A046521 -id:A046521 - cp's OEIS frontend

A111959 Renewal array for aerated central binomial coefficients.

1, 0, 1, 2, 0, 1, 0, 4, 0, 1, 6, 0, 6, 0, 1, 0, 16, 0, 8, 0, 1, 20, 0, 30, 0, 10, 0, 1, 0, 64, 0, 48, 0, 12, 0, 1, 70, 0, 140, 0, 70, 0, 14, 0, 1, 0, 256, 0, 256, 0, 96, 0, 16, 0, 1, 252, 0, 630, 0, 420, 0, 126, 0, 18, 0, 1, 0, 1024, 0, 1280, 0, 640, 0, 160, 0, 20, 0, 1, 924, 0, 2772, 0

Offset: 0

Paul Barry, Aug 23 2005

Row sums are A098615.
Binomial transform (product with C(n,k)) is A111960.
Diagonal sums are A026671 (with interpolated zeros).
Inverse is (1/sqrt(1+4x^2),x/sqrt(1+4x^2)), or (sqrt(-1))^(n-k)*T(n,k). [corrected by Peter Bala, Aug 13 2021]
The Riordan array (1,x/sqrt(1-4*x^2)) is the same array with an additional column of zeros (besides the top element 1) added to the left. - Vladimir Kruchinin, Feb 17 2011

			From _Peter Bala_, Aug 13 2021: (Start)
Triangle begins
  1;
  0,  1;
  2,  0, 1;
  0,  4, 0, 1;
  6,  0, 6, 0, 1;
  0, 16, 0, 8, 0, 1;
Infinitesimal generator begins
  0;
  0, 0;
  2, 0, 0;
  0, 4, 0, 0;
  0, 0, 6, 0, 0;
  0, 0, 0, 8, 0, 0; (End)

Paul Barry, On the duals of the Fibonacci and Catalan-Fibonacci polynomials and Motzkin paths, arXiv:2101.10218 [math.CO], 2021.
Vladimir Kruchinin, Composition of ordinary generating functions, arXiv:1009.2565 [math.CO], 2010.

Cf. A054335, A038231, A046521.

Riordan array (1/sqrt(1-4x^2), x/sqrt(1-4x^2)); number triangle T(n, k)=(1+(-1)^(n-k))*binomial((n-1)/2, (n-k)/2)*2^(n-k)/2.
G.f.: 1/(1-xy-2x^2/(1-x^2/(1-x^2/(1-x^2/(1-.... (continued fraction). - Paul Barry, Jan 28 2009
From Peter Bala, Aug 13 2021: (Start)
T(2*n,2*k) = A046521(n,k); T(2*n+1,2*k+1) = A038231(n,k).
The row entries, read from right to left, are the coefficients in the n-th order Taylor polynomial of (sqrt(1 + 4*x^2))^((n-1)/2) at x = 0.
The infinitesimal generator of this array has the sequence [2, 4, 6, 8, 10, ...] on the second subdiagonal below the main diagonal and zeros elsewhere.
The m-th power of the array is the Riordan array (1/sqrt(1 - 4*m*x^2), x/sqrt(1 - 4*m*x^2)) with entries given by sqrt(m)^(n-k)*T(n,k). (End)

A319203 Triangular Riordan matrix T = R^(-1) for triangular Riordan matrix R = (1/(1 - x^2 - x^3), x/(1 - x^2 - x^3)) = A104578.

Original entry on oeis.org

1, 0, 1, -1, 0, 1, -1, -2, 0, 1, 2, -2, -3, 0, 1, 5, 5, -3, -4, 0, 1, -2, 12, 9, -4, -5, 0, 1, -21, -7, 21, 14, -5, -6, 0, 1, -14, -56, -16, 32, 20, -6, -7, 0, 1, 72, -30, -108, -30, 45, 27, -7, -8, 0, 1, 138, 210, -45, -180, -50, 60, 35, -8, -9, 0, 1

Offset: 0

Wolfdieter Lang, Oct 29 2018

This is the lower triangular Riordan matrix (f(t), t*f(t)), with f(t) = F^{[-1]}(t)/t, where F(x) = x/(1 - x^2 - x^3). The expansion of f(t) is given in A319201, the sequence of column k = 0.
This gives the inverse Matrix (with upper diagonals filled with 0's) of the Riordan matrix from A104578 for any finite dimension.
The row sums give A321204, and the alternating row sums give A321205.
The A- and Z-sequences of this inverse Riordan triangle of (F(x)/x, F(x)) are A = [1, 0, -1, -1] generated by 1/(F(x)/x), and Z = [0,-1, -1] generated from 1/F(x) - 1/x. See the link W. Lang link for A- and Z- sequences in A006232 with references.
For the Boas-Buck recurrence of Riordan triangles see the Aug 10 2017 remark in A046521, also for the reference. For this Bell-type triangle the sequence b is generated by B(t) = (log(f(t)))' = (1/(1/f(t) + t^2*f(t) + 2*t^3*f(t)^2) - 1)/t, and is given in A319204.

