontend

Row sums = the Lucas numbers, A000032 starting (1, 3, 4, 7, 11, 18,...).

Row sums = A074878: (1, 2, 6, 14, 32, 70, 150,...).

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

Powers of 2 doubled up. The usual OEIS policy is to omit the duplicates in such cases (when this would become A000079). This is an exception.

Number of symmetric compositions of n: e.g., 5 = 2+1+2 = 1+3+1 = 1+1+1+1+1 so a(5) = 4; 6 = 3+3 = 2+2+2 = 1+4+1 = 2+1+1+2 = 1+2+2+1 = 1+1+2+1+1 = 1+1+1+1+1+1 so a(6) = 8. - Henry Bottomley, Dec 10 2001

This sequence is the number of digits of each term of A061519. - Dmitry Kamenetsky, Jan 17 2009

Starting with offset 1 = binomial transform of [1, 1, -1, 3, -7, 17, -41, ...]; where A001333 = (1, 1, 3, 7, 17, 41, ...). - Gary W. Adamson, Mar 25 2009

a(n+1) is the number of symmetric subsets of [n]={1,2,...,n}. A subset S of [n] is symmetric if k is an element of S implies (n-k+1) is an element of S. - Dennis P. Walsh, Oct 27 2009

INVERT and inverse INVERT transforms give A006138, A039834(n-1).

The Kn21 sums, see A180662, of triangle A065941 equal the terms of this sequence. - Johannes W. Meijer, Aug 15 2011

First differences of A027383. - Jason Kimberley, Nov 01 2011

Run lengths in A079944. - Jeremy Gardiner, Nov 21 2011

Number of binary palindromes (A006995) between 2^(n-1) and 2^n (for n>1). - Hieronymus Fischer, Feb 17 2012

Pisano period lengths: 1, 1, 4, 1, 8, 4, 6, 1, 12, 8, 20, 4, 24, 6, 8, 1, 16, 12, 36, 8, ... . - R. J. Mathar, Aug 10 2012

Range of row n of the Circular Pascal array of order 4. - Shaun V. Ault, May 30 2014

a(n) is the number of permutations of length n avoiding both 213 and 312 in the classical sense which are breadth-first search reading words of increasing unary-binary trees. For more details, see the entry for permutations avoiding 231 at A245898. - Manda Riehl, Aug 05 2014

Also, the decimal representation of the diagonal from the origin to the corner (and from the corner to the origin except for the initial term) of the n-th stage of growth of the two-dimensional cellular automaton defined by "Rule 190", based on the 5-celled von Neumann neighborhood when initialized with a single black (ON) cell at stage zero. - Robert Price, May 10 2017

a(n + 1) + n - 1, n > 0, is the number of maximal subsemigroups of the monoid of partial order-preserving or -reversing mappings on a set with n elements. See the East et al. link. - James Mitchell and Wilf A. Wilson, Jul 21 2017

Number of symmetric stairs with n cells. A stair is a snake polyomino allowing only two directions for adjacent cells: east and north. See A005418. - Christian Barrientos, May 11 2018

For n >= 4, a(n) is the exponent of the group of the Gaussian integers in a reduced system modulo (1+i)^(n+2). See A302254. - Jianing Song, Jun 27 2018

a(n) is the number of length-(n+1) binary sequences, denoted , with s(1)=1 and with s(i+1)=s(i) for odd i. - Dennis P. Walsh, Sep 06 2018

a(n+1) is the number of subsets of {1,2,..,n} in which all differences between successive elements of subsets are even. For example, for n = 7, a(6) = 8 and the 8 subsets are {7}, {1,7}, {3,7}, {5,7}, {1,3,7}, {1,5,7}, {3,5,7}, {1,3,5,7}. For odd differences between elements see Comment in A000045 (Fibonacci numbers). - Enrique Navarrete, Jul 01 2020

Also, the number of walks of length n on the graph x--y--z, starting at x. - Sean A. Irvine, May 30 2025

In other words, a(n), n >= 2, is the least k such that prime(n) divides 2^k-1.

