cp's OEIS Frontend

G.f. for row polynomials S(n,x) (signed triangle): 1/(1-x*z+z^2). Unsigned triangle |a(n,m)| has Fibonacci polynomials F(n+1,x) as row polynomials with g.f. 1/(1-x*z-z^2). |a(n,m)| triangle has rows of Pascal's triangle A007318 in the even-numbered diagonals (odd-numbered ones have only 0's).

Row sums (unsigned triangle) A000045(n+1) (Fibonacci). Row sums (signed triangle) S(n,1) sequence = periodic(1,1,0,-1,-1,0) = A010892.

Alternating row sums A049347(n) = S(n,-1) = periodic(1,-1,0). - Wolfdieter Lang, Nov 04 2011

S(n,x) is the characteristic polynomial of the adjacency matrix of the n-path. - Michael Somos, Jun 24 2002

S(n,x) is also the matching polynomial of the n-path. - Eric W. Weisstein, Apr 10 2017

|T(n,k)| = number of compositions of n+1 into k+1 odd parts. Example: |T(7,3)| = 10 because we have (1,1,3,3), (1,3,1,3), (1,3,3,1), (3,1,1,3), (3,1,3,1), (3,3,1,1), (1,1,1,5), (1,1,5,1), (1,5,1,1) and (5,1,1,1). - Emeric Deutsch, Apr 09 2005

S(n,x)= R(n,x) + S(n-2,x), n >= 2, S(-1,x)=0, S(0,x)=1, R(n,x):=2*T(n,x/2) = Sum_{m=0..n} A127672(n,m)*x^m (monic integer Chebyshev T-Polynomials). This is the rewritten so-called trace of the transfer matrix formula for the T-polynomials. - Wolfdieter Lang, Dec 02 2010

In a regular N-gon inscribed in a unit circle, the side length is d(N,1) = 2*sin(Pi/N). The length ratio R(N,k):=d(N,k)/d(N,1) for the (k-1)-th diagonal, with k from {2,3,...,floor(N/2)}, N >= 4, equals S(k-1,x) = sin(k*Pi/N)/sin(Pi/N) with x=rho(N):=R(N,2) = 2*cos(Pi/N). Example: N=7 (heptagon), rho=R(7,2), sigma:=R(N,3) = S(2,rho) = rho^2 - 1. Motivated by the quoted paper by P. Steinbach. - Wolfdieter Lang, Dec 02 2010

From Wolfdieter Lang, Jul 12 2011: (Start)

In q- or basic analysis, q-numbers are [n]_q := S(n-1,q+1/q) = (q^n-(1/q)^n)/(q-1/q), with the row polynomials S(n,x), n >= 0.

The zeros of the row polynomials S(n-1,x) are (from those of Chebyshev U-polynomials):

x(n-1;k) = +- t(k,rho(n)), k = 1..ceiling((n-1)/2), n >= 2, with t(n,x) the row polynomials of A127672 and rho(n):= 2*cos(Pi/n). The simple vanishing zero for even n appears here as +0 and -0.

Factorization of the row polynomials S(n-1,x), x >= 1, in terms of the minimal polynomials of cos(2 Pi/2), called Psi(n,x), with coefficients given by A181875/A181876:

S(n-1,x) = (2^(n-1))*Product_{n>=1}(Psi(d,x/2), 2 < d | 2n).

(From the rewritten eq. (3) of the Watkins and Zeitlin reference, given under A181872.) [See the W. Lang ArXiv link, Proposition 9, eq. (62). - Wolfdieter Lang, Apr 14 2018]

(End)

The discriminants of the S(n,x) polynomials are found in A127670. - Wolfdieter Lang, Aug 03 2011

This is an example for a subclass of Riordan convolution arrays (lower triangular matrices) called Bell arrays. See the L. W. Shapiro et al. reference under A007318. If a Riordan array is named (G(z),F(z)) with F(z)=z*Fhat(z), the o.g.f. for the row polynomials is G(z)/(1-x*z*Fhat(z)), and it becomes a Bell array if G(z)=Fhat(z). For the present Bell type triangle G(z)=1/(1+z^2) (see the o.g.f. comment above). This leads to the o.g.f. for the column no. k, k >= 0, x^k/(1+x^2)^(k+1) (see the formula section), the one for the row sums and for the alternating row sums (see comments above). The Riordan (Bell) A- and Z-sequences (defined in a W. Lang link under A006232, with references) have o.g.f.s 1-x*c(x^2) and -x*c(x^2), with the o.g.f. of the Catalan numbers A000108. Together they lead to a recurrence given in the formula section. - Wolfdieter Lang, Nov 04 2011

The determinant of the N x N matrix S(N,[x[1], ..., x[N]]) with elements S(m-1,x[n]), for n, m = 1, 2, ..., N, and for any x[n], is identical with the determinant of V(N,[x[1], ..., x[N]]) with elements x[n]^(m-1) (a Vandermondian, which equals Product_{1 <= i < j<= N} (x[j] - x[i])). This is a special instance of a theorem valid for any N >= 1 and any monic polynomial system p(m,x), m>=0, with p(0,x) = 1. For this theorem see the Vein-Dale reference, p. 59. Thanks to L. Edson Jeffery for an email asking for a proof of the non-singularity of the matrix S(N,[x[1], ...., x[N]]) if and only if the x[j], j = 1..N, are pairwise distinct. - Wolfdieter Lang, Aug 26 2013

These S polynomials also appear in the context of modular forms. The rescaled Hecke operator T*n = n^((1-k)/2)*T_n acting on modular forms of weight k satisfies T*(p^n) = S(n, T*p), for each prime p and positive integer n. See the Koecher-Krieg reference, p. 223. - _Wolfdieter Lang, Jan 22 2016

For a shifted o.g.f. (mod signs), its compositional inverse, and connections to Motzkin and Fibonacci polynomials, non-crossing partitions and other combinatorial structures, see A097610. - Tom Copeland, Jan 23 2016

From M. Sinan Kul, Jan 30 2016; edited by Wolfdieter Lang, Jan 31 2016 and Feb 01 2016: (Start)

Solutions of the Diophantine equation u^2 + v^2 - k*u*v = 1 for integer k given by (u(k,n), v(k,n)) = (S(n,k), S(n-1,k)) because of the Cassini-Simson identity: S(n,x)^2 - S(n+1,x)*S(n-1, x) = 1, after use of the S-recurrence. Note that S(-n, x) = -S(-n-2, x), n >= 1, and the periodicity of some S(n, k) sequences.

Hence another way to obtain the row polynomials would be to take powers of the matrix [x, -1; 1,0]: S(n, x) = (([x, -1; 1, 0])^n)[1,1], n >= 0.

See also a Feb 01 2016 comment on A115139 for a well-known S(n, x) sum formula.

Then we have with the present T triangle

A039834(n) = -i^(n+1)*T(n-1, k) where i is the imaginary unit and n >= 0.

A051286(n) = Sum_{i=0..n} T(n,i)^2 (see the Philippe Deléham, Nov 21 2005 formula),

A181545(n) = Sum_{i=0..n+1} abs(T(n,i)^3),

A181546(n) = Sum_{i=0..n+1} T(n,i)^4,

A181547(n) = Sum_{i=0..n+1} abs(T(n,i)^5).