			The triangle T(n, k) begins:
n\k     0   1    2    3   4  5  6  7  8  9 10 ...
-------------------------------------------------
0:      1
1:      0   1
2:     -1   0    1
3:     -1  -2    0    1
4:      2  -2   -3    0   1
5;      5   5   -3   -4   0  1
6:     -2  12    9   -4  -5  0  1
7:    -21  -7   21   14  -5 -6  0  1
8:    -14 -56  -16   32  20 -6 -7  0  1
9:     72 -30 -108  -30  45 27 -7 -8  0  1
10:   138 210  -45 -180 -50 60 35 -8 -9  0  1
...
Recurrence from A- and Z-sequence: 5 =  T(5, 0) = -(-2 + (-3)); 9 = T(6, 2) = 5 - (- 4 + 0).
Recurrence of Boas-Buck type, with B = [0,-2,-3, 6, ...] = A319204: 9 = T(6, 2) = ((2+1)/(6-2))*(6*1 + (-3)*0 + (-2)*(-3) + 0*(-3)) = (3/4)*12 = 9.

Cf. A104578, A319201, A319204, A321204, A321205.

(* The function RiordanArray is defined in A256893. *)
nmax = 10;
R = RiordanArray[1/(1 - #^2 - #^3)&, #/(1 - #^2 - #^3)&, nmax+1];
M = Inverse[PadRight[#, nmax+1]& /@ R];
T[n_, k_] := M[[n+1, k+1]];
Table[T[n, k], {n, 0, nmax}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 19 2019 *)

Recurrence from the Z- and A-sequence: T(n, k) = 0 if n < k; T(0, 0) = 1;
T(n, 0) = -(T(n-1, 1) + T(n-1, 2)), for n >= 1; and T(n, m) = T(n-1, k-1) - T(n-1, k+1) - T(n-1, k+2), for n>=1 and k >= 1.
Boas-Buck recurrence with B(n) = A319204(n): T(n, k) = ((k+1)/(n-k))*Sum_{j=k..n-1} b(n-1-j)*T(j, k), for n >= 1, k = 0,1, ..., n-1, and input T(n,n) = 1, for n >= 0.
G.f. of row polynomials R(n,x) = Sum_{k=0..n} T(n, k)*x^k is G(x,z) = f(z)/(1-x*z*f(z)) with the expansion of f given in A319201.
G.f. of column sequences Gcol(k, x) = x^k*f(x)^{k+1}, for k >= 0.

A020932 Expansion of 1/(1-4*x)^(21/2).

Original entry on oeis.org

1, 42, 966, 16100, 217350, 2521260, 26053020, 245642760, 2149374150, 17672631900, 137846528820, 1027583214840, 7364346373020, 50983936428600, 342320716020600, 2236495344667920, 14257657822257990, 88900689950549820, 543281994142248900, 3259691964853493400

Offset: 0

N. J. A. Sloane

Harvey P. Dale, Table of n, a(n) for n = 0..1000

Cf. A000984, A001622, A020930, A046521 (eleventh column).

List([0..20], n-> Binomial(2*(n+10),n+10)*Binomial(n+10, 10)/Binomial(20,10)); # G. C. Greubel, Jul 21 2019

[&*[2*n+i: i in [1..19 by 2]]*Binomial(2*n, n)/654729075: n in [0..20]]; // Vincenzo Librandi, Jul 05 2013

CoefficientList[Series[1/(1-4x)^(21/2),{x,0,30}],x] (* Harvey P. Dale, Oct 10 2011 *)

vector(20, n, n--; m=n+10; binomial(2*m,m)*binomial(m, 10)/binomial(20,10) ) \\ G. C. Greubel, Jul 21 2019

[binomial(2*(n+10), n+10)*binomial(n+10, 10)/binomial(20,10) for n in (0..20)] # G. C. Greubel, Jul 21 2019

a(n) = binomial(n+10, 10)*A000984(n+10)/A000984(10), where A000984 are the central binomial coefficients. - Wolfdieter Lang
a(n) = ((2*n+19)*(2*n+17)*(2*n+15)*(2*n+13)*(2*n+11)*(2*n+9)*(2*n+7)*(2*n+5)*(2*n+3)*(2*n+1)/654729075)*Binomial(2*n, n). - Vincenzo Librandi, Jul 05 2013
Boas-Buck recurrence: a(n) = (42/n)*Sum_{k=0..n-1} 4^(n-k-1)*a(k), n >= 1, a(0) = 1. Proof from a(n) = A046521(n+10, 10). See a comment there. - Wolfdieter Lang, Aug 10 2017
From Amiram Eldar, Mar 27 2022: (Start)
Sum_{n>=0} 1/a(n) = 38476615836/85085 - 83106*sqrt(3)*Pi.
Sum_{n>=0} (-1)^n/a(n) = 29687500*sqrt(5)*log(phi) - 4892382624460/153153, where phi is the golden ratio (A001622). (End)

A283150 Riordan array (1/(1-9x)^(1/3), x/(9x-1)).