Concerning the complexity of computing this sequence, see for example Bach and Shallit, p. 115, exercise 8.

Also A002326((p_n-1)/2). Conjecture: If p_n is not a Wieferich prime (1093, 3511, ...) then A002326(((p_n)^k-1)/2) = a(n)*(p_n)^(k-1). - Vladimir Shevelev, May 26 2008

If for distinct i,j,...,k we have a(i)=a(j)=...=a(k) then the number N = p_i*p_j*...*p_k is in A001262 and moreover A137576((N-1)/2) = N. For example, a(16)=a(37)=a(255)=52. Therefore we could take N = p_16*p_37*p_255 = 53*157*1613 = 13421773. - Vladimir Shevelev, Jun 14 2008

Also degree of the irreducible polynomial factors for the polynomial (x^p+1)/(x+1) over GF(2), where p is the n-th prime. - V. Raman, Oct 04 2012

Is this the same as the smallest k > 1 not already in the sequence such that p = prime(n) is a factor of 2^k-1 (A270600)? If the answer is yes, is the sequence a permutation of the positive integers > 1? - Felix Fröhlich, Feb 21 2016. Answer: No, it is easy to prove that 6 is missing and obviously 11 appears twice. - N. J. A. Sloane, Feb 21 2016

pi(A112927(m)) is the index at which a given number m first appears in this sequence. - M. F. Hasler, Feb 21 2016

Coefficient array for Morgan-Voyce polynomial b(n,x). A053122 (unsigned) is the coefficient array for B(n,x). Reversal of A054142. - Paul Barry, Jan 19 2004

This triangle is formed from even-numbered rows of triangle A011973 read in reverse order. - Philippe Deléham, Feb 16 2004

T(n,k) is the number of nondecreasing Dyck paths of semilength n+1, having k+1 peaks. T(n,k) is the number of nondecreasing Dyck paths of semilength n+1, having k peaks at height >= 2. T(n,k) is the number of directed column-convex polyominoes of area n+1, having k+1 columns. - Emeric Deutsch, May 31 2004

Riordan array (1/(1-x), x/(1-x)^2). - Paul Barry, May 09 2005

The triangular matrix a(n,k) = (-1)^(n+k)*T(n,k) is the matrix inverse of A039599. - Philippe Deléham, May 26 2005

The n-th row gives absolute values of coefficients of reciprocal of g.f. of bottom-line of n-wave sequence. - Floor van Lamoen (fvlamoen(AT)planet.nl), Sep 24 2006

Unsigned version of A129818. - Philippe Deléham, Oct 25 2007

T(n, k) is also the number of idempotent order-preserving full transformations (of an n-chain) of height k >=1 (height(alpha) = |Im(alpha)|) and of waist n (waist(alpha) = max(Im(alpha))). - Abdullahi Umar, Oct 02 2008

A085478 is jointly generated with A078812 as a triangular array of coefficients of polynomials u(n,x): initially, u(1,x) = v(1,x) = 1; for n>1, u(n,x) = u(n-1,x)+x*v(n-1)x and v(n,x) = u(n-1,x)+(x+1)*v(n-1,x). See the Mathematica section. - Clark Kimberling, Feb 25 2012

Per Kimberling's recursion relations, see A102426. - Tom Copeland, Jan 19 2016

Subtriangle of the triangle given by (0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Mar 26 2012

T(n,k) is also the number of compositions (ordered partitions) of 2*n+1 into 2*k+1 parts which are all odd. Proof: The o.g.f. of column k, x^k/(1-x)^(2*k+1) for k >= 0, is the o.g.f. of the odd-indexed members of the sequence with o.g.f. (x/(1-x^2))^(2*k+1) (bisection, odd part). Thus T(n,k) is obtained from the sum of the multinomial numbers A048996 for the partitions of 2*n+1 into 2*k+1 parts, all of which are odd. E.g., T(3,1) = 3 + 3 from the numbers for the partitions [1,1,5] and [1,3,3], namely 3!/(2!*1!) and 3!/(1!*2!), respectively. The number triangle with the number of these partitions as entries is A152157. - Wolfdieter Lang, Jul 09 2012