S(n, 0) = A056594(n), and for k = 1..10 the sequences S(n-1, k) with offset n = 0 are A128834, A001477, A001906, A001353, A004254, A001109, A004187, A001090, A018913, A004189.

(End)

For more on the Diophantine equation presented by Kul, see the Ismail paper. - Tom Copeland, Jan 31 2016

The o.g.f. for the Legendre polynomials L(n,x) is 1 / sqrt(1- 2x*z + z^2), and squaring it gives the o.g.f. of U(n,x), A053117, so Sum_{k=0..n} L(k,x/2) L(n-k,x/2) = S(n,x). This gives S(n,x) = L(n/2,x/2)^2 + 2*Sum_{k=0..n/2-1} L(k,x/2) L(n-k,x/2) for n even and S(n,x) = 2*Sum_{k=0..(n-1)/2} L(k,x/2) L(n-k,x/2) for odd n. For a connection to elliptic curves and modular forms, see A053117. For the normalized Legendre polynomials, see A100258. For other properties and relations to other polynomials, see Allouche et al. - Tom Copeland, Feb 04 2016

LG(x,h1,h2) = -log(1 - h1*x + h2*x^2) = Sum_{n>0} F(n,-h1,h2,0,..,0) x^n/n is a log series generator of the bivariate row polynomials of A127672 with A127672(0,0) = 0 and where F(n,b1,b2,..,bn) are the Faber polynomials of A263916. Exp(LG(x,h1,h2)) = 1 / (1 - h1*x + h2*x^2 ) is the o.g.f. of the bivariate row polynomials of this entry. - Tom Copeland, Feb 15 2016 (Instances of the bivariate o.g.f. for this entry are on pp. 5 and 18 of Sunada. - Tom Copeland, Jan 18 2021)

For distinct odd primes p and q the Legendre symbol can be written as Legendre(q,p) = Product_{k=1..P} S(q-1, 2*cos(2*Pi*k/p)), with P = (p-1)/2. See the Lemmermeyer reference, eq. (8.1) on p. 236. Using the zeros of S(q-1, x) (see above) one has S(q-1, x) = Product_{l=1..Q} (x^2 - (2*cos(Pi*l/q))^2), with Q = (q-1)/2. Thus S(q-1, 2*cos(2*Pi*k/p)) = ((-4)^Q)*Product_{l=1..Q} (sin^2(2*Pi*k/p) - sin^2(Pi*l/q)) = ((-4)^Q)*Product_{m=1..Q} (sin^2(2*Pi*k/p) - sin^2(2*Pi*m/q)). For the proof of the last equality see a W. Lang comment on the triangle A057059 for n = Q and an obvious function f. This leads to Eisenstein's proof of the quadratic reciprocity law Legendre(q,p) = ((-1)^(P*Q)) * Legendre(p,q), See the Lemmermeyer reference, pp. 236-237. - Wolfdieter Lang, Aug 28 2016

For connections to generalized Fibonacci polynomials, compare their generating function on p. 5 of the Amdeberhan et al. link with the o.g.f. given above for the bivariate row polynomials of this entry. - Tom Copeland, Jan 08 2017

The formula for Ramanujan's tau function (see A000594) for prime powers is tau(p^k) = p^(11*k/2)*S(k, p^(-11/2)*tau(p)) for k >= 1, and p = A000040(n), n >= 1. See the Hardy reference, p. 164, eqs. (10.3.4) and (10.3.6) rewritten in terms of S. - Wolfdieter Lang, Jan 27 2017

From Wolfdieter Lang, May 08 2017: (Start)

The number of zeros Z(n) of the S(n, x) polynomials in the open interval (-1,+1) is 2*b(n) for even n >= 0 and 1 + 2*b(n) for odd n >= 1, where b(n) = floor(n/2) - floor((n+1)/3). This b(n) is the number of integers k in the interval (n+1)/3 < k <= floor(n/2). See a comment on the zeros of S(n, x) above, and b(n) = A008615(n-2), n >= 0. The numbers Z(n) have been proposed (with a conjecture related to A008611) by Michel Lagneau, as the number of zeros of Fibonacci polynomials on the imaginary axis (-I,+I), with I=sqrt(-1). They are Z(n) = A008611(n-1), n >= 0, with A008611(-1) = 0. Also Z(n) = A194960(n-4), n >= 0. Proof using the A008611 version. A194960 follows from this.

In general the number of zeros Z(a;n) of S(n, x) for n >= 0 in the open interval (-a,+a) for a from the interval (0,2) (x >= 2 never has zeros, and a=0 is trivial: Z(0;n) = 0) is with b(a;n) = floor(n//2) - floor((n+1)*arccos(a/2)/Pi), as above Z(a;n) = 2*b(a;n) for even n >= 0 and 1 + 2*b(a;n) for odd n >= 1. For the closed interval [-a,+a] Z(0;n) = 1 and for a from (0,1) one uses for Z(a;n) the values b(a;n) = floor(n/2) - ceiling((n+1)*arccos(a/2)/Pi) + 1. (End)

The Riordan row polynomials S(n, x) (Chebyshev S) belong to the Boas-Buck class (see a comment and references in A046521), hence they satisfy the Boas-Buck identity: (E_x - n*1)*S(n, x) = (E_x + 1)*Sum_{p=0..n-1} (1 - (-1)^p)*(-1)^((p+1)/2)*S(n-1-p, x), for n >= 0, where E_x = x*d/dx (Euler operator). For the triangle T(n, k) this entails a recurrence for the sequence of column k, given in the formula section. - Wolfdieter Lang, Aug 11 2017

The e.g.f. E(x,t) := Sum_{n>=0} (t^n/n!)*S(n,x) for the row polynomials is obtained via inverse Laplace transformation from the above given o.g.f. as E(x,t) = ((1/xm)*exp(t/xm) - (1/xp)*exp(t/xp) )/(xp - xm) with xp = (x + sqrt(x^2-4))/2 and xm = (x - sqrt(x^2-4))/2. - Wolfdieter Lang, Nov 08 2017

From Wolfdieter Lang, Apr 12 2018: (Start)

Factorization of row polynomials S(n, x), for n >= 1, in terms of C polynomials (not Chebyshev C) with coefficients given in A187360. This is obtained from the factorization into Psi polynomials (see the Jul 12 2011 comment above) but written in terms of minimal polynomials of 2*cos(2*Pi/n) with coefficients in A232624:

S(2*k, x) = Product_{2 <= d | (2*k+1)} C(d, x)*(-1)^deg(d)*C(d, -x), with deg(d) = A055034(d) the degree of C(d, x).

S(2*k+1, x) = Product_{2 <= d | 2*(k+1)} C(d, x) * Product_{3 <= 2*d + 1 | (k+1)} (-1)^(deg(2*d+1))*C(2*d+1, -x).

Note that (-1)^(deg(2*d+1))*C(2*d+1, -x)*C(2*d+1, x) pairs always appear.

The number of C factors of S(2*k, x), for k >= 0, is 2*(tau(2*k+1) - 1) = 2*(A099774(k+1) - 1) = 2*A095374(k), and for S(2*k+1, x), for k >= 0, it is tau(2*(k+1)) + tau_{odd}(k+1) - 2 = A302707(k), with tau(2*k+1) = A099774(k+1), tau(n) = A000005 and tau(2*(k+1)) = A099777(k+1).

