A107264 -id:A107264 - cp's OEIS frontend

A007564 Shifts left when INVERT transform applied thrice.

1, 1, 4, 19, 100, 562, 3304, 20071, 124996, 793774, 5120632, 33463102, 221060008, 1473830308, 9904186192, 67015401391, 456192667396, 3122028222934, 21467769499864, 148246598341018, 1027656663676600, 7148588698592956, 49884553176689584

Offset: 0

N. J. A. Sloane

More generally, coefficients of (1+m*x-sqrt(m^2*x^2-(2*m+2)*x+1))/(2*m*x) are given by a(n) = Sum_{k=0..n} (m+1)^k*N(n,k) where N(n,k) = (1/n)*binomial(n,k)*binomial(n,k+1) are the Narayana numbers (A001263). - Benoit Cloitre, May 24 2003
If y = x*A(x) then 3*y^2 - (1+2*x)*y + x = 0 and x = y*(1-3*y)/(1-2*y). - Michael Somos, Sep 28 2003
The sequence 0,1,4,19,... with g.f. (1-4*x-sqrt(1-8*x+4*x^2))/(6*x) and has a(n) = Sum_{k=0..floor((n-1)/2)} binomial(n-1,2k)*C(k)*4^(n-1-2*k)*3^k. a(n+1) = Sum_{k=0..floor(n/2)} binomial(n,2*k)*C(k)*4^(n-2*k)*3^k counts Motzkin paths of length n in which the level steps have 4 colors and the up steps have 3. It is the binomial transform of A107264 and corresponds to the series reversion of x/(1+4*x+3*x^2). - Paul Barry, May 18 2005
The Hankel transform of this sequence is 3^binomial(n+1,2). - Philippe Deléham, Oct 29 2007
a(n) is the number of Schroder paths of semilength n in which there are no (2,0)-steps at level 0 and at a higher level they come in 2 colors. Example: a(2)=4 because we have UDUD, UUDD, UBD, and URD, where U=(1,1), D=(1,-1), while B (R) is a blue (red) (2,0)-step. - Emeric Deutsch, May 02 2011
a(n) is the number of Schroder paths of semilength n-1 in which the (2,0)-steps at level 0 come in 3 colors and those at a higher level come in 2 colors. Example: a(3)=19 because, denoting  U=(1,1), H=(1,0), and D=(1,-1), we have 3^2 = 9 paths of shape HH, 3 paths of shape HUD, 3 paths of shape UDH, 2 paths of shape UHD, and 1 path of each of the shapes UDUD and UUDD. - Emeric Deutsch, May 02 2011
From David Callan, Jun 21 2013: (Start)
a(n) = number of (left) planted binary trees with n edges in which each vertex has a designated favorite neighbor. Planted binary trees are counted by the Catalan numbers A000108.
Example: for n=2, there are 2 planted binary trees: edges LL and LR from the root (L=left, R=right). Each has just one vertex with 2 neighbors, and so a(2)=4.
Proof outline: each vertex has 1,2 or 3 neighbors. Let X (resp. Y) denote the number of vertices with 2 (resp. 3) neighbors. Then X + 2Y = n - 1 (split the non-root edges into pairs with a common parent vertex and singletons). Thus the number of choices for designating favorite neighbors is 2^X * 3^Y = 2^(n-1)(3/4)^Y. The distribution for Y is known because, under the rotation correspondence, a.k.a. the deBruijn-Morselt bijection, vertices with 2 children in an n-edge planted binary tree correspond to DDUs in a Dyck path, and DDUs have the Touchard distribution (A091894) with gf F(x,y) = (1-2x+2xy - sqrt(1-4x+4x^2-4x^2 y))/(2xy). The desired g.f., Sum_{n>=1} a(n)*x^n, is therefore 1/2*(F(2x,3/4)-1). (End)

			G.f. = 1 + x + 4*x^2 + 19*x^3 + 100*x^4 + 562*x^5 + 3304*x^6 + 20071*x^7 + 124996*x^8 + ...