The matrix elements of the inverse are T^(-1)(n,k) = (-1)^(n+k)*A039599(n,k). - R. J. Mathar, Mar 12 2013

T(n,k) = A258993(n+1,k) for k = 0..n-1. - Reinhard Zumkeller, Jun 22 2015

The n-th row polynomial in descending powers of x is the n-th Taylor polynomial of the algebraic function F(x)*G(x)^n about 0, where F(x) = (1 + sqrt(1 + 4*x))/(2*sqrt(1 + 4*x)) and G(x) = ((1 + sqrt(1 + 4*x))/2)^2. For example, for n = 4, (1 + sqrt(1 + 4*x))/(2*sqrt(1 + 4*x)) * ((1 + sqrt(1 + 4*x))/2)^8 = (x^4 + 10*x^3 + 15*x^2 + 7*x + 1) + O(x^5). - Peter Bala, Feb 23 2018

Row n also gives the coefficients of the characteristc polynomial of the tridiagonal n X n matrix M_n given in A332602: Phi(n, x) := Det(M_n - x*1_n) = Sum_{k=0..n} T(n, k)*(-x)^k, for n >= 0, with Phi(0, x) := 1. - Wolfdieter Lang, Mar 25 2020

It appears that the largest root of the n-th degree polynomial is equal to the sum of the distinct diagonals of a (2*n+1)-gon including the edge, 1. The largest root of x^3 - 6*x^2 + 5*x - 1 is 5.048917... = the sum of (1 + 1.80193... + 2.24697...). Alternatively, the largest root of the n-th degree polynomial is equal to the square of sigma(2*n+1). Check: 5.048917... is the square of sigma(7), 2.24697.... Given N = 2*n+1, sigma(N) (N odd) can be defined as 1/(2*sin(Pi/(2*N))). Relating to the 9-gon, the largest root of x^4 - 10*x^3 + 15*x^2 - 7*x + 1 is 8.290859..., = the sum of (1 + 1.879385... + 2.532088... + 2.879385...), and is the square of sigma(9), 2.879385... Refer to A231187 for a further clarification of sigma(7). - Gary W. Adamson, Jun 28 2022

For n >=1, the n-th row is given by the coefficients of the minimal polynomial of -4*sin(Pi/(4*n + 2))^2. - Eric W. Weisstein, Jul 12 2023

Denoting this lower triangular array by L, then L * diag(binomial(2*k,k)^2) * transpose(L) is the LDU factorization of A143007, the square array of crystal ball sequences for the A_n X A_n lattices. - Peter Bala, Feb 06 2024

T(n, k) is the number of occurrences of the periodic substring (01)^k in the periodic string (01)^n (see Proposition 4.7 at page 7 in Fang). - Stefano Spezia, Jun 09 2024

Multiplicative suborder of 2 (mod 2n+1) (or sord(2, 2n+1)).

This is called quasi-order of 2 mod b, with b = 2*n+1, for n >= 1, in the Hilton/Pederson reference.

For the complexity of computing this, see A002326.

Also, the order of the so-called "milk shuffle" of a deck of n cards, which maps cards (1,2,...,n) to (1,n,2,n-1,3,n-2,...). See the paper of Lévy. - Jeffrey Shallit, Jun 09 2019

It appears that under iteration of the base-n Kaprekar map, for even n > 2 (A165012, A165051, A165090, A151949 in bases 4, 6, 8, 10), almost all cycles are of length a(n/2 - 1); proved under the additional constraint that the cycle contains at least one element satisfying "number of digits (n-1) - number of digits 0 = o(total number of digits)". - Joseph Myers, Sep 05 2009