For the reverse problem, the factorization of C polynomials into S polynomials, see A255237. (End)

The S polynomials with general initial conditions S(a,b;n,x) = x*S(a,b;n-1,x) - S(a,b;n-2,x), for n >= 1, with S(a,b;-1,x) = a and S(a,b;0,x) = b are S(a,b;n,x) = b*S(n, x) - a*S(n-1, x), for n >= -1. Recall that S(-2, x) = -1 and S(-1, x) = 0. The o.g.f. is G(a,b;z,x) = (b - a*z)/(1 - x*z + z^2). - Wolfdieter Lang, Oct 18 2019

Also the convolution triangle of A101455. - Peter Luschny, Oct 06 2022

From Wolfdieter Lang, Apr 26 2023: (Start)

Multi-section of S-polynomials: S(m*n+k, x) = S(m+k, x)*S(n-1, R(m, x)) - S(k, x)*S(n-2, R(m, x)), with R(n, x) = S(n, x) - S(n-2, x) (see A127672), S(-2, x) = -1, and S(-1, x) = 0, for n >= 0, m >= 1, and k = 0, 1, ..., m-1.

O.g.f. of {S(m*n+k, y)}_{n>=0}: G(m,k,y,x) = (S(k, y) - (S(k, y)*R(m, y) - S(m+k, y))*x)/(1 - R(m,y)*x + x^2).

See eqs. (40) and (49), with r = x or y and s =-1, of the G. Detlefs and W. Lang link at A034807. (End)

S(n, x) for complex n and complex x: S(n, x) = ((-i/2)/sqrt(1 - (x/2)^2))*(q(x/2)*exp(+n*log(q(x/2))) - (1/q(x/2))*exp(-n*log(q(x/2)))), with q(x) = x + sqrt(1 - x^2)*i. Here log(z) = |z| + Arg(z)*i, with Arg(z) from [-Pi,+Pi) (principal branch). This satisfies the recurrence relation for S because it is derived from the Binet - de Moivre formula for S. Examples: S(n/m, 0) = cos((n/m)*Pi/4), for n >= 0 and m >= 1. S(n*i, 0) = (1/2)*(1 + exp(n*Pi))*exp(-(n/2)*Pi), for n >= 0. S(1+i, 2+i) = 0.6397424847... + 1.0355669490...*i. Thanks to Roberto Alfano for asking a question leading to this formula. - Wolfdieter Lang, Jun 05 2023

Lim_{n->oo} S(n, x)/S(n-1, x) = r(x) = (x - sqrt(x^2 -4))/2, for |x| >= 2. For x = +-2, this limit is +-1. - Wolfdieter Lang, Nov 15 2023

Sperner's theorem says that this is the maximal number of subsets of an n-set such that no one contains another.

When computed from index -1, [seq(binomial(n,floor(n/2)), n = -1..30)]; -> [1,1,1,2,3,6,10,20,35,70,126,...] and convolved with aerated Catalan numbers [seq(((n+1) mod 2)*binomial(n,n/2)/((n/2)+1), n = 0..30)]; -> [1,0,1,0,2,0,5,0,14,0,42,0,132,0,...] shifts left by one: [1,1,2,3,6,10,20,35,70,126,252,...] and if again convolved with aerated Catalan numbers, gives A037952 apart from the initial term. - Antti Karttunen, Jun 05 2001 [This is correct because the g.f.'s satisfy (1+x*g001405(x))*g126120(x) = g001405(x) and g001405(x)*g126120(x) = g037952(x)/x. - R. J. Mathar, Sep 23 2021]

Number of ordered trees with n+1 edges, having nonroot nodes of outdegree 0 or 2. - Emeric Deutsch, Aug 02 2002

Gives for n >= 1 the maximum absolute column sum norm of the inverse of the Vandermonde matrix (a_ij) i=0..n-1, j=0..n-1 with a_00=1 and a_ij=i^j for (i,j) != (0,0). - Torsten Muetze, Feb 06 2004

Image of Catalan numbers A000108 under the Riordan array (1/(1-2x),-x/(1-2x)) or A065109. - Paul Barry, Jan 27 2005

Number of left factors of Dyck paths, consisting of n steps. Example: a(4)=6 because we have UDUD, UDUU, UUDD, UUDU, UUUD and UUUU, where U=(1,1) and D=(1,-1). - Emeric Deutsch, Apr 23 2005

Number of dispersed Dyck paths of length n; they are defined as concatenations of Dyck paths and (1,0)-steps on the x-axis; equivalently, Motzkin paths with no (1,0)-steps at positive height. Example: a(4)=6 because we have HHHH, HHUD, HUDH, UDHH, UDUD, and UUDD, where U=(1,1), H=(1,0), and D=(1,-1). - Emeric Deutsch, Jun 04 2011

a(n) is odd iff n=2^k-1. - Jon Perry, May 05 2005

An inverse Chebyshev transform of binomial(1,n)=(1,1,0,0,0,...) where g(x)->(1/sqrt(1-4*x^2))*g(x*c(x^2)), with c(x) the g.f. of A000108. - Paul Barry, May 13 2005

In a random walk on the number line, starting at 0 and with 0 absorbing after the first step, number of ways of ending up at a positive integer after n steps. - Joshua Zucker, Jul 31 2005

Maximum number of sums of the form Sum_{i=1..n} e(i)*a(i) that are congruent to 0 mod q, where e_i=0 or 1 and gcd(a_i,q)=1, provided that q > ceiling(n/2). - Ralf Stephan, Apr 27 2003

Also the number of standard tableaux of height <= 2. - Mike Zabrocki, Mar 24 2007

Hankel transform of this sequence forms A000012 = [1,1,1,1,1,1,1,...]. - Philippe Deléham, Oct 24 2007

A001263 * [1, -2, 3, -4, 5, ...] = [1, -1, -2, 3, 6, -10, -20, 35, 70, -126, ...]. - Gary W. Adamson, Jan 02 2008

Equals right border of triangle A153585. - Gary W. Adamson, Dec 28 2008

Second binomial transform of A168491. - Philippe Deléham, Nov 27 2009

a(n) is also the number of distinct strings of length n, each of which is a prefix of a string of balanced parentheses; see example. - Lee A. Newberg, Apr 26 2010

Number of symmetric balanced strings of n pairs of parentheses; see example. - Joerg Arndt, Jul 25 2011

a(n) is the number of permutation patterns modulo 2. - Olivier Gérard, Feb 25 2011

For n >= 2, a(n-1) is the number of incongruent two-color bracelets of 2*n-1 beads, n of which are black (A007123), having a diameter of symmetry. - Vladimir Shevelev, May 03 2011

The number of permutations of n elements where p(k-2) < p(k) for all k. - Joerg Arndt, Jul 23 2011

Also size of the equivalence class of S_{n+1} containing the identity permutation under transformations of positionally adjacent elements of the form abc <--> cba where a < b < c, cf. A210668. - Tom Roby, May 15 2012

a(n) is the number of symmetric Dyck paths of length 2n. - Matt Watson, Sep 26 2012

a(n) is divisible by A000108(floor(n/2)) = abs(A129996(n-2)). - Paul Curtz, Oct 23 2012

a(n) is the number of permutations of length n avoiding both 213 and 231 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

Number of symmetric standard Young tableaux of shape (n,n). - Ran Pan, Apr 10 2015

From Luciano Ancora, May 09 2015: (Start)

Also "stepped path" in the array formed by partial sums of the all 1's sequence (or a Pascal's triangle displayed as a square). Example:

[1], [1], 1, 1, 1, 1, 1, ... A000012

1, [2], [3], 4, 5, 6, 7, ...