Original entry on oeis.org

1, 3, -1, 18, -12, 1, 126, -126, 21, -1, 945, -1260, 315, -30, 1, 7371, -12285, 4095, -585, 39, -1, 58968, -117936, 49140, -9360, 936, -48, 1, 480168, -1120392, 560196, -133380, 17784, -1368, 57, -1, 3961386, -10563696, 6162156, -1760616, 293436, -30096, 1881, -66, 1, 33011550, -99034650, 66023100

Offset: 0

Tom Richardson, Mar 01 2017

Triangle read by rows. This is an example of a Riordan group involution. Dual Riordan array of A283151. With A283151 and A248324, forms doubly infinite Riordan array. For b and c the sequences A283151 and A248324, respectively, and i,j >= 0, the doubly infinite array with d(i,j) = a(i,j), d(-j,-i) = b(i,j), d(i,-j) = c(j,i), and d(-i,j) = 0 (except d(0,0)=1) is a doubly infinite Riordan array.

Matrix inverse of a(m,n) is a(m,n). - Werner Schulte, Aug 05 2017

			The triangle begins
        1;
        3,        -1;
       18,       -12,       1;
      126,      -126,      21,       -1;
      945,     -1260,     315,      -30,      1;
     7371,    -12285,    4095,     -585,     39,     -1;
    58968,   -117936,   49140,    -9360,    936,    -48,    1;
   480168,  -1120392,  560196,  -133380,  17784,  -1368,   57,  -1;
  3961386, -10563696, 6162156, -1760616, 293436, -30096, 1881, -66, 1;

Peter Bala, A 4-parameter family of embedded Riordan arrays
Peter Bala, A note on the diagonals of a proper Riordan Array
H. Prodinger, Some information about the binomial transform, The Fibonacci Quarterly, 32, 1994, 412-415.
Thomas M. Richardson, The three 'R's and Dual Riordan Arrays, arXiv:1609.01193 [math.CO], 2016.

Cf. A004987, A248324, A283151, A046521, A155579.

T := (n, k) -> (-1)^k*binomial(n - 2/3, n - k)*9^(n - k):
for n from 0 to 6 do seq(T(n, k), k = 0..n) od; # Peter Luschny, Sep 03 2021

a(m,n) = binomial(-n-1/3, m-n)*(-1)^m*9^(m-n);
tabl(nn) = for(n=0, nn, for (k=0, n, print1(a(n, k), ", ")); print); \\ Michel Marcus, Aug 07 2017

a(m,n) = binomial(-n-1/3, m-n)*(-1)^m*9^(m-n).
G.f.: (1-9x)^(2/3)/(xy-9x+1).
Recurrence: a(m,n) = a(m, n-1)*(n-1-m)/(9*n-6) for 0 < n <= m. - Werner Schulte, Aug 05 2017
From Peter Bala, Mar 05 2018 (Start):
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. Then (-1)^n*P(n,x) is the n-th degree Taylor polynomial of (1 - 9*x)^(n-2/3) about 0. For example, for n = 4 we have (1 - 9*x)^(10/3) = 945*x^4 - 1260*x^3 + 315*x^2 - 30*x + 1 + O(x^5).
Let R(n,x) denote the n-th row polynomial of this triangle. The polynomial R(n,9*x) has the e.g.f. Sum_{k = 0..n} T(n,k)*(9*x)^k/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,9*x). For example, when n = 3 we have exp(-x)*(126 - 126*(9*x) + 21*(9*x)^2/2! - (9*x)^3/3!) = 126 - 1260*x + 4095*x^2/2! - 9360*x^3/3! + 17784*x^4/4! - ....
Let F(x) = (1 - ( 1 - 9*x)^(2/3))/(3*x) denote the o.g.f. of A155579. The derivatives of F(x) are related to the row polynomials P(n,x) by the identity x^n/n! * (d/dx)^n(F(x)) = 1/(3*x)*( (-1)^n - P(n,x)/(1 - 9*x)^(n-2/3) ), n = 0,1,2,.... Cf. A283151 and A046521. (End)
From Peter Bala, Aug 18 2021: (Start)
T(n,k) = (-1)^k*binomial(n-2/3, n-k)*9^(n-k).
Analogous to the binomial transform we have the following sequence transformation formula: g(n) = Sum_{k = 0..n} T(n,k)*b^(n-k)*f(k) iff f(n) = Sum_{k = 0..n} T(n,k)*b^(n-k)*g(k). See Prodinger, bottom of p. 413, with b replaced with 9*b, c = -1 and d = 1/3.
Equivalently, if F(x) = Sum_{n >= 0} f(n)*x^n and G(x) = Sum_{n >= 0} g(n)*x^n are a pair of formal power series then
G(x) = (1/(1 - 9*b*x)^(1/3)) * F(x/(1 - 9*b*x)) iff F(x) = (1/(1 + 9*b*x)^(1/3)) * G(x/(1 + 9*b*x)).
The infinitesimal generator of the unsigned array has the sequence (9*n+3) n>=0 on the main subdiagonal and zeros elsewhere. The m-th power of the unsigned array has entries m^(n-k)*|T(n,k)|. (End)

Offset corrected by Werner Schulte, Aug 05 2017

A283151 Triangle read by rows: Riordan array (1/(1-9x)^(2/3), x/(9x-1)).