From Gary W. Adamson, Sep 20 2011: (Start)

a(n) can be determined by the cycle lengths of iterates using x^2 - 2, seed 2*cos(2*Pi/N); as shown in the A065941 comment of Sep 06 2011. The iterative map of the logistic equation 4x*(1-x) is likewise chaotic with the same cycle lengths but initiating the trajectory with sin^2(2*Pi/N), N = 2n+1 [Kappraff & Adamson, 2004]. Chaotic terms with the identical cycle lengths can be obtained by applying Newton's method to i = sqrt(-1) [Strang, also Kappraff and Adamson, 2003], resulting in the morphism for the cot(2*Pi/N) trajectory: (x^2-1)/2x. (End)

From Gary W. Adamson, Sep 11 2019: (Start)

Using x^2 - 2 with seed 2*cos(Pi/7), we obtain the period-three trajectory 1.8019377...-> 1.24697...-> -0.445041... For an odd prime N, the trajectory terms represent diagonal lengths of regular star 2N-gons, with edge the shortest value (0.445... in this case.) (Cf. "Polygons and Chaos", p. 9, Fig 4.) We can normalize such lengths by dividing through with the lowest value, giving 3 diagonals of the 14-gon: (1, 2.801937..., 4.048917...). Label the terms ranked in magnitude with odd integers (1, 3, 5), and we find that the diagonal lengths are in agreement with the diagonal formula (sin(j*Pi)/14)/(sin(Pi/14)), with j = (1,3,5). (End)

Roots of signed n-th row A054142 polynomials are chaotic with respect to the operation (-2, x^2), with cycle lengths a(n). Example: starting with a root to x^3 - 5x^2 + 6x - 1 = 0; (2 + 2*cos(2*Pi/N) = 3.24697...); we obtain the trajectory (3.24697...-> 1.55495...-> 0.198062...); the roots to the polynomial with cycle length 3 matching a(3) = 3. - Gary W. Adamson, Sep 21 2011

From Juhani Heino, Oct 26 2015: (Start)

Start a sequence with numbers 1 and n. For next numbers, add previous numbers going backwards until the sum is even. Then the new number is sum/2. I conjecture that the sequence returns to 1,n and a(n) is the cycle length.

For example:

1,7,4,2,1,7,... so a(7) = 4.

1,6,3,5,4,2,1,6,... so a(6) = 6. (End)

From Juhani Heino, Nov 06 2015: (Start)

Proof of the above conjecture: Let n = -1/2; thus 2n + 1 = 0, so operations are performed mod (2n + 1). When the member is even, it is divided by 2. When it is odd, multiply by n, so effectively divide by -2. This is all well-defined in the sense that new members m are 1 <= m <= n. Now see what happens starting from an odd member m. The next member is -m/2. As long as there are even members, divide by 2 and end up with an odd -m/(2^k). Now add all the members starting with m. The sum is m/(2^k). It's divided by 2, so the next member is m/(2^(k+1)). That is the same as (-m/(2^k))/(-2), as with the definition.

So actually start from 1 and always divide by 2, although the sign sometimes changes. Eventually 1 is reached again. The chain can be traversed backwards and then 2^(cycle length) == +-1 (mod 2n + 1).

To conclude, we take care of a(0): sequence 1,0 continues with zeros and never returns to 1. So let us declare that cycle length 0 means unavailable. (End)

From Gary W. Adamson, Aug 20 2019: (Start)

Terms in the sequence can be obtained by applying the doubling sequence mod (2n + 1), then counting the terms until the next term is == +1 (mod 2n + 1). Example: given 25, the trajectory is (1, 2, 4, 8, 16, 7, 14, 3, 6, 12).

The cycle ends since the next term is 24 == -1 (mod 25) and has a period of 10. (End)

From Gary W. Adamson, Sep 04 2019: (Start)

Conjecture of Kappraff and Adamson in "Polygons and Chaos", p. 13 Section 7, "Chaos and Number": Given the cycle length for N = 2n + 1, the same cycle length is present in bases 4, 9, 16, 25, ..., m^2, for the expansion of 1/N.