1, 3, [6], [10], 15, 21, 28, ...

1, 4, 10, [20], [35], 56, 84, ...

1, 5, 15, 35, [70], [126], 210, ...

Sequences in second formula are the mixed diagonals shown in this array. (End)

a(n) = A265848(n,n). - Reinhard Zumkeller, Dec 24 2015

The constant Sum_{n >= 0} a(n)/n! is 1 + A130820. - Peter Bala, Jul 02 2016

Number of meanders (walks starting at the origin and ending at any altitude >= 0 that may touch but never go below the x-axis) with n steps from {-1,1}. - David Nguyen, Dec 20 2016

a(n) is also the number of paths of n steps (either up or down by 1) that end at the maximal value achieved along the path. - Winston Luo, Jun 01 2017

Number of binary n-tuples such that the number of 1's in the even positions is the same as the number of 1's in the odd positions. - Juan A. Olmos, Dec 21 2017

Equivalently, a(n) is the number of subsets of {1,...,n} containing as many even numbers as odd numbers. - Gus Wiseman, Mar 17 2018

a(n) is the number of Dyck paths with semilength = n+1, returns to the x-axis = floor((n+3)/2) and up movements in odd positions = floor((n+3)/2). Example: a(4)=6, U=up movement in odd position, u=up movement in even position, d=down movement, -=return to x-axis: Uududd-Ud-Ud-, Ud-Uudd-Uudd-, Uudd-Uudd-Ud-, Ud-Ud-Uududd-, Uudd-Ud-Uudd-, Ud-Uududd-Ud-. - Roger Ford, Dec 29 2017

Let C_n(R, H) denote the transition matrix from the ribbon basis to the homogeneous basis of the graded component of the algebra of noncommutative symmetric functions of order n. Letting I(2^(n-1)) denote the identity matrix of order 2^(n-1), it has been conjectured that the dimension of the kernel of C_n(R, H) - I(2^(n-1)) is always equal to a(n-1). - John M. Campbell, Mar 30 2018

The number of U-equivalence classes of Łukasiewicz paths. Łukasiewicz paths are U-equivalent iff the positions of pattern U are identical in these paths. - Sergey Kirgizov, Apr 08 2018

All binary self-dual codes of length 2n, for n > 0, must contain at least a(n) codewords of weight n. More to the point, there will always be at least one, perhaps unique, binary self-dual code of length 2n that will contain exactly a(n) codewords that have a hamming weight equal to half the length of the code (n). This code can be constructed by direct summing the unique binary self-dual code of length 2 (up to permutation equivalence) to itself n times. A permutation equivalent code can be constructed by augmenting two identity matrices of length n together. - Nathan J. Russell, Nov 25 2018

Closed under addition. - Torlach Rush, Apr 18 2019

The sequence starting (1, 2, 3, 6, ...) is the invert transform of A097331: (1, 1, 0, 1, 0, 2, 0, 5, 0, 14, 0, 42, ...). - Gary W. Adamson, Feb 22 2020

From Gary W. Adamson, Feb 24 2020: (Start)

The sequence is the culminating limit of an infinite set of sequences with convergents of 2*cos(Pi/N), N = (3, 5, 7, 9, ...).

The first few such sequences are:

N = 3: (1, 1, 1, 1, 1, 1, 1, 1, ...)

N = 5: (1, 1, 2, 3, 5, 8, 13, 21, ...) = A000045

N = 7: (1, 1, 2, 3, 6, 10, 19, 33, ...) = A028495, a(n)/a(n-1) tends to 1.801937...

N = 9 (1, 1, 2, 3, 6, 10, 20, 35, ...) = A061551, a(n)/a(n_1) tends to 1.879385...

...

In the limit one gets the current sequence with ratio 2. (End)

a(n) is also the number of monotone lattice paths from (0,0) to (floor(n/2),ceiling(n/2)). These are the number of Grand Dyck paths when n is even. - Nachum Dershowitz, Aug 12 2020

The maximum number of preimages that a permutation of length n+1 can have under the consecutive-132-avoiding stack-sorting map. - Colin Defant, Aug 28 2020

Counts faro permutations of length n. Faro permutations are permutations avoiding the three consecutive patterns 231, 321 and 312. They are obtained by a perfect faro shuffle of two nondecreasing words of lengths differing by at most one. - Sergey Kirgizov, Jan 12 2021

Per "Sperner's Theorem", the largest possible familes of finite sets none of which contain any other sets in the family. - Renzo Benedetti, May 26 2021

a(n-1) are the incomplete, primitive Dyck paths of n steps without a first return: paths of U and D steps starting at the origin, never touching the horizontal axis later on, and ending above the horizontal axis. n=1: {U}, n=2: {UU}, n=3: {UUU, UUD}, n=4: {UUUU, UUUD, UUDU}, n=5: {UUUUU, UUUUD, UUUDD, UUDUU, UUUDU, UUDUD}. For comparison: A037952 counts incomplete Dyck paths with n steps with any number of intermediate returns to the horizontal axis, ending above the horizontal axis. - R. J. Mathar, Sep 24 2021

a(n) is the number of noncrossing partitions of [n] whose nontrivial blocks are of type {a,b}, with a <= n/2, b > n/2. - Francesca Aicardi, May 29 2022

Maximal coefficient of (1+x)^n. - Vaclav Kotesovec, Dec 30 2022

Sums of lower-left-to-upper-right diagonals of the Catalan Triangle A001263. - Howard A. Landman, Sep 16 2024

T(n,k) is the number of lattice paths from (0,0) to (n,n) with steps E = (1,0) and N = (0,1) which touch but do not cross the line x - y = k and only situated above this line; example: T(3,2) = 5 because we have EENNNE, EENNEN, EENENN, ENEENN, NEEENN. - Philippe Deléham, May 23 2005

The matrix inverse of this triangle is the triangular matrix T(n,k) = (-1)^(n+k)* A085478(n,k). - Philippe Deléham, May 26 2005

Essentially the same as A050155 except with a leading diagonal A000108 (Catalan numbers) 1, 1, 2, 5, 14, 42, 132, 429, .... - Philippe Deléham, May 31 2005

Number of Grand Dyck paths of semilength n and having k downward returns to the x-axis. (A Grand Dyck path of semilength n is a path in the half-plane x>=0, starting at (0,0), ending at (2n,0) and consisting of steps u=(1,1) and d=(1,-1)). Example: T(3,2)=5 because we have u(d)uud(d),uud(d)u(d),u(d)u(d)du,u(d)duu(d) and duu(d)u(d) (the downward returns to the x-axis are shown between parentheses). - Emeric Deutsch, May 06 2006

Riordan array (c(x),x*c(x)^2) where c(x) is the g.f. of A000108; inverse array is (1/(1+x),x/(1+x)^2). - Philippe Deléham, Feb 12 2007

The triangle may also be generated from M^n*[1,0,0,0,0,0,0,0,...], where M is the infinite tridiagonal matrix with all 1's in the super and subdiagonals and [1,2,2,2,2,2,2,...] in the main diagonal. - Philippe Deléham, Feb 26 2007