Original entry on oeis.org

1, 6, -1, 45, -15, 1, 360, -180, 24, -1, 2970, -1980, 396, -33, 1, 24948, -20790, 5544, -693, 42, -1, 212058, -212058, 70686, -11781, 1071, -51, 1, 1817640, -2120580, 848232, -176715, 21420, -1530, 60, -1, 15677145, -20902860, 9754668, -2438667, 369495, -35190, 2070, -69, 1, 135868590, -203802885

Offset: 0

Tom Richardson, Mar 01 2017

This is an example of a Riordan group involution.
Dual Riordan array of A283150.
With A283150 and A248324, forms doubly infinite Riordan array. For b and c the sequences A283150 and A248324, respectively, and i,j >= 0, the doubly infinite array with d(i,j) = a(i,j), d(-j,-i) = b(i,j), d(i,-j) = c(i,j), and d(-i,j) = 0 (except d(0,0)=1) is a doubly infinite Riordan array.

			Triangle begins
         1;
         6,        -1;
        45,       -15,       1;
       360,      -180,      24,       -1;
      2970,     -1980,     396,      -33,      1;
     24948,    -20790,    5544,     -693,     42,     -1;
    212058,   -212058,   70686,   -11781,   1071,    -51,    1;
   1817640,  -2120580,  848232,  -176715,  21420,  -1530,   60,  -1;
  15677145, -20902860, 9754668, -2438667, 369495, -35190, 2070, -69, 1;

Peter Bala, A 4-parameter family of embedded Riordan arrays
Peter Bala, A note on the diagonals of a proper Riordan Array
H. Prodinger, Some information about the binomial transform, The Fibonacci Quarterly, 32, 1994, 412-415.
Thomas M. Richardson, The three 'R's and Dual Riordan Arrays, arXiv:1609.01193 [math.CO], 2016.

Cf. A004988, A248324, A283150, A025748, A046521.

a(m,n) = binomial(-n-2/3, m-n)*(-1)^m*9^(m-n).
G.f.: (1-9x)^(1/3)/(xy-9x+1).
Recurrence: a(m,n) = a(m,n-1)*(n-1-m)/(9*n-3) for 0 < n <= m; matrix inverse of a(m,n) is a(m,n). - Werner Schulte, Aug 05 2017
From Peter Bala, Mar 05 2018 (Start):
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. Then (-1)^n*P(n,x) is the n-th degree Taylor polynomial of (1 - 9*x)^(n-1/3) about 0. For example, for n = 4 we have (1 - 9*x)^(11/3) = 2970*x^4 - 1980*x^3 + 396*x^2 - 33*x + 1 + O(x^5).
Let R(n,x) denote the n-th row polynomial of this triangle. The polynomial R(n,9*x) has the e.g.f. Sum_{k = 0..n} T(n,k)*(9*x)^k/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,9*x). For example, when n = 3 we have exp(-x)*(360 - 180*(9*x) + 24*(9*x)^2/2! - (9*x)^3/3!) = 360 - 1980*x + 5544*x^2/2! - 11781*x^3/3! + 21420*x^4/4! - ....
Let F(x) = (1 - ( 1 - 9*x)^(1/3))/(3*x). See A025748. The derivatives of F(x) are related to the row polynomials P(n,x) by the identity x^n/n! * (d/dx)^n(F(x)) = 1/(3*x)*( (-1)^n - P(n,x)/(1 - 9*x)^(n-1/3) ), n = 0,1,2,.... Cf. A283151 and A046521. (End)
From Peter Bala, Aug 18 2021: (Start)
T(n,k) = (-1)^k*binomial(n-1/3, n-k)*9^(n-k).
Analogous to the binomial transform we have the following sequence transformation formula: g(n) = Sum_{k = 0..n} T(n,k)*b^(n-k)*f(k) iff f(n) = Sum_{k = 0..n} T(n,k)*b^(n-k)*g(k). See Prodinger, bottom of p. 413, with b replaced with 9*b, c = -1 and d = 2/3.
Equivalently, if F(x) = Sum_{n >= 0} f(n)*x^n and G(x) = Sum_{n >= 0} g(n)*x^n are a pair of formal power series then
G(x) = (1/(1 - 9*b*x)^(2/3)) * F(x/(1 - 9*b*x)) iff F(x) = (1/(1 + 9*b*x)^(2/3)) * G(x/(1 + 9*b*x)).
The infinitesimal generator of the unsigned array has the sequence (9*n+6) n>=0 on the main subdiagonal and zeros elsewhere. The m-th power of the unsigned array has entries m^(n-k)*|T(n,k)|. (End)

Offset corrected by Werner Schulte, Aug 05 2017

A321196 Riordan triangle T = (1/(1 + x^2 - x^3), x/(1 + x^2 - x^3)).