Examples: The cycle length for 7 is 3, likewise for 1/7 in base 4: 0.021021021.... In base 9 the expansion of 1/7 is 0.125125125... Check: The first few terms are 1/9 + 2/81 + 5/729 = 104/279 = 0.1426611... (close to 1/7 = 0.142857...). (End)

From Gary W. Adamson, Sep 24 2019: (Start)

An exception to the rule for 1/N in bases m^2: (when N divides m^2 as in 1/7 in base 49, = 7/49, rational). When all terms in the cycle are the same, the identity reduces to 1/N in (some bases) = .a, a, a, .... The minimal values of "a" for 1/N are provided as examples, with the generalization 1/N in base (N-1)^2 = .a, a, a, ... for N odd:

1/3 in base 4 = .1, 1, 1, ...

1/5 in base 16 = .3, 3, 3, ...

1/7 in base 36 = .5, 5, 5, ...

1/9 in base 64 = .7, 7, 7, ...

1/11 in base 100 = .9, 9, 9, ... (Check: the first three terms are 9/100 +9/(100^2) + 9/(100^3) = 0.090909 where 1/11 = 0.09090909...). (End)

For N = 2n+1, the corresponding entry is equal to the degree of the polynomial for N shown in (Lang, Table 2, p. 46). As shown, x^3 - 3x - 1 is the minimal polynomial for N = 9, with roots (1.87938..., -1.53208..., 0.347296...); matching the (abs) values of the 2*cos(Pi/9) trajectory using x^2 - 2. Thus, a(4) = 3. If N is prime, the polynomials shown in Table 2 are the same as those for the same N in A065941. If different, the minimal polynomials shown in Table 2 are factors of those in A065941. - Gary W. Adamson, Oct 01 2019

The terms in the 2*cos(Pi/N) trajectory (roots to the minimal polynomials in A187360 and (Lang)), are quickly obtained from the doubling trajectory (mod N) by using the operation L(m) 2*cos(x)--> 2*cos(m*x), where L(2), the second degree Lucas polynomial (A034807) is x^2 - 2. Relating to the heptagon and using seed 2*cos(Pi/7), we obtain the trajectory 1.8019..., 1.24697..., and 0.445041....; cyclic with period 3. All such roots can be derived from the N-th roots of Unity and can be mapped on the Vesica Piscis. Given the roots of Unity (Polar 1Angle(k*2*Pi/N), k = 1, 2, ..., (N-1)/2) the Vesica Piscis maps these points on the left (L) circle to the (R) circle by adding 1A(0) or (a + b*I) = (1 + 0i). But this operation is the same as vector addition in which the resultant vector is 1 + 1A(k*(2*Pi/N)). Example: given the radius at 2*Pi/7 on the left circle, this maps to (1 + 1A(2*Pi/7)) on the right circle; or 1A(2*Pi/7) --> (1.8019377...A(Pi/7). Similarly, 1 + 1A((2)*2*Pi/7)) maps to (1.24697...A (2*Pi/7); and 1 + 1A(3*2*Pi/7) maps to (.0445041...A(3*Pi/7). - Gary W. Adamson, Oct 23 2019

From Gary W. Adamson, Dec 01 2021: (Start)

As to segregating the two sets: (A014659 terms are those N = (2*n+1), N divides (2^m - 1), and (A014657 terms are those N that divide (2^m + 1)); it appears that the following criteria apply: Given IcoS(N, 1) (cf. Lang link "On the Equivalence...", p. 16, Definition 20), if the number of odd terms is odd, then N belongs to A014659, otherwise A014657. In IcoS(11, 1): (1, 2, 4, 3, 5), three odd terms indicate that 11 is a term in A014657. IcoS(15, 1) has the orbit (1, 2, 4, 7) with two odd terms indicating that 15 is a term in A014659.