Inverse binomial matrix applied to A124733. Binomial matrix applied to A089942. - Philippe Deléham, Feb 26 2007

Number of standard tableaux of shape (n+k,n-k). - Philippe Deléham, Mar 22 2007

From Philippe Deléham, Mar 30 2007: (Start)

This triangle belongs to the family of triangles defined by: T(0,0)=1, T(n,k)=0 if k<0 or if k>n, T(n,0)=x*T(n-1,0)+T(n-1,1), T(n,k)=T(n-1,k-1)+y*T(n-1,k)+T(n-1,k+1) for k>=1. Other triangles arise by choosing different values for (x,y):

(0,0) -> A053121; (0,1) -> A089942; (0,2) -> A126093; (0,3) -> A126970

(1,0) -> A061554; (1,1) -> A064189; (1,2) -> A039599; (1,3) -> A110877;

(1,4) -> A124576; (2,0) -> A126075; (2,1) -> A038622; (2,2) -> A039598;

(2,3) -> A124733; (2,4) -> A124575; (3,0) -> A126953; (3,1) -> A126954;

(3,2) -> A111418; (3,3) -> A091965; (3,4) -> A124574; (4,3) -> A126791;

(4,4) -> A052179; (4,5) -> A126331; (5,5) -> A125906. (End)

The table U(n,k) = Sum_{j=0..n} T(n,j)*k^j is given in A098474. - Philippe Deléham, Mar 29 2007

Sequence read mod 2 gives A127872. - Philippe Deléham, Apr 12 2007

Number of 2n step walks from (0,0) to (2n,2k) and consisting of step u=(1,1) and d=(1,-1) and the path stays in the nonnegative quadrant. Example: T(3,0)=5 because we have uuuddd, uududd, ududud, uduudd, uuddud; T(3,1)=9 because we have uuuudd, uuuddu, uuudud, ududuu, uuduud, uduudu, uudduu, uduuud, uududu; T(3,2)=5 because we have uuuuud, uuuudu, uuuduu, uuduuu, uduuuu; T(3,3)=1 because we have uuuuuu. - Philippe Deléham, Apr 16 2007, Apr 17 2007, Apr 18 2007

Triangular matrix, read by rows, equal to the matrix inverse of triangle A129818. - Philippe Deléham, Jun 19 2007

Let Sum_{n>=0} a(n)*x^n = (1+x)/(1-mx+x^2) = o.g.f. of A_m, then Sum_{k=0..n} T(n,k)*a(k) = (m+2)^n. Related expansions of A_m are: A099493, A033999, A057078, A057077, A057079, A005408, A002878, A001834, A030221, A002315, A033890, A057080, A057081, A054320, A097783, A077416, A126866, A028230, A161591, for m=-3,-2,-1,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15, respectively. - Philippe Deléham, Nov 16 2009

The Kn11, Kn12, Fi1 and Fi2 triangle sums link the triangle given above with three sequences; see the crossrefs. For the definitions of these triangle sums, see A180662. - Johannes W. Meijer, Apr 20 2011

4^n = (n-th row terms) dot (first n+1 odd integer terms). Example: 4^4 = 256 = (14, 28, 20, 7, 1) dot (1, 3, 5, 7, 9) = (14 + 84 + 100 + 49 + 9) = 256. - Gary W. Adamson, Jun 13 2011

The linear system of n equations with coefficients defined by the first n rows solve for diagonal lengths of regular polygons with N= 2n+1 edges; the constants c^0, c^1, c^2, ... are on the right hand side, where c = 2 + 2*cos(2*Pi/N). Example: take the first 4 rows relating to the 9-gon (nonagon), N = 2*4 + 1; with c = 2 + 2*cos(2*Pi/9) = 3.5320888.... The equations are (1,0,0,0) = 1; (1,1,0,0) = c; (2,3,1,0) = c^2; (5,9,5,1) = c^3. The solutions are 1, 2.53208..., 2.87938..., and 1.87938...; the four distinct diagonal lengths of the 9-gon (nonagon) with edge = 1. (Cf. comment in A089942 which uses the analogous operations but with c = 1 + 2*cos(2*Pi/9).) - Gary W. Adamson, Sep 21 2011

Also called the Lobb numbers, after Andrew Lobb, are a natural generalization of the Catalan numbers, given by L(m,n)=(2m+1)*Binomial(2n,m+n)/(m+n+1), where n >= m >= 0. For m=0, we get the n-th Catalan number. See added reference. - Jayanta Basu, Apr 30 2013

From Wolfdieter Lang, Sep 20 2013: (Start)

T(n, k) = A053121(2*n, 2*k). T(n, k) appears in the formula for the (2*n)-th power of the algebraic number rho(N):= 2*cos(Pi/N) = R(N, 2) in terms of the odd-indexed diagonal/side length ratios R(N, 2*k+1) = S(2*k, rho(N)) in the regular N-gon inscribed in the unit circle (length unit 1). S(n, x) are Chebyshev's S polynomials (see A049310):

rho(N)^(2*n) = Sum_{k=0..n} T(n, k)*R(N, 2*k+1), n >= 0, identical in N > = 1. For a proof see the Sep 21 2013 comment under A053121. Note that this is the unreduced version if R(N, j) with j > delta(N), the degree of the algebraic number rho(N) (see A055034), appears.

For the odd powers of rho(n) see A039598. (End)

Unsigned coefficients of polynomial numerators of Eqn. 2.1 of the Chakravarty and Kodama paper, defining the polynomials of A067311. - Tom Copeland, May 26 2016

The triangle is the Riordan square of the Catalan numbers in the sense of A321620. - Peter Luschny, Feb 14 2023

The entries in this triangle (in its many forms) are often called ballot numbers.

T(n,k) = number of standard tableaux of shape (n,k) (n > 0, 0 <= k <= n). Example: T(3,1) = 3 because we have 134/2, 124/3 and 123/4. - Emeric Deutsch, May 18 2004

T(n,k) is the number of full binary trees with n+1 internal nodes and jump-length k. In the preorder traversal of a full binary tree, any transition from a node at a deeper level to a node on a strictly higher level is called a jump; the positive difference of the levels is called the jump distance; the sum of the jump distances in a given ordered tree is called the jump-length. - Emeric Deutsch, Jan 18 2007

The k-th diagonal from the right (k=1, 2, ...) gives the sequence obtained by asking in how many ways can we toss a fair coin until we first get k more heads than tails. The k-th diagonal has formula k(2m+k-1)!/(m!(m+k)!) and g.f. (C(x))^k where C(x) is the generating function for the Catalan numbers, (1-sqrt(1-4*x))/(2*x). - Anthony C Robin, Jul 12 2007

T(n,k) is also the number of order-decreasing and order-preserving full transformations (of an n-element chain) of waist k (waist (alpha) = max(Im(alpha))). - Abdullahi Umar, Aug 25 2008

Formatted as an upper right triangle (see tables) a(c,r) is the number of different triangulated planar polygons with c+2 vertices, with triangle degree c-r+1 for the same vertex X (c=column number, r=row number, with c >= r >= 1). - Patrick Labarque, Jul 28 2010

The triangle sums, see A180662 for their definitions, link Catalan's triangle, its mirror is A033184, with several sequences, see crossrefs. - Johannes W. Meijer, Sep 22 2010