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

G. C. Greubel, Table of n, a(n) for n = 0..1000 (terms 0 to 100 by T. D. Noe)
C. Banderier, M. Bousquet-Mélou, A. Denise, P. Flajolet, D. Gardy and D. Gouyou-Beauchamps, Generating functions for generating trees, Discrete Mathematics 246(1-3), March 2002, pp. 29-55.
Paul Barry, On Integer-Sequence-Based Constructions of Generalized Pascal Triangles, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.4.
Paul Barry and Aoife Hennessy, A Note on Narayana Triangles and Related Polynomials, Riordan Arrays, and MIMO Capacity Calculations, J. Int. Seq. 14 (2011), Article 11.3.8.
Paul Barry and Aoife Hennessy, Generalized Narayana Polynomials, Riordan Arrays, and Lattice Paths, Journal of Integer Sequences, Vol. 15 (2012), Article 12.4.8.
M. Bernstein and N. J. A. Sloane, Some canonical sequences of integers, Linear Alg. Applications, 226-228 (1995), 57-72; erratum 320 (2000), 210. [Link to arXiv version]
M. Bernstein and N. J. A. Sloane, Some canonical sequences of integers, Linear Alg. Applications, 226-228 (1995), 57-72; erratum 320 (2000), 210. [Link to Lin. Alg. Applic. version together with omitted figures]
Z. Chen and H. Pan, Identities involving weighted Catalan-Schroder and Motzkin Paths, arXiv:1608.02448 arXiv:1608.02448 [math.CO], 2016. Eq. (1.13) a=1, b=3.
C. Coker, A family of eigensequences, Discrete Math. 282 (2004), 249-250.
Shishuo Fu and Yaling Wang, Bijective recurrences concerning two Schröder triangles, arXiv:1908.03912 [math.CO], 2019.
Aoife Hennessy, A Study of Riordan Arrays with Applications to Continued Fractions, Orthogonal Polynomials and Lattice Paths, Ph. D. Thesis, Waterford Institute of Technology, Oct. 2011.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 443
Huyile Liang, Jeffrey Remmel, and Sainan Zheng, Stieltjes moment sequences of polynomials, arXiv:1710.05795 [math.CO], 2017, see page 11.
N. J. A. Sloane, Transforms

Cf. A047891, A054872.

A007564_list := proc(n) local j, a, w; a := array(0..n); a[0] := 1;
for w from 1 to n do a[w] := a[w-1]+3*add(a[j]*a[w-j-1],j=1..w-1) od;
convert(a, list) end: A007564_list(21); # Peter Luschny, May 19 2011

a[0]=1; a[1]=1; a[n_]/;n>=2 := a[n] = a[n-1] + 3 Sum[a[k-1]a[n-k],{k,n-1}] ; Table[a[n],{n,0,10}] (* David Callan, Aug 25 2009 *)
Table[Hypergeometric2F1[-n, 1 - n, 2, 3], {n, 0, 22}] (* Arkadiusz Wesolowski, Aug 13 2012 *)
Table[(2^n (LegendreP[n+1, 2] - LegendreP[n-1, 2]) + 2 KroneckerDelta[n])/(6n+3), {n, 0, 20}] (* Vladimir Reshetnikov, Nov 01 2015 *)
CoefficientList[Series[(1+2x-Sqrt[1-8x+4x^2])/(6x),{x,0,30}],x] (* Harvey P. Dale, Feb 07 2016 *)

{a(n) = if( n<1, n==0, sum( k=0, n, 3^k * binomial( n, k) * binomial( n, k+1)) / n)} /* Michael Somos, Sep 28 2003 */