It appears that if sin(2^m * Pi/N) has a negative sign, then N is in A014659; otherwise N is in A014657. With N = 15, m is 4 and sin(16 * Pi/15) is -0.2079116... If N is 11, m is 5 and sin(32 * Pi/11) is 0.2817325. (End)

On the iterative map using x^2 - 2, (Devaney, p. 126) states that we must find the function that takes 2*cos(Pi) -> 2*cos(2*Pi). "However, we may write 2*cos(2*Pi) = 2*(2*cos^2(Pi) - 1) = (2*cos(Pi))^2 - 2. So the required function is x^2 - 2." On the period 3 implies chaos theorem of James Yorke and T.Y. Li, proved in 1975; Devaney (p. 133) states that if F is continuous and we find a cycle of period 3, there are infinitely many other cycles for this map with every possible period. Check: The x^2 - 2 orbit for 7 has a period of 3, so this entry has periodic points of all other periods. - Gary W. Adamson, Jan 04 2023

It appears that a(n) is the length of the cycle starting at 2/(2*n+1) for the map x->1 - abs(2*x-1). - Michel Marcus, Jul 16 2025

Row sums are Fibonacci(n+2). Diagonal sums are A016116. - Paul Barry, Jul 07 2004

Riordan array (1/(1-x), x/(1-x^2)). Matrix inverse is A106180. - Paul Barry, Apr 24 2005

As an infinite lower triangular matrix * [1,2,3,...] = A055244. - Gary W. Adamson, Dec 23 2008

From Emeric Deutsch, Jun 18 2010: (Start)

T(n,k) is the number of alternating parity increasing subsequences of {1,2,...,n} of size k, starting with an odd number (Terquem's problem, see the Riordan reference, p. 17). Example: T(8,5)=6 because we have 12345, 12347, 12367, 12567, 14567, and 34567.

T(n,k) is the number of alternating parity increasing subsequences of {1,2,...,n,n+1} of size k, starting with an even number. Example: T(7,4)=5 because we have 2345, 2347, 2367, 2567, and 4567. (End)

From L. Edson Jeffery, Mar 01 2011: (Start)

This triangle can be constructed as follows. Interlace two copies of the table of binomial coefficients to get the preliminary table

1

1

1 1

1 1

1 2 1

1 2 1

1 3 3 1

1 3 3 1

...,

then shift each entire r-th column up r rows, r=0,1,2,.... Also, a signed version of this sequence (A187660 in tabular form) begins with

1;

1, -1;

1, -1, -1;

1, -2, -1, 1;

1, -2, -3, 1, 1;

...

(compare with A066170, A130777). Let T(N,k) denote the k-th entry in row N of the signed table. Then, for N>1, row N gives the coefficients of the characteristic function p_N(x) = Sum_{k=0..N} T(N,k)*x^(N-k) = 0 of the N X N matrix U_N=[(0 ... 0 1);(0 ... 0 1 1);...;(0 1 ... 1);(1 ... 1)]. Now let Q_r(t) be a polynomial with recurrence relation Q_r(t)=t*Q_(r-1)(t)-Q_(r-2)(t) (r>1), with Q_0(t)=1 and Q_1(t)=t. Then p_N(x)=0 has solutions Q_(N-1)(phi_j), where phi_j=2*(-1)^(j-1)*cos(j*Pi/(2*N+1)), j=1,2,...,N.

For example, row N=3 is {1,-2,-1,1}, giving the coefficients of the characteristic function p_3(x) = x^3-2*x^2-x+1 = 0 for the 3 X 3 matrix U_3=[(0 0 1);(0 1 1);(1 1 1)], with eigenvalues Q_2(phi_j)=[2*(-1)^(j-1)*cos(j*Pi/7)]^2-1, j=1,2,3. (End)

Given the signed polynomials (+--++--,...) of the triangle, the largest root of the n-th row polynomial is the longest (2n+1) regular polygon diagonal length, with edge = 1. Example: the largest root to x^3 - 2x^2 - x + 1 = 0 is 2.24697...; the longest heptagon diagonal, sin(3*Pi/7)/sin(Pi/7). - Gary W. Adamson, Sep 06 2011