The m-th row of Catalan's triangle consists of the unique nonnegative differences of the form binomial(m+k,m)-binomial(m+k,m+1) with 0 <= k <= m (See Links). - R. J. Cano, Jul 22 2014

T(n,k) is also the number of nondecreasing parking functions of length n+1 whose maximum element is k+1. For example T(3,2) = 5 because we have (1,1,1,3), (1,1,2,3), (1,2,2,3), (1,1,3,3), (1,2,3,3). - Ran Pan, Nov 16 2015

T(n,k) is the number of Dyck paths from (0,0) to (n+2,n+2) which start with n-k+2 east steps and touch the diagonal y=x only on the last north step. - Felipe Rueda, Sep 18 2019

T(n-1,k) for k < n is number of well-formed strings of n parenthesis pairs with prefix of exactly n-k opening parenthesis; T(n,n) = T(n,n-1). - Hermann Stamm-Wilbrandt, May 02 2021

T(n,k) is the number of leaves at level k+1 in all ordered trees with n+1 edges. - Emeric Deutsch, Jan 15 2005

Riordan array ((1-2x-sqrt(1-4x))/(2x^2),(1-2x-sqrt(1-4x))/(2x)). Inverse array is A053122. - Paul Barry, Mar 17 2005

T(n,k) is the number of walks of n steps, each in direction N, S, W, or E, starting at the origin, remaining in the upper half-plane and ending at height k (see the R. K. Guy reference, p. 5). Example: T(3,2)=6 because we have ENN, WNN, NEN, NWN, NNE and NNW. - Emeric Deutsch, Apr 15 2005

Triangle T(n,k), 0<=k<=n, read by rows given by T(0,0)=1, T(n,k)=0 if k<0 or if k>n, T(n,0) = 2*T(n-1,0) + T(n-1,1), T(n,k) = T(n-1,k-1) + 2*T(n-1,k) + T(n-1,k+1) for k>=1. - Philippe Deléham, Mar 30 2007

Number of (2n+1)-step walks from (0,0) to (2n+1,2k+1) and consisting of steps u=(1,1) and d=(1,-1) in which the path stays in the nonnegative quadrant. Examples: T(2,0)=5 because we have uuudd, uudud, uuddu, uduud, ududu; T(2,1)=4 because we have uuuud, uuudu, uuduu, uduuu; T(2,2)=1 because we have uuuuu. - Philippe Deléham, Apr 16 2007, Apr 18 2007

Triangle read by rows: T(n,k)=number of lattice paths from (0,0) to (n,k) that do not go below the line y=0 and consist of steps U=(1,1), D=(1,-1) and two types of steps H=(1,0); example: T(3,1)=14 because we have UDU, UUD, 4 HHU paths, 4 HUH paths and 4 UHH paths. - Philippe Deléham, Sep 25 2007

This triangle belongs to the family of triangles defined by T(0,0)=1, T(n,k)=0 if k<0 or if k>n, T(n,0) = x*T(n-1,0) + T(n-1,1), T(n,k) = T(n-1,k-1) + y*T(n-1,k) + T(n-1,k+1) for k>=1. Other triangles arise by choosing different values for (x,y): (0,0) -> A053121; (0,1) -> A089942; (0,2) -> A126093; (0,3) -> A126970; (1,0) -> A061554; (1,1) -> A064189; (1,2) -> A039599; (1,3) -> A110877; (1,4) -> A124576; (2,0) -> A126075; (2,1) -> A038622; (2,2) -> A039598; (2,3) -> A124733; (2,4) -> A124575; (3,0) -> A126953; (3,1) -> A126954; (3,2) -> A111418; (3,3) -> A091965; (3,4) -> A124574; (4,3) -> A126791; (4,4) -> A052179; (4,5) -> A126331; (5,5) -> A125906. - Philippe Deléham, Sep 25 2007

With offset [1,1] this is the (ordinary) convolution triangle a(n,m) with o.g.f. of column m given by (c(x)-1)^m, where c(x) is the o.g.f. for Catalan numbers A000108. See the Riordan comment by Paul Barry.

T(n, k) is also the number of order-preserving full transformations (of an n-chain) with exactly k fixed points. - Abdullahi Umar, Oct 02 2008

T(n,k)/2^(2n+1) = coefficients of the maximally flat lowpass digital differentiator of the order N=2n+3. - Pavel Holoborodko (pavel(AT)holoborodko.com), Dec 19 2008

The signed triangle S(n,k) := (-1)^(n-k)*T(n,k) provides the transformation matrix between f(n,l) := L(2*l)*5^n*F(2*l)^(2*n+1) (F=Fibonacci numbers A000045, L=Lucas numbers A000032) and F(4*l*(k+1)), k = 0, ..., n, for each l>=0: f(n,l) = Sum_{k=0..n} S(n,k)*F(4*l*(k+1)), n>=0, l>=0. Proof: the o.g.f. of the l.h.s., G(l;x) := Sum_{n>=0} f(n,l)*x^n = F(4*l)/(1 - 5*F(2*l)^2*x) is shown to match the o.g.f. of the r.h.s.: after an interchange of the n- and k-summation, the Riordan property of S = (C(x)/x,C(x)) (compare with the above comments by Paul Barry), with C(x) := 1 - c(-x), with the o.g.f. c(x) of A000108 (Catalan numbers), is used, to obtain, after an index shift, first Sum_{k>=0} F(4*l*(k))*GS(k;x), with the o.g.f of column k of triangle S which is GS(k;x) := Sum_{n>=k} S(n,k)*x^n = C(x)^(k+1)/x. The result is GF(l;C(x))/x with the o.g.f. GF(l,x) := Sum_{k>=0} F(4*l*k)*x^k = x*F(4*l)/(1-L(4*l)*x+x^2) (see a comment on A049670, and A028412). If one uses then the identity L(4*n) - 5*F(2*n)^2 = 2 (in Koshy's book [reference under A065563] this is No. 15, p. 88, attributed to Lucas, 1876), the proof that one recovers the o.g.f. of the l.h.s. from above boils down to a trivial identity on the Catalan o.g.f., namely 1/c^2(-x) = 1 + 2*x - (x*c(-x))^2. - Wolfdieter Lang, Aug 27 2012

O.g.f. for row polynomials R(x) := Sum_{k=0..n} a(n,k)*x^k:

((1+x) - C(z))/(x - (1+x)^2*z) with C the o.g.f. of A000108 (Catalan numbers). From Riordan ((C(x)-1)/x,C(x)-1), compare with a Paul Barry comment above. This coincides with the o.g.f. given by Emeric Deutsch in the formula section. - Wolfdieter Lang, Nov 13 2012

The A-sequence for this Riordan triangle is [1,2,1] and the Z-sequence is [2,1]. See a W. Lang link under A006232 with details and references. - Wolfdieter Lang, Nov 13 2012

From Wolfdieter Lang, Sep 20 2013: (Start)

T(n, k) = A053121(2*n+1, 2*k+1). T(n, k) appears in the formula for the (2*n+1)-th power of the algebraic number rho(N) := 2*cos(Pi/N) = R(N, 2) in terms of the even-indexed diagonal/side length ratios R(N, 2*(k+1)) = S(2*k+1, rho(N)) in the regular N-gon inscribed in the unit circle (length unit 1). S(n, x) are Chebyshev's S polynomials (see A049310): rho(N)^(2*n+1) = Sum_{k=0..n} T(n, k)*R(N, 2*(k+1)), n >= 0, identical in N >= 1. For a proof see the Sep 21 2013 comment under A053121. Note that this is the unreduced version if R(N, j) with j > delta(N), the degree of the algebraic number rho(N) (see A055034), appears. For the even powers of rho(n) see A039599. (End)

The tridiagonal Toeplitz production matrix P in the Example section corresponds to the unsigned Cartan matrix for the simple Lie algebra A_n as n tends to infinity (cf. Damianou ref. in A053122). - Tom Copeland, Dec 11 2015 (revised Dec 28 2015)

T(n,k) is the number of pairs of non-intersecting walks of n steps, each in direction N or E, starting at the origin, and such that the end points of the two paths are separated by a horizontal distance of k. See Shapiro 1976. - Peter Bala, Apr 12 2017

Also the convolution triangle of the Catalan numbers A000108. - Peter Luschny, Oct 07 2022

Inverse binomial transform of A001006.