Original entry on oeis.org

1, 0, 1, -1, 0, 1, 1, -2, 0, 1, 1, 2, -3, 0, 1, -2, 3, 3, -4, 0, 1, 0, -6, 6, 4, -5, 0, 1, 3, -1, -12, 10, 5, -6, 0, 1, -2, 12, -4, -20, 15, 6, -7, 0, 1, -3, -7, 30, -10, -30, 21, 7, -8, 0, 1, 5, -16, -15, 60, -20, -42, 28, 8, -9, 0, 1

Offset: 0

Wolfdieter Lang, Nov 09 2018

This is the (ordinary) convolution triangle based on A077961 (the column k = 0 of T).
The row polynomials R(n, x) := Sum_{k=0..n} T(n, k)*x^k, with R(-1, x) = 0, appear in the Cayley-Hamilton formula for nonnegative powers of a 3 X 3 matrix with Det M = sigma(3; 3) = x1*x2*x3 = +1, sigma(3; 2) := x1*x2 + x1*x*3 + x2*x^3 = +1 and Tr M = sigma(3; 1) = x1 + x2 = x, where x1, x2, and x3 are the eigenvalues of M, and sigma the elementary symmetric functions, as M^n = R(n-2, x)*M^2 + (-R(n-3, x) + R(n-4, x))*M + R(n-3, x)*1_3, for n >= 3, where M^0 = 1_3 is the 3 X 3 unit matrix.
For the Cayley-Hamilton formula for 3 X 3 matrices with Det M = +1, sigma(3,2) = -1 and Tr(M) = x see A104578.
The row sums give A133872 (repeat(1, 1, 0, 0)). The alternating row sums give A057597(n+2), for n >= 0.
The Riordan triangle (1/(1 + x^2 + x^3), x/(1 + x^2 + x^3)) has entries t(n, m) = (-1)^(n-m)*T(n, m) (from the g.f. G(-x, -z), where the g.f. G of T is given below).
The inverse of Riordan T is T^{-1}, given in A321198.

			The triangle T(n, k) begins:
n\k  0   1   2   3   4   5  6  7  8  9 10 ...
---------------------------------------------
0:   1
1:   0   1
2:  -1   0   1
3:   1  -2   0   1
4:   1   2  -3   0   1
5:  -2   3   3  -4   0   1
6:   0  -6   6   4  -5   0  1
7:   3  -1 -12  10   5  -6  0  1
8:  -2  12  -4 -20  15   6 -7  0  1
9:  -3  -7  30 -10 -30  21  7 -8  0  1
10:  5 -16 -15  60 -20 -42 28  8 -9  0  1
...
Cayley-Hamilton formula for the matrix TS(x) =[[x,-1,1], [1,0,0], [0,1,0]] with Det(TS(x)) = +1, sigma(3, 2) = +1, and Tr(TS(x)) = x. For n = 3: TS(x)^3 = R(1, x)*TS(x)^2 + (-R(0, x) + R(-1, x))*TS(x) + R(0, x)*1_3 = x*TS(x)^2 - TS(x) + 1_3. Compare this for x = -1 with r^3 = R(3)*r^2 + (-R(2) + R(1))*r + R(2)*1 = r^2 - r + 1, where r = 1/t = A192918, with the tribonacci constant t = A058265, and R(n) = A057597(n) = R(n-2, -1).
Recurrence: T(5, 2) = T(4, 1) - T(3, 2) + T(2, 2) = 1 -(-1) + 1 = 3.
Boas-Buck type recurrence with B = {0, -2, 3, ...}:
  T(5, 2) = ((2+1)/(5-2))*(3*1 + (-2)*0 + 0*(-3)) = 1*3 = 3.
Z- and A-recurrence with A(n) = {1, 0, -1, 1, -1, ...} and Z(n) = A(n+1):
  T(4, 0) = 0*T(3, 0) - 1*T(3, 1) + 1*T(3, 2) - 1*T(3, 3) = 0 + 2 + 0 - 1 = 1.
  T(5, 2) = 1*T(4, 1) + 0*T(4, 2) - 1*T(4, 3) + 1*T(4, 4) = 2 + 0 + 0 + 1 = 3.

Cf. A057597, A058265, A077961, A001005, A104578, A112455, A133872, A192918, A321197, A321198.