{a(n) = if( n<0, 0, n++; polcoeff( serreverse( x * (1 - 3*x) / (1 - 2*x) + x * O(x^n)), n))} /* Michael Somos, Sep 28 2003 */

a(n) = (2^n*(pollegendre(n+1,2)-pollegendre(n-1,2)) + 2*(n==0))/(6*n+3); \\ Michel Marcus, Nov 02 2015

x='x+O('x^100); Vec((1+2*x-sqrt(1-8*x+4*x^2))/(6*x)) \\ Altug Alkan, Nov 02 2015

G.f.: (1+2*x-sqrt(1-8*x+4*x^2))/(6*x). - Emeric Deutsch, Nov 03 2001
a(0)=1; for n>=1, a(n) = Sum_{k=0..n} 3^k*N(n,k) where N(n,k) = (1/n)*binomial(n, k)*binomial(n, k+1) are the Narayana numbers (A001263). - Benoit Cloitre, May 24 2003
a(n) = Sum_{k=0..n} A088617(n, k)*3^k*(-2)^(n-k). - Philippe Deléham, Jan 21 2004
With offset 1: a(1) = 1, a(n) = -2*a(n-1) + 3*Sum_{i=1..n-1} a(i)*a(n-i). - Benoit Cloitre, Mar 16 2004
D-finite with recurrence a(n) = (4*(2n-1)*a(n-1) - 4*(n-2)*a(n-2)) / (n+1) for n>=2, a(0) = a(1) = 1. - Philippe Deléham, Aug 19 2005
From Paul Barry, Dec 15 2008: (Start)
G.f.: 1/(1-x/(1-3x/(1-x/(1-3x/(1-x/(1-3x/(1-x/(1-3x........ (continued fraction).
The g.f. of a(n+1) is 1/(1-4x-3x^2/(1-4x-3x^2/(1-4x-3x^2/(1-4x-3x^2.... (continued fraction). (End)
a(0) = 1, for n>=1, 3a(n) = A047891(n). - Aoife Hennessy (aoife.hennessy(AT)gmail.com), Dec 02 2009
a(n) = upper left term in M^n, M = the production matrix:
1, 1
3, 3, 3
1, 1, 1, 1
3, 3, 3, 3, 3
1, 1, 1, 1, 1, 1
...
- Gary W. Adamson, Jul 08 2011
G.f.: A(x)= (1+2*x-sqrt(1-8*x+4*x^2))/(6*x)= 1/G(0); G(k)= 1 + 2*x - 3*x/G(k+1); (continued fraction, 1-step ). - Sergei N. Gladkovskii, Jan 05 2012
a(n) ~ sqrt(6+4*sqrt(3))*(4+2*sqrt(3))^n/(6*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 07 2012
a(n) = 2^n/sqrt(3)*LegendreP(n,-1,2) for n >= 1, where LegendreP is the associated Legendre function of the first kind, in Maple's notation. - Robert Israel, Mar 24 2015

A107267 A square array of Motzkin related transforms, read by antidiagonals.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 2, 2, 1, 0, 4, 6, 3, 1, 0, 9, 20, 12, 4, 1, 0, 21, 72, 54, 20, 5, 1, 0, 51, 272, 261, 112, 30, 6, 1, 0, 127, 1064, 1323, 672, 200, 42, 7, 1, 0, 323, 4272, 6939, 4224, 1425, 324, 56, 8, 1, 0, 835, 17504, 37341, 27456, 10625, 2664, 490, 72, 9, 1

Offset: 0

Paul Barry, May 15 2005

Rows are transforms of k^n, k>=0, under the matrix A107131. As a number triangle, with T(n,k)=if(k<=n,sum{j=0..n-k, (1/(j+1))C(j+1,n-k-j+1)C(n-k,j)k^j},0), row sums are A107268 and diagonal sums are A107269. Rows are series reversions of x/(1+kx+kx^2), k>=0. Conjecture: rows count weighted Motzkin paths.

Row k counts colored Motzkin paths, where H(1,0) and U(1,1) each have k colors and D(1,-1) one color. - Paul Barry, May 16 2005

			Array begins
  1, 0,  0,   0,    0,     0,      0, ...
  1, 1,  2,   4,    9,    21,     51, ...
  1, 2,  6,  20,   72,   272,   1064, ...
  1, 3, 12,  54,  261,  1323,   6939, ...
  1, 4, 20, 112,  672,  4224,  27456, ...
  1, 5, 30, 200, 1425, 10625,  81875, ...
  1, 6, 42, 324, 2664, 22896, 203256, ...