The Hankel transform of this sequence gives A000012 = [1,1,1,1,1,...].

Counts returning walks (excursions) of length n on a 1-d integer lattice with step set {+1,-1} which stay in the chamber x >= 0. - Andrew V. Sutherland, Feb 29 2008

Moment sequence of the trace of a random matrix in G=USp(2)=SU(2). If X=tr(A) is a random variable (A distributed according to the Haar measure on G) then a(n) = E[X^n]. - Andrew V. Sutherland, Feb 29 2008

Essentially the same as A097331. - R. J. Mathar, Jun 15 2008

Number of distinct proper binary trees with n nodes. - Chris R. Sims (chris.r.sims(AT)gmail.com), Jun 30 2010

-a(n-1), with a(-1):=0, n>=0, is the Z-sequence for the Riordan array A049310 (Chebyshev S). For the definition see that triangle. - Wolfdieter Lang, Nov 04 2011

See A180874 (also A238390 and A097610) and A263916 for relations to the general Bell A036040, cycle index A036039, and cumulant expansion polynomials A127671 through the Faber polynomials. - Tom Copeland, Jan 26 2016

A signed version is generated by evaluating polynomials in A126216 that are essentially the face polynomials of the associahedra. This entry's sequence is related to an inversion relation on p. 34 of Mizera, related to Feynman diagrams. - Tom Copeland, Dec 09 2019

Motzkin triangle read in reverse order.

T(n,k) = number of lattice paths from (0,0) to (n,k), staying weakly above the x-axis and consisting of steps U=(1,1), D=(1,-1) and H=(1,0). Example: T(3,1) = 5 because we have HHU, UDU, HUH, UHH and UUD. Columns 0,1,2 and 3 give A001006 (Motzkin numbers), A002026 (first differences of Motzkin numbers), A005322 and A005323, respectively. - Emeric Deutsch, Feb 29 2004

Riordan array ((1-x-sqrt(1-2x-3x^2))/(2x^2), (1-x-sqrt(1-2x-3x^2))/(2x)). Inverse is the array (1/(1+x+x^2), x/(1+x+x^2)) (A104562). - Paul Barry, Mar 15 2005

Inverse binomial matrix applied to A039598. - Philippe Deléham, Feb 28 2007

Triangle T(n,k), 0 <= k <= n, read by rows given by: T(0,0)=1, T(n,k)=0 if k < 0 or if k > n, T(n,0) = T(n-1,0) + T(n-1,1), T(n,k) = T(n-1,k-1) + T(n-1,k) + T(n-1,k+1) for k >= 1. - Philippe Deléham, Mar 27 2007

This triangle belongs to the family of triangles defined by: T(0,0)=1, T(n,k)=0 if k < 0 or if k > n, T(n,0) = x*T(n-1,0) + T(n-1,1), T(n,k) = T(n-1,k-1) + y*T(n-1,k) + T(n-1,k+1) for k >= 1. Other triangles arise from choosing different values for (x,y): (0,0) -> A053121; (0,1) -> A089942; (0,2) -> A126093; (0,3) -> A126970; (1,0)-> A061554; (1,1) -> A064189; (1,2) -> A039599; (1,3) -> A110877; (1,4) -> A124576; (2,0) -> A126075; (2,1) -> A038622; (2,2) -> A039598; (2,3) -> A124733; (2,4) -> A124575; (3,0) -> A126953; (3,1) -> A126954; (3,2) -> A111418; (3,3) -> A091965; (3,4) -> A124574; (4,3) -> A126791; (4,4) -> A052179; (4,5) -> A126331; (5,5) -> A125906. - Philippe Deléham, Sep 25 2007

Equals binomial transform of triangle A053121. - Gary W. Adamson, Oct 25 2008

Consider a semi-infinite chessboard with squares labeled (n,k), ranks or rows n >= 0, files or columns k >= 0; the number of king-paths of length n from (0,0) to (n,k), 0 <= k <= n, is T(n,k). The recurrence relation given above relates to the movements of the king. This is essentially the comment made by Harrie Grondijs for the Motzkin triangle A026300. - Johannes W. Meijer, Oct 10 2010

a(n) = 2^A048881 = 2^{maximal power of 2 dividing the n-th Catalan number (A000108)}. [Comment corrected by N. J. A. Sloane, Apr 30 2018]

Row sums of triangle A128937. - Philippe Deléham, May 02 2007

a(n) = sum of (n+1)-th row terms of triangle A167364. - Gary W. Adamson, Nov 01 2009

a(n), n >= 1: Numerators of Maclaurin series for 1 - ((sin x)/x)^2, A117972(n), n >= 2: Denominators of Maclaurin series for 1 - ((sin x)/x)^2, the correlation function in Montgomery's pair correlation conjecture. - Daniel Forgues, Oct 16 2011

For n > 0: a(n) = A007954(A007931(n)). - Reinhard Zumkeller, Oct 26 2012

a(n) = A261363(2*(n+1), n+1). - Reinhard Zumkeller, Aug 16 2015

From Gus Wiseman, Oct 30 2022: (Start)

Also the number of coarsenings of the (n+1)-th composition in standard order. The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions. See link for sequences related to standard compositions. For example, the a(10) = 4 coarsenings of (2,1,1) are: (2,1,1), (2,2), (3,1), (4).

Also the number of times n+1 appears in A357134. For example, 11 appears at positions 11, 20, 33, and 1024, so a(10) = 4.