T[n_, k_] /; 0 <= k <= n := T[n, k] = T[n - 1, k - 1] - T[n - 2, k] + T[n - 3, k]; T[0, 0] = 1; T[, ] = 0;
Table[T[n, k], {n, 0, 10}, {k, 0, n}] (* Jean-François Alcover, Jul 06 2019 *)

# uses[riordan_array from A256893]
riordan_array(1/(1 + x^2 - x^3), x/(1 + x^2 - x^3), 11) # Peter Luschny, Nov 13 2018

T(n, k) = T(n-1, k-1) - T(n-2, k) + T(n-3, k), T(0, 0) = 1, T(n,k) = 0 if n < k or if k < 0. (Cf. A104578.)
The Riordan property T = (G(x), x*G(x)) with G(x) = 1/(1 + x^2 - x^3) implies the following.
G.f. of row polynomials R(n, x) is G(x, z) = 1/(1 - x*z + z^2 - z^3).
G.f. of column sequence k: x^k/(1 + x^2 - x^3)^(k+1), k >= 0.
Boas-Buck recurrence (see the Aug 10 2017 remark in A046521, also for two references):
T(n, k) = ((k+1)/(n-k))*Sum_{j=k..n-1} B(n-1-j)*T(j, k), for n >= 1, k = 0,1, ..., n-1, and input T(n, n) = 1, for n >= 0. Here B(n) = [x^n]*(d/dx)log(G(x)) = x*(-2 + 3*x)/(1 + x^2 - x^3) = (-1)^n*A112455(n+1), for n >= 0.
Recurrences from the A- and Z- sequences (see the W. Lang link under A006232 with references), which are A(n) = A321197(n) and Z(n) = A(n+1).
  T(0, 0) = 1, T(n, k) = 0 for n < k, and
  T(n, 0) = Sum_{j=0..n-1} Z(j)*T(n-1, j), for n >= 1, and
  T(n, k) = Sum_{j=0..n-k} A(j)*T(n-1, k-1+j), for n >= m >= 1.

A321198 Triangular Riordan matrix T = R^(-1) for triangular Riordan matrix R = (1/(1 + x^2 - x^3), x/(1 + x^2 - x^3)) given in A321196.

Original entry on oeis.org

1, 0, 1, 1, 0, 1, -1, 2, 0, 1, 2, -2, 3, 0, 1, -5, 5, -3, 4, 0, 1, 8, -12, 9, -4, 5, 0, 1, -21, 21, -21, 14, -5, 6, 0, 1, 42, -56, 40, -32, 20, -6, 7, 0, 1, -96, 114, -108, 66, -45, 27, -7, 8, 0, 1, 222, -270, 225, -180, 100, -60, 35, -8, 9, 0, 1

Offset: 0

Wolfdieter Lang, Nov 12 2018

Riordan triangle T = (f(t), t*f(t)), with f(t) = F^{[-1]}(t)/t, where F^{[-1]}(t) is the compositional inverse of t = F(x) = x/(1 + x^2 - x^3). The expansion of f(t) is given by {(-1)^n*A001005(n)}_{n >= 0}, the sequence of column k = 0.
This gives the inverse matrix (with upper diagonals filled with zeros) of the Riordan matrix from A321196 for any finite dimension.
The inverse of the Riordan matrix (1/(1 + x^2 + x^3), x/(1 + x^2 + x^3)) is obtained from the triangle t(n, k) = (-1)^(n-k)*T(n, k), with vanishing upper diagonals.
The row sums give A321199. The alternating row sums give A321200.
The finite A- and Z-sequences of this inverse Riordan triangle of (F(x)/x, F(x)) are A = [1, 0, +1, -1] generated by 1/(F(x)/x), and Z = [0, +1, -1] generated by 1/F(x) - 1/x. See the W. Lang link for A- and Z- sequences in A006232 with references.
For the Boas-Buck column recurrences of Riordan triangles see the Aug 10 2017 remark in A046521, also for two references. For this Bell-type Riordan triangle the Boas-Buck sequence b is generated by B(t) = (log(f(t)))' = (1/(1/f(t) - t^2*f(t) + 2*t^3*f(t)^2) - 1)/t, and b(n) = (-1)^(n)*A176806(n+1), for n >= 0, because the parity of e_3 in the rows n of A321201 coincides with the one of n. See A321203 for the multinomials with negative signs for odd row numbers.

			The triangle begins:
n\k     0     1     2     3    4    5   6   7  8  9  10 ...
-----------------------------------------------------------
0:      1
1:      0     1
2:      1     0     1
3:     -1     2     0     1
4:      2    -2     3     0    1
5:     -5     5    -3     4    0    1
6:      8   -12     9    -4    5    0   1
7:    -21    21   -21    14   -5    6   0   1
8:     42   -56    40   -32   20   -6   7   0  1
9:    -96   114  -108    66  -45   27  -7   8  0  1
10:   222  -270   225  -180  100  -60  35  -8  9  0   1
...
------------------------------------------------------------
Recurrence (from A- and Z-sequence): Z: T(5, 0) = T(4, 1) - T(4, 2) = -2 - 3 = -5. A: T(5, 2) = T(4, 1) + T(4, 3) - T(4, 4) = - 2 + 0 - 1 = -3.
Recurrence column k = 2 (Boas-Buck type sequence b = (-1)^(n+1)* = {0, 2, -3, 6, ...}): T(5, 2) = (3/3)*(b(2)*T(2, 2) + b(1)*T(3, 2) + b(0)*T(4, 2)) = -3*1 + 2*0 + 0*4 = -3.