Given the signed polynomials from Gary W. Adamson's comment, the largest root of the n-th polynomial also equals the length from the center to a corner (vertex) of a regular 2*(2*n+1)-sided polygon with side (edge) length = 1. - L. Edson Jeffery, Jan 01 2012

Put f(x,0) = 1 and f(x,n) = x + 1/f(x,n-1). Then f(x,n) = u(x,n)/v(x,n), where u(x,n) and v(x,n) are polynomials. The flattened triangles of coefficients of u and v are both essentially A046854, as indicated by the Mathematica program headed "Polynomials". - Clark Kimberling, Oct 12 2014

From Jeremy Dover, Jun 07 2016: (Start)

T(n,k) is the number of binary strings of length n+1 starting with 0 that have exactly k pairs of consecutive 0's and no pairs of consecutive 1's.

T(n,k) is the number of binary strings of length n+2 starting with 1 that have exactly k pairs of consecutive 0's and no pairs of consecutive 1's. (End)

Matrix inverse of A124645.

Let L(n,x) = Sum_{k=0..n} T(n,k)*x^(n-k) and Pi=3.14...:

L(n,x) = Product_{k=1..n} (x - 2*cos((2*k-1)*Pi/(2*n+1)));

Sum_{k=0..n} T(n,k) = L(n,1) = A010892(n+1);

Sum_{k=0..n} abs(T(n,k)) = A000045(n+2);

abs(T(n,k)) = A065941(n,k), T(n,k) = A065941(n,k)*A087960(k);

T(2*n,k) + T(2*n+1,k+1) = 0 for 0 <= k <= 2*n;

T(n,0) = A000012(n) = 1; T(n,1) = -1 for n > 0;

T(n,2) = -(n-1) for n > 1; T(n,3) = A000027(n)=n for n > 2;

T(n,4) = A000217(n-3) for n > 3; T(n,5) = -A000217(n-4) for n > 4;

T(n,6) = -A000292(n-5) for n > 5; T(n,7) = A000292(n-6) for n > 6;

T(n,n-3) = A058187(n-3)*(-1)^floor(n/2) for n > 2;

T(n,n-2) = A008805(n-2)*(-1)^floor((n+1)/2) for n > 1;

T(n,n-1) = A008619(n-1)*(-1)^floor(n/2) for n > 0;

T(n,n) = L(n,0) = (-1)^floor((n+1)/2);

L(n,1) = A010892(n+1); L(n,-1) = A061347(n+2);

L(n,2) = 1; L(n,-2) = A005408(n)*(-1)^n;

L(n,3) = A001519(n); L(n,-3) = A002878(n)*(-1)^n;

L(n,4) = A001835(n+1); L(n,-4) = A001834(n)*(-1)^n;

L(n,5) = A004253(n); L(n,-5) = A030221(n)*(-1)^n;

L(n,6) = A001653(n); L(n,-6) = A002315(n)*(-1)^n;

L(n,7) = A049685(n); L(n,-7) = A033890(n)*(-1)^n;

L(n,8) = A070997(n); L(n,-8) = A057080(n)*(-1)^n;

L(n,9) = A070998(n); L(n,-9) = A057081(n)*(-1)^n;

L(n,10) = A072256(n+1); L(n,-10) = A054320(n)*(-1)^n;

L(n,11) = A078922(n+1); L(n,-11) = A097783(n)*(-1)^n;

L(n,12) = A077417(n); L(n,-12) = A077416(n)*(-1)^n;

L(n,13) = A085260(n);

L(n,14) = A001570(n); L(n,-14) = A028230(n)*(-1)^n;

L(n,n) = A108366(n); L(n,-n) = A108367(n).

Row n of the matrix inverse (A124645) has g.f.: x^floor(n/2)*(1-x)^(n-floor(n/2)). - Paul D. Hanna, Jun 12 2005

From L. Edson Jeffery, Mar 12 2011: (Start)

Conjecture: Let N=2*n+1, with n > 2. Then T(n,k) (0 <= k <= n) gives the k-th coefficient in the characteristic function p_N(x)=0, of degree n in x, for the n X n tridiagonal unit-primitive matrix G_N (see [Jeffery]) of the form

G_N=A_{N,1}=

(0 1 0 ... 0)

(1 0 1 0 ... 0)

(0 1 0 1 0 ... 0)

...