Seiichi Manyama, Antidiagonals n = 0..139, flattened

Rows n=0..6 give A000007, A001006, A071356, A107264, A003645, A107265, A107266.
Main diagonal gives A292716.
Cf. A000108.

Number array T(n,k) = Sum_{j=0..k} n^j * binomial(k,j) * binomial(j+1,k-j+1)/(j+1).
G.f. of row k: 1/(1 - k*x - k*x^2/(1 - k*x - k*x^2/(1 - k*x - k*x^2/(1 - k*x - k*x^2/(1 - ...))))), a continued fraction. - Ilya Gutkovskiy, Sep 21 2017
From Seiichi Manyama, May 05 2019: (Start)
T(n,k) = Sum_{j=0..floor(k/2)} n^(k-j) * binomial(k,2*j) * binomial(2*j,j)/(j+1) = Sum_{j=0..floor(k/2)} n^(k-j) * binomial(k,2*j) * A000108(j).
(k+2) * T(n,k) = n * (2*k+1) * T(n,k-1) - n * (n-4) * (k-1) * T(n,k-2). (End)

A156016 Expansion of (1-x-sqrt(1-6x-3x^2))/(2x).

Original entry on oeis.org

1, 3, 9, 36, 162, 783, 3969, 20817, 112023, 615033, 3431403, 19398690, 110880900, 639730305, 3720657807, 21790419444, 128398625658, 760668489729, 4528069760691, 27070491820644, 162464919528222, 978463778897637

Offset: 0

Paul Barry, Feb 01 2009

Transform of Catalan numbers by Riordan array ((1+x)/(1-x), x(1+x)/(1-x)^2).

Vincenzo Librandi, Table of n, a(n) for n = 0..200
M. Dziemianczuk, On Directed Lattice Paths With Additional Vertical Steps, arXiv preprint arXiv:1410.5747 [math.CO], 2014.
M. Dziemianczuk, On Directed Lattice Paths With Additional Vertical Steps, Discrete Mathematics, Volume 339, Issue 3, 6 March 2016, Pages 1116-1139.

Cf. A000108, A100239, A107264.

CoefficientList[Series[(1-x-Sqrt[1-6x-3x^2])/(2x),{x,0,30}],x] (* Harvey P. Dale, Jul 27 2014 *)

a(n) = Sum_{k=0..n} Sum_{j=0..k+1} C(k+1,j)*C(n+k-j,n-k-j)*A000108(k).
a(n+1) = 3*A107264(n-1).  - Philippe Deléham, Feb 04 2009
D-finite with recurrence: (n+1)*a(n) + 3*(-2*n+1)*a(n-1) + 3*(-n+2)*a(n-2) = 0. - R. J. Mathar, Dec 03 2014
G.f. A(x) satisfies: A(x) = 1 + x * (1 + A(x) + A(x)^2). - Ilya Gutkovskiy, Jul 01 2020

A100239 G.f. A(x) satisfies: 3^n + 1 = Sum_{k=0..n} [x^k]A(x)^n and also satisfies: (3+z)^n + (1+z)^n - z^n = Sum_{k=0..n} [x^k](A(x) + z*x)^n for all z, where [x^k]A(x)^n denotes the coefficient of x^k in A(x)^n.