Cf. A001005, A176806, A321196, A321199, A321200, A321201, A321203.

T[n_, k_] := T[n - 1, k - 1] + T[n - 1, k + 1] - T[n - 1, k + 2] /; n >= k >= 0;Table[T[n, k], {n, 0, 10}, {k, 0, n}] (* James C. McMahon, Dec 10 2023 *)

Recurrence (from A-and Z-sequences): T(n, k) = 0 for n < k, T(0, 0) = 1. Z: T(n, 0) = T(n-1, 1) - T(n-1, 2), n >= 1; A: T(n, k) = T(n-1, k-1) + T(n-1, k+1) - T(n-1, k+2), n >= k >= 0.
Recurrence for column k (Boas-Buck type): T(n, n) = 1; T(n, k) = ((k+1)/(n-k))*Sum_{j=k..n-1} b(n-1-j)*T(j, k), n >= m+1 >= 1. For b see the Boas-Buck comment above.
G.f of row polynomials R(n, x) := Sum_{k=0..n} T(n, k)*x^k: G(x, z) = f(z)/(1 - x*z*f(z)), with f(z) = F^{[-1]}(z)/z, where F^{[-1]}(z) is the compositional inverse of z = F(y) = y/(1 + y^2 - y^3).
G.f of column k: Gcol(k, x) = x^k*f(x)^{k+1}.

A063016 a(n) is the product of Catalan(n) and (2^(n+1) - 1).

Original entry on oeis.org

1, 3, 14, 75, 434, 2646, 16764, 109395, 730730, 4973826, 34381412, 240728670, 1703826292, 12170930700, 87633375480, 635351667075, 4634365164570, 33985474184970, 250419761106900, 1853107999454250, 13765951702923420, 102618937160787060, 767411273728449480

Offset: 0

Olivier Gérard, Jul 04 2001

From Michael Wallner, Nov 03 2021: (Start)
a(n) is also the number of n X 2 Young tableaux with (possibly) vertical walls. The entries in cells that are separated by such a wall do not have to obey any order constraints. See Banderier, Wallner 2021 and Banderier et al. 2018.
a(n) is also the number of binary trees with n vertices and marked leaves, where at least 1 leaf has to be marked. Banderier, Wallner 2021 give a bijection to n X 2 Young tableaux with vertical walls. (End)

Harry J. Smith, Table of n, a(n) for n = 0..200
C. Banderier, P. Marchal, and M. Wallner, Rectangular Young tableaux with local decreases and the density method for uniform random generation, GASCOM 2018. Vol. 2113. CEUR Workshop Proceedings, 2018.
C. Banderier and M. Wallner, Young tableaux with periodic walls: counting with the density method Sém. Lothar. Combin. 85B (2021), Art. 47, 12 pp.

Cf. A000108, A063017, A126646, A007317, A046521.

Table[CatalanNumber[n]*(2^(n+1)-1),{n,0,20}] (* Harvey P. Dale, Oct 20 2014 *)

a(n) = (2^(n + 1) - 1)*binomial(2*n, n)/(n + 1); \\ Harry J. Smith, Aug 16 2009

def A063016(n) :
    return (8^(n+1)-4^(n+1))*factorial(n-1/2)/(4*sqrt(pi)*factorial(n+1))
[A063016(i) for i in (0..20)] # Peter Luschny, Jul 24 2012

a(n) = Catalan(n)*(2^(n+1) - 1) = A000108(n) * A126646(n).
D-finite with recurrence: a(n) = 2*(2*n-1)*(3*n*a(n-1)-4*(2*n-3)*a(n-2))/((n+1)*n). - Georg Fischer, Jun 06 2021
G.f.: A(x) = (sqrt(1-4*x) - sqrt(1-8*x))/(2*x).
G.f.: G(0)/(2*x) where G(k) = 1 - 2^k/(1 - 2*x*(2*k-1)/(2*x*(2*k-1) - 2^k*(k+1)/G(k+1))); (continued fraction, 3rd kind, 3-step). - Sergei N. Gladkovskii, Jul 24 2012
From Peter Bala, Aug 17 2021: (Start)
a(n) = Sum_{k = 0..n} A046521(n,k)*Catalan(k).
G.f.: A(x) = 1/sqrt(1 - 4*x)*c(x/(1 - 4*x)), where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the g.f. of the Catalan numbers A000108. Inversely, c(x) = 1/sqrt(1 + 4*x)*A(x/(1 + 4*x)).
Series reversion of x*A(x) = x*(1 - 3*x + 4*x^2*c(-2*x^2)). (End)