(End)

Equivalently, a triangle read by rows, where the rows are obtained by sorting the elements of the rows of Pascal's triangle (A007318) into descending order. - Philippe Deléham, May 21 2005

Equivalently, as a triangle read by rows, this is T(n,k)=binomial(n,floor((n-k)/2)); column k then has e.g.f. Bessel_I(k,2x)+Bessel_I(k+1,2x). - Paul Barry, Feb 28 2006

Antidiagonal sums are A037952(n+1) = C(n+1,[n/2]). Matrix inverse is the row reversal of triangle A066170. Eigensequence is A125094(n) = Sum_{k=0..n-1} A125093(n-1,k)*A125094(k). - Paul D. Hanna, Nov 20 2006

Riordan array (1/(1-x-x^2*c(x^2)),x*c(x^2)); where c(x)=g.f.for Catalan numbers A000108. - Philippe Deléham, Mar 17 2007

Triangle T(n,k), 0<=k<=n, read by rows given by: T(0,0)=1, T(n,k)=0 if k<0 or if k>n, T(n,0)=T(n-1,0)+T(n-1,1), T(n,k)=T(n-1,k-1)+T(n-1,k+1) for k>=1. - Philippe Deléham, Mar 27 2007

This triangle belongs to the family of triangles defined by: T(0,0)=1, T(n,k)=0 if k<0 or if k>n, T(n,0)=x*T(n-1,0)+T(n-1,1), T(n,k)=T(n-1,k-1)+y*T(n-1,k)+T(n-1,k+1) for k>=1. Other triangles arise by choosing different values for (x,y): (0,0) -> A053121; (0,1) -> A089942; (0,2) -> A126093; (0,3) -> A126970; (1,0)-> A061554; (1,1) -> A064189; (1,2) -> A039599; (1,3) -> A110877; (1,4) -> A124576; (2,0) -> A126075; (2,1) -> A038622; (2,2) -> A039598; (2,3) -> A124733; (2,4) -> A124575; (3,0) -> A126953; (3,1) -> A126954; (3,2) -> A111418; (3,3) -> A091965; (3,4) -> A124574; (4,3) -> A126791; (4,4) -> A052179; (4,5) -> A126331; (5,5) -> A125906. - Philippe Deléham, Sep 25 2007

T(n,k) is the number of paths from (0,k) to some (n,m) which never dip below y=0, touch y=0 at least once and are made up only of the steps (1,1) and (1,-1). This can be proved using the recurrence supplied by Deléham. - Gerald McGarvey, Oct 15 2008

Triangle read by rows = partial sums of A053121 terms starting from the right. - Gary W. Adamson, Oct 24 2008

As a subset of the "family of triangles" (Deleham comment of Sep 25 2007), beginning with a signed variant of A061554, M = (-1,0) = (1; -1, 1; 2, -1, 1; -3, 3, -1, 1; ...) successive binomial transforms of M yield (0,1) - A089942; (1,2) - A039599; (2,3) - A124733; (3,4) - A124574; (4,5) - A126331; ... such that the binomial transform of the triangle generated from (n,n+1) = the triangle generated from (n+1,n+2). Similarly, another subset beginning with A053121 - (0,0), and taking successive binomial transforms yields (1,1) - A064189; (2,2) - A039598; (3,3) - A091965, ... By rows, the triangle generated from (n,n) can be obtained by taking pairwise sums from the (n-1,n) triangle starting from the right. For example, row 2 of (1,2) - A039599 = (2, 3, 1); and taking pairwise sums from the right we obtain (5, 4, 1) = row 2 of (2,2) - A039598. - Gary W. Adamson, Aug 04 2011

The triangle by rows (n) with alternating signs (+-+...) from the top as a set of simultaneous equations solves for diagonal lengths of odd N (N = 2n+1) regular polygons. The constants in each case are powers of c = 2*cos(2*Pi/N). By way of example, the first 3 rows relate to the heptagon and the simultaneous equations are (1,0,0) = 1; (-1,1,0) = c = 1.24697...; and (2,-1,1) = c^2. The answers are 1, 2.24697..., and 1.801...; the 3 distinct diagonal lengths of the heptagon with edge = 1. - Gary W. Adamson, Sep 07 2011

Or, a triangle related to A000984 (central binomial) and A000302 (powers of 4).

This is an example of a Riordan matrix. See the Shapiro et al. reference quoted under A053121 and Notes 1 and 2 of the Wolfdieter Lang reference, p. 306.

As a number triangle, this is the Riordan array (1/sqrt(1-4x),x/(1-4x)). - Paul Barry, May 30 2005

The A- and Z- sequences for this Riordan matrix are (see the Wolfdieter Lang link under A006232 for the D. G. Rogers, D. Merlini et al. and R. Sprugnoli references on Riordan A- and Z-sequences with a summary): A-sequence [1,4,0,0,0,...] and Z-sequence 4+2*A000108(n)*(-1)^(n+1)=[2, 2, -4, 10, -28, 84, -264, 858, -2860, 9724, -33592, 117572, -416024, 1485800, -5348880, 19389690, -70715340, 259289580, -955277400, 3534526380], n >= 0. The o.g.f. for the Z-sequence is 4-2*c(-x) with the Catalan number o.g.f. c(x). - Wolfdieter Lang, Jun 01 2007

As a triangle, T(2n,n) is A001448. Row sums are A046748. Diagonal sums are A176280. - Paul Barry, Apr 14 2010

From Wolfdieter Lang, Aug 10 2017: (Start)

The row polynomials R(n, x) of Riordan triangles R = (G(x), F(x)), with F(x)= x*Fhat(x), belong to the class of Boas-Buck polynomials (see the reference). Hence they satisfy the Boas-Buck identity (we use the notation of Rainville, Theorem 50, p. 141):

(E_x - n*1)*R(n, x) = -Sum_{k=0..n-1} (alpha(k)*1 + beta(k)*E_x)*R(n-1.k, x), for n >= 0, where E_x = x*d/dx (Euler operator). The Boas-Buck sequences are given by alpha(k) := [x^k] ((d/dx)log(G(x))) and beta(k) := [x^k] (d/dx)log(Fhat(x)).

This entails a recurrence for the sequence of column m of the Riordan triangle T, n > m >= 0: T(n, m) = (1/(n-m))*Sum_{k=m..n-1} (alpha(n-1-k) + m*beta(n-1-k))*T(k, m), with input T(m,m).

For the present case the Boas-Buck identity for the row polynomials is (E_x - n*1)*R(n, x) = -Sum_{k=0..n-1} 2^(2*k+1)*(1 + 2*E_x)*R(n-1-k, x), for n >= 0. For the ensuing recurrence for the columns m of the triangle T see the formula and example section. (End)

From Peter Bala, Mar 04 2018: (Start)

The following two remarks are particular cases of more general results for Riordan arrays of the form (f(x), x/(1 - k*x)).

1) Let R(n,x) denote the n-th row polynomial of this triangle. The polynomial R(n,4*x) has the e.g.f. Sum_{k = 0..n} T(n,k)*(4*x)/k!. The e.g.f. for the n-th diagonal of the triangle (starting at n = 0 for the main diagonal) equals exp(x) * the e.g.f. for the polynomial R(n,4*x). For example, when n = 3 we have exp(x)*(20 + 30*(4*x) + 10*(4*x)^2/2! + (4*x)^3/3!) = 20 + 140*x + 420*x^2/2! + 924*x^3/3! + 1716*x^4/4! + ....

2) Let P(n,x) = Sum_{k = 0..n} T(n,k)*x^(n-k) denote the n-th row polynomial in descending powers of x. P(n,x) is the n-th degree Taylor polynomial of (1 + 4*x)^(n-1/2) about 0. For example, for n = 4 we have (1 + 4*x)^(7/2) = 70*x^4 + 140*x^3 + 70*x^2 + 14*x + 1 + O(x^5).

Let C(x) = (1 - sqrt(1 - 4*x))/(2*x) denote the o.g.f. of the Catalan numbers A000108. The derivatives of C(x) are determined by the identity (-1)^n * x^n/n! * (d/dx)^n(C(x)) = 1/(2*x)*( 1 - P(n,-x)/(1 - 4*x)^(n-1/2) ), n = 0,1,2,.... See Lang 2002. Cf. A283150 and A283151. (End)