Original entry on oeis.org

1, 3, -3, 9, -36, 162, -783, 3969, -20817, 112023, -615033, 3431403, -19398690, 110880900, -639730305, 3720657807, -21790419444, 128398625658, -760668489729, 4528069760691, -27070491820644, 162464919528222, -978463778897637, 5911727071716891, -35821932198013809

Offset: 0

Paul D. Hanna, Nov 30 2004

sign

			From the table of powers of A(x), we see that
3^n+1 = Sum of coefficients [x^0] through [x^n] in A(x)^n:
A^1 = [1,  3], -3,    9,  -36,  162, -783, 3969, -20817, 112023, ...
A^2 = [1,  6,   3],   0,   -9,   54, -297, 1620,  -8910,  49572, ...
A^3 = [1,  9,  18,    0],   0,    0,  -27,  243,  -1701,  10935, ...
A^4 = [1, 12,  42,   36,   -9],   0,    0,    0,    -81,    972, ...
A^5 = [1, 15,  75,  135,   45,  -27],   0,    0,      0,      0, ...
A^6 = [1, 18, 117,  324,  324,    0,  -54],   0,      0,      0, ...
A^7 = [1, 21, 168,  630, 1071,  567, -189,  -81],     0,      0, ...
A^8 = [1, 24, 228, 1080, 2610, 2808,  540, -648,    -81],     0, ...
the main diagonal of which is:
[x^n]A(x)^(n+1) = (n+1)*A057083(n) for n>=0.

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. A057083, A100226, A100239, A107264.

a[n_]:= a[n]= 3^n*Boole[n<2] + 3*(-1)^(n+1)*Sum[Binomial[k+1, n-k-1]*Binomial[n-2,k]*3^k/(k+1), {k,0,n-2}];
Table[a[n], {n,0,40}] (* G. C. Greubel, May 21 2022 *)

a(n)=if(n==0, 1, (3^n+1-sum(k=0, n, polcoeff(sum(j=0, min(k, n-1), a(j)*x^j)^n + x*O(x^k), k)))/n)

a(n)=polcoeff((1+3*x+sqrt(1+6*x-3*x^2+x^2*O(x^n)))/2,n)

def A100239_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( (1+3*x+sqrt(1+6*x-3*x^2))/2 ).list()
A100239_list(40) # G. C. Greubel, May 21 2022

G.f.: A(x) = (1+3*x+sqrt(1+6*x-3*x^2))/2.
Given g.f. A(x), then B(x) = A(x) - (1+2*x) series reversion is -B(-x). - Michael Somos, Sep 07 2005
Given g.f. A(x) and C(x) = g.f. of A025226, then B(x)=A(x)-1-2x satisfies B(x) = x - C(x*B(x)). - Michael Somos, Sep 07 2005
a(n) = 3^n*[n<2] + 3*(-1)^(n+1)*A107264(n-2). - G. C. Greubel, May 21 2022

A120983 Triangle read by rows: T(n,k) is the number of ternary trees with n edges and having k vertices of outdegree 3 (n >= 0, k >= 0).

Original entry on oeis.org

1, 3, 12, 54, 1, 261, 12, 1323, 105, 6939, 810, 3, 37341, 5859, 63, 205011, 40824, 840, 1143801, 277830, 9072, 12, 6466230, 1861380, 86670, 360, 36960300, 12335895, 764478, 6435, 213243435, 81120204, 6377778, 89100, 55, 1240219269, 530408736

Offset: 0

Emeric Deutsch, Jul 21 2006

A ternary tree is a rooted tree in which each vertex has at most three children and each child of a vertex is designated as its left or middle or right child.

			T(3,1)=1 because we have (Q,L,M,R), where Q denotes the root and L (M,R) denotes a left (middle, right) child of Q.
Triangle starts:
     1;
     3;
    12;
    54,   1;
   261,  12;
  1323, 105;
  6939, 810, 3;

Cf. A001764, A107264, A120429, A120981, A120982, A004321.

T:=(n,k)->(1/(n+1))*binomial(n+1,k)*sum(3^j*binomial(n+1-k,j)*binomial(j,n-3*k-j),j=0..n+1-k): for n from 0 to 14 do seq(T(n,k),k=0..floor(n/3)) od; # yields sequence in triangular form

T(n,k) = (1/(n+1))*binomial(n+1,k)*Sum_{j=0..n+1-k} 3^j*binomial(n+1-k, j)*binomial(j, n-3k-j).
G.f.: G = G(t,z) satisfies G = 1 + 3zG + 3z^2*G^2 + tz^3*G^3.
Row n has 1+floor(n/3) terms.
Row sums yield A001764.
T(n,0) = A107264(n).
Sum_{k>=1} k*T(n,k) = binomial(3n, n-3) = A004321(n).