Initial term 0 removed by Harry J. Smith, Aug 16 2009

A277604 Array of coefficients T(k,n) of the formal power series A(k,x) read by upwards antidiagonals, where A(k,x) = sqrt(1 + 2xA(k,x) + (4k+1)x^2*(A(k,x))^2), k >= 0.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 5, 5, 1, 1, 1, 7, 9, 13, 1, 1, 1, 9, 13, 37, 25, 1, 1, 1, 11, 17, 73, 81, 61, 1, 1, 1, 13, 21, 121, 169, 301, 125, 1, 1, 1, 15, 25, 181, 289, 841, 729, 295, 1, 1, 1, 17, 29, 253, 441, 1801, 2197, 2549, 625, 1, 1, 1, 19, 33, 337, 625, 3301, 4913, 10123, 6561, 1447, 1

Offset: 0

Werner Schulte, Oct 29 2016

For k = 0 see A000012, for k = 1 see A098615, and for k = 2 see A200376.
It will be interesting using the formulae for k < 0 (attention: signed terms!). Especially for k = -1 see A157674.
If G is the g.f. of central binomial coefficients (see A000984) and B(k,x) = G(k*x^2), then B(k,x) = A(k,x)/(1+x*A(k,x)) and A(k,x) = B(k,x) / (1-x*B(k,x)) for k >= 0. - Werner Schulte, Aug 07 2017

			The terms define the array T(k,n) for k >= 0 and n >= 0, i.e.,
k\n  0  1   2   3    4     5      6      7       8        9  . . .
0:   1  1   1   1    1     1      1      1       1        1  . . .
1:   1  1   3   5   13    25     61    125     295      625  . . .
2:   1  1   5   9   37    81    301    729    2549     6561  . . .
3:   1  1   7  13   73   169    841   2197   10123    28561  . . .
4:   1  1   9  17  121   289   1801   4913   28057    83521  . . .
5:   1  1  11  21  181   441   3301   9261   63071   194481  . . .
6:   1  1  13  25  253   625   5461  15625  123565   390625  . . .
7:   1  1  15  29  337   841   8401  24389  219619   707281  . . .
8:   1  1  17  33  433  1089  12241  35937  362993  1185921  . . .
9:   1  1  19  37  541  1369  17101  50653  567127  1874161  . . .
etc.

Cf. A000012, A000045, A000108, A000984, A015441, A046521, A098615, A111959, A200376.

A(k,x) = (x + sqrt(1 - 4*k*x^2))/(1 - (4*k+1)*x^2) for k >= 0.
T(k,0) = 1 and T(k,2*n+2) = (4*k+1)^(n+1)-2*(Sum_{i=0..n} A000108(i)*k^(i+1)* (4*k+1)^(n-i)), and T(k,2*n+1) = (4*k+1)^n for k >= 0 and n >= 0.
A(k,x) = 1/(1 - x - 2*k*x^2*C(k*x^2)), k >= 0, where C is the g.f. of A000108.
Conjecture: If B(k,n) satisfy B(k,0) = B(k,1) = 1 and B(k,n+2) = B(k,n+1) + k*B(k,n) for k >= 0 and n >= 0 (generalized Fibonacci numbers, see A015441) and G(k,x) = Sum_{n>=0} A000108(n)*B(k,n)*x^n for k >= 0, then you will have (1): A(k,x*G(k,x)) = G(k,x) and (2): G(k,x/A(k,x)) = A(k,x) for k >= 0. Especially for k = 1 see A098615 and for k = 2 see A200376.
Conjecture: T(k,2*n) = Sum_{i=0..n} A046521(n,i)*k^(n-i) for k, n >= 0. - Werner Schulte, Aug 02 2017
Recurrence: T(k,2*n+2) = (4*k+1)*T(k,2*n)-2*k^(n+1)*A000108(n) with initial value T(k,0) = 1 for k >= 0 and n >= 0. - Werner Schulte, Aug 09 2017
T(k,n) = Sum_{i=0..n} A111959(n,i)*k^((n-i)/2) for k >= 0 and n >= 0. - Werner Schulte, Aug 09 2017

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A111959 Renewal array for aerated central binomial coefficients.

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A319203 Triangular Riordan matrix T = R^(-1) for triangular Riordan matrix R = (1/(1 - x^2 - x^3), x/(1 - x^2 - x^3)) = A104578.

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A020932 Expansion of 1/(1-4*x)^(21/2).

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A283150 Riordan array (1/(1-9x)^(1/3), x/(9x-1)).

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A283151 Triangle read by rows: Riordan array (1/(1-9x)^(2/3), x/(9x-1)).

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A321196 Riordan triangle T = (1/(1 + x^2 - x^3), x/(1 + x^2 - x^3)).

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A321198 Triangular Riordan matrix T = R^(-1) for triangular Riordan matrix R = (1/(1 + x^2 - x^3), x/(1 + x^2 - x^3)) given in A321196.

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A063016 a(n) is the product of Catalan(n) and (2^(n+1) - 1).

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A277604 Array of coefficients T(k,n) of the formal power series A(k,x) read by upwards antidiagonals, where A(k,x) = sqrt(1 + 2xA(k,x) + (4k+1)x^2*(A(k,x))^2), k >= 0.