(0 ... 0 1 0 1)

(0 ... 0 1 1),

with solutions phi_j = 2*cos((2*j-1)*Pi/N), j=1,2,...,n. For example, for n=3,

G_7=A_{7,1}=

(0 1 0)

(1 0 1)

(0 1 1).

We have {T(3,k)}=(1,-1,-2,1), while the characteristic function of G_7 is p(x) = x^3-x^2-2*x+1 = 0, with solutions phi_j = 2*cos((2*j-1)*Pi/7), j=1,2,3. (End)

The triangle sums, see A180662 for their definitions, link A108299 with several sequences, see the crossrefs. - Johannes W. Meijer, Aug 08 2011

The roots to the polynomials are chaotic using iterates of the operation (x^2 - 2), with cycle lengths L and initial seeds returning to the same term or (-1)* the seed. Periodic cycle lengths L are shown in A003558 such that for the polynomial represented by row r, the cycle length L is A003558(r-1). The matrices corresponding to the rows as characteristic polynomials are likewise chaotic [cf. Kappraff et al., 2005] with the same cycle lengths but substituting 2*I for the "2" in (x^2 - 2), where I = the Identity matrix. For example, the roots to x^3 - x^2 - 2x + 1 = 0 are 1.801937..., -1.246979..., and 0.445041... With 1.801937... as the initial seed and using (x^2 - 2), we obtain the 3-period trajectory of 8.801937... -> 1.246979... -> -0.445041... (returning to -1.801937...). We note that A003558(2) = 3. The corresponding matrix M is: [0,1,0; 1,0,1; 0,1,1,]. Using seed M with (x^2 - 2*I), we obtain the 3-period with the cycle completed at (-1)*M. - Gary W. Adamson, Feb 07 2012

Number of subsets with at least 3 elements of an n-element set.

For n>4, number of simple rank-(n-1) matroids over S_n.

Number of non-interval subsets of {1,2,3,...,n} (cf. A000124). - Jose Luis Arregui (arregui(AT)unizar.es), Jun 27 2006

The partial sums of the second diagonal of A008292 or third column of A123125. - Tom Copeland, Sep 09 2008

a(n) is the number of binary sequences of length n having at least three 0's. - Geoffrey Critzer, Feb 11 2009

Starting with "1" = eigensequence of a triangle with the tetrahedral numbers (1, 4, 10, 20, ...) as the left border and the rest 1's. - Gary W. Adamson, Jul 24 2010

a(n) is also the number of crossing set partitions of [n+1] with two blocks. - Peter Luschny, Apr 29 2011

The Kn24 sums, see A180662, of triangle A065941 equal the terms (doubled) of this sequence minus the three leading zeros. - Johannes W. Meijer, Aug 14 2011

From L. Edson Jeffery, Dec 28 2011: (Start)

Nonzero terms of this sequence can be found from the row sums of the fourth sub-triangle extracted from Pascal's triangle as indicated below by braces:

1;

1, 1;

1, 2, 1;

{1}, 3, 3, 1;

{1, 4}, 6, 4, 1;

{1, 5, 10}, 10, 5, 1;

{1, 6, 15, 20}, 15, 6, 1;

... (End)

Partial sums of A000295 (Eulerian Numbers, Column 2).

Second differences equal 2^(n-2) - 1, for n >= 4. - Richard R. Forberg, Jul 11 2013

Starting (0, 0, 1, 5, 16, ...) is the binomial transform of (0, 0, 1, 2, 2, 2, ...). - Gary W. Adamson, Jul 27 2015

a(n - 1) is the rank of the divisor class group of the moduli space of stable rational curves with n marked points, see Keel p. 550. - Harry Richman, Aug 10 2024