A122871 Expansion of (1 - 2x - sqrt(1 - 4x - 8x^2))/(6x^2).

Original entry on oeis.org

1, 2, 7, 26, 106, 452, 1999, 9074, 42046, 198044, 945430, 4564100, 22243060, 109285256, 540738943, 2692103714, 13475973238, 67784600108, 342439638418, 1736727343436, 8839203054604, 45132514680248, 231121351433158

Offset: 0

Paul Barry, Sep 16 2006

Series reversion of x/(1+2x+3x^2). Binomial transform is A107264. Counts colored Motzkin paths. Second binomial transform of 1,0,3,0,18,0,... or 3^n*binomial(n) (A005159) with interpolated zeros.

Hankel transform is 3^binomial(n+1,2). - Paul Barry, Oct 01 2009

G. C. Greubel, Table of n, a(n) for n = 0..1000 (terms 0..200 from Vincenzo Librandi)
Paul Barry, Characterizations of the Borel triangle and Borel polynomials, arXiv:2001.08799 [math.CO], 2020.
Serkan Demiriz, Adem Şahin, and Sezer Erdem, Some topological and geometric properties of novel generalized Motzkin sequence spaces, Rendiconti Circ. Mat. Palermo Ser. 2 (2025) Vol. 74, No. 136. See p. 4.
Aoife Hennessy, A Study of Riordan Arrays with Applications to Continued Fractions, Orthogonal Polynomials and Lattice Paths, Ph. D. Thesis, Waterford Institute of Technology, Oct. 2011.

CoefficientList[Series[(1-2*x-Sqrt[1-4*x-8*x^2])/(6*x^2), {x, 0, 20}], x] (* Vaclav Kotesovec, Oct 19 2012 *)

x='x+O('x^50); Vec((1 - 2*x - sqrt(1 - 4*x - 8*x^2))/(6*x^2)) \\ G. C. Greubel, Mar 19 2017

def A122871_list(n):  # n>=1
    T = [0]*(n+1); R = [1]
    for m in (1..n-1):
        a,b,c = 1,0,0
        for k in range(m,-1,-1):
            r = a + 2*b + 3*c
            if k < m : T[k+2] = u;
            a,b,c = T[k-1],a,b
            u = r
        T[1] = u; R.append(u)
    return R
A122871_list(23)  # Peter Luschny, Nov 01 2012

E.g.f.: exp(2*x)*Bessel_I(1, sqrt(3)*2*x)/(sqrt(3)x).
a(n) = Sum_{k=0..floor(n/2)} binomial(n,2*k)*binomial(k)3^k*2^(n-2k).
G.f.: 1/(1-2x-3x^2/(1-2x-3x^2/(1-2x-3x^2/(1-2x-3x^2/(1-.... (continued fraction). - Paul Barry, Oct 01 2009
D-finite with recurrence: (n+2)*a(n) - 2*(2n+1)*a(n-1) + 8*(1-n)*a(n-2) = 0. - R. J. Mathar, Nov 14 2011
a(n) ~ 2*sqrt(9+5*sqrt(3))*(2+2*sqrt(3))^n/(3*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 19 2012

A369156 Expansion of (1/x) * Series_Reversion( x / ((1+x)^4-x^4) ).

Original entry on oeis.org

1, 4, 22, 140, 968, 7064, 53544, 417456, 3326304, 26967040, 221733568, 1844667136, 15498804480, 131325820032, 1120928667264, 9628975973120, 83181462291968, 722175844640768, 6297942966129664, 55143987250677760, 484589284705202176, 4272491458636754944

Offset: 0

Seiichi Manyama, Jan 15 2024

nonn

Index entries for reversions of series

Cf. A107264, A369157.

Cf. A317133, A369126.

my(N=30, x='x+O('x^N)); Vec(serreverse(x/((1+x)^4-x^4))/x)

a(n) = sum(k=0, n\4, (-1)^k*binomial(n+1, k)*binomial(4*n-4*k+4, n-4*k))/(n+1);

a(n) = (1/(n+1)) * Sum_{k=0..floor(n/4)} (-1)^k * binomial(n+1,k) * binomial(4*n-4*k+4,n-4*k).

A348189 Pseudo-involutory Riordan companion of 1 + 2xM(x), where M(x) is the g.f. of A001006.

Original entry on oeis.org

1, 0, 0, 2, 0, 6, 8, 24, 60, 148, 396, 1026, 2744, 7350, 19872, 54102, 148104, 407682, 1127328, 3130542, 8726256, 24407634, 68482776, 192698124, 543642476, 1537443024, 4357677516, 12376868254, 35221087656, 100409367690, 286730523104, 820078634232, 2348966799132

Offset: 1

Alexander Burstein, Oct 06 2021

nonn

Alexander Burstein and Louis W. Shapiro, Pseudo-involutions in the Riordan group, arXiv:2112.11595 [math.CO], 2021.
Alexander Burstein and Louis W. Shapiro, Pseudo-involutions in the Riordan group and Chebyshev polynomials, arXiv:2502.13673 [math.CO], 2025.

Cf. A001006, A007971, A086246.

a[n_] := SeriesCoefficient[(1 - Sqrt[1-2*x-3*x^2])/(x * (2 + x - Sqrt[1-2*x-3*x^2])), {x, 0, n}]; Array[a, 33, 0] (* Amiram Eldar, Oct 06 2021 *)

my(x='x+O('x^35)); Vec((1-sqrt(1-2*x-3*x^2))/(x*(2+x-sqrt(1-2*x-3*x^2)))) \\ Michel Marcus, Oct 06 2021

G.f.: A(x) = (1 - sqrt(1 - 2*x - 3*x^2))/(x*(2 + x - sqrt(1 - 2*x - 3*x^2))).
If M(x) is the g.f. of A001006, then A(x) = (1 + 2*x*M(x))/(1 + 2*x + 2*x^2*M(x)).
Let M(x) be the g.f. of A001006 and F(x) = 1 + 2*x*M(x) (equivalently, x*F(x) = g.f. of A007971). Then F(-x*A(x)) = 1/F(x).
A(-x*A(x)) = 1/A(x).
G.f.: Let F(x) be the g.f. of A107264, then A(x) = 1 + 2*x^3*A(x)^2*F(x^2*A(x)). - Alexander Burstein, Feb 14 2022

cp's OEIS Frontend

A201638 Triangle read by rows, T(n,k) for 0<=k<=n, generalizes the colored Motzkin paths of A107264.

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A007564 Shifts left when INVERT transform applied thrice.

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A107267 A square array of Motzkin related transforms, read by antidiagonals.

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A156016 Expansion of (1-x-sqrt(1-6x-3x^2))/(2x).

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A100239 G.f. A(x) satisfies: 3^n + 1 = Sum_{k=0..n} [x^k]A(x)^n and also satisfies: (3+z)^n + (1+z)^n - z^n = Sum_{k=0..n} [x^k](A(x) + z*x)^n for all z, where [x^k]A(x)^n denotes the coefficient of x^k in A(x)^n.

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A120983 Triangle read by rows: T(n,k) is the number of ternary trees with n edges and having k vertices of outdegree 3 (n >= 0, k >= 0).

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A122871 Expansion of (1 - 2*x - sqrt(1 - 4*x - 8*x^2))/(6*x^2).

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A122871 Expansion of (1 - 2x - sqrt(1 - 4x - 8x^2))/(6x^2).

A348189 Pseudo-involutory Riordan companion of 1 + 2xM(x), where M(x) is the g.f. of A001006.