A219534 -id:A219534 - cp's OEIS frontend

A025227 a(n) = a(1)a(n-1) + a(2)a(n-2) + ... + a(n-1)*a(1) for n >= 3.

0, 1, 2, 4, 12, 40, 144, 544, 2128, 8544, 35008, 145792, 615296, 2625792, 11311616, 49124352, 214838528, 945350144, 4182412288, 18593224704, 83015133184, 372090122240, 1673660915712, 7552262979584, 34178799378432, 155096251351040, 705533929816064

Offset: 0

Clark Kimberling

Series reversion of g.f. A(x) is -A(-x). - Michael Somos, Jul 27 2003
a(n) is the number of royal paths (A006318) from (0,0) to (n-1,n-1) such that every northeast (diagonal) step is either immediately followed by a north step or ends the path. For example a(3)=4 counts EDN, EENN, END, ENEN (E=east, D=diagonal, N=north). - David Callan, Jul 03 2006
From David Callan, Sep 25 2006: (Start)
a(n) is the number of ordered trees with n leaves in which (i) every node (= non-root non-leaf vertex) has at least 2 children and (ii) each leaf is either the leftmost or rightmost child of its parent. For example, a(3)=4 counts
              |
     /\      / \
      /\       /\
and their mirror images. (End)
From William Sit (wyscc(AT)sci.ccny.cuny.edu), Jun 26 2010: (Start)
a(n+1), n >= 0, is also the number of Rota-Baxter words in one idempotent generator x and one operator of arity n.
Alternatively, a(n+1) is the number of ways of adding pairs of parentheses to a string of n x's (the number m of parentheses pairs necessarily satisfies m <= n <= 2m+1 for a nonzero count), such that no two pairs of parentheses are immediately nested and no two x's remain adjacent. (End)
a(n) is the number of colored binary trees on n-1 vertices where leaves have 2 possible colors and internal nodes have 1 color. - Alexander Burstein, Mar 07 2020

			For n=2, a(3) = 4 has the following words: x(x), (x)x, (x(x)), ((x)x) corresponding to A(1,2)=2, and A(2,2)=2. - William Sit (wyscc(AT)sci.ccny.cuny.edu), Jun 26 2010

L. Guo and W. Sit, Enumeration of Rota-Baxter Words (extended abstract), ISSAC 2006 Proceedings, 123-131. [From William Sit (wyscc(AT)sci.ccny.cuny.edu), Jun 26 2010]
L. Guo and W. Sit, Enumeration of Rota-Baxter Words, to appear in Mathematics in Computer Science, Special Issue on AADIOS special session, ACA, 2009. [From William Sit (wyscc(AT)sci.ccny.cuny.edu), Jun 26 2010]

Michael De Vlieger, Table of n, a(n) for n = 0..1470
Paul Barry, Riordan Pseudo-Involutions, Continued Fractions and Somos 4 Sequences, arXiv:1807.05794 [math.CO], 2018.
Alexander Burstein and Louis W. Shapiro, Pseudo-involutions in the Riordan group, arXiv:2112.11595 [math.CO], 2021.
Bérénice Delcroix-Oger and Clément Dupont, Lie-operads and operadic modules from poset cohomology, arXiv:2505.06094 [math.CO], 2025. See p. 22.
Maciej Dziemianczuk, On Directed Lattice Paths With Additional Vertical Steps, arXiv preprint arXiv:1410.5747 [math.CO], 2014.
Li Guo and William Y. Sit, Enumeration and generating functions of Rota-Baxter Words, Math. Comput. Sci. 4 (2010) 313-337.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 655
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 657
Donatella Merlini, Douglas G. Rogers, Renzo Sprugnoli, and M. Cecilia Verri, On some alternative characterizations of Riordan arrays, Canad. J. Math., 49 (1997), 301-320.

Cf. A052709, A052709, A100238.
Cf. A182399, A219534, A000108, A006318.
Cf. A068763.

Table[CatalanNumber[n-1] Hypergeometric2F1[(1-n)/2, -n/2, 3/2-n, -1] + KroneckerDelta[n], {n, 0, 20}] (* Vladimir Reshetnikov, May 17 2016 *)

a(n)=polcoeff((1-sqrt(1-4*x-4*x^2+x*O(x^n)))/2,n)

a(n) = A052709(n) + A052709(n-1).
A100238(n) = -(-1)^n*a(n), for n>1.
a(n) = Sum_{k=0..floor(n/2)} C(n-k-1)*binomial(n-k, k), where C(q)=binomial(2q, q)/(q+1) are the Catalan numbers (A000108). - Emeric Deutsch, Nov 14 2001 [{a(n+1)}A068763.%20-%20_Wolfdieter%20Lang">{n>=0} = row sum of A068763. - _Wolfdieter Lang, Jan 21 2023]
D-finite with recurrence n*a(n) = (4n-6)*a(n-1)+(4n-12)*a(n-2), n>2. a(1)=1, a(2)=2.
G.f. satisfies A(x)-A(x)^2 = x+x^2. - Ralf Stephan, Jun 30 2003
a(n) = Sum_{k=0..n-1} C(k)*C(k+1, n-k-1). - Paul Barry, Feb 23 2005
G.f. A(x) satisfies A(x)=x+C(2x*A(x)) where C(x) is g.f. of Catalan numbers A000108 offset 1. - Michael Somos, Sep 08 2005
G.f.: (1-sqrt(1-4x-4x^2))/2 = 2(x+x^2)/(1+sqrt(1-4x-4x^2)). - Michael Somos, Jun 08 2000
Given an integer t >= 1 and initial values u = [a_0, a_1, ..., a_{t-1}], we may define an infinite sequence Phi(u) by setting a_n = a_{n-1} + a_0*a_{n-1} + a_1*a_{n-2} + ... + a_{n-2}*a_1 for n >= t. For example Phi([1]) is the Catalan numbers A000108. The present sequence is (essentially) Phi([1,2]). - Gary W. Adamson, Oct 27 2008
From William Sit (wyscc(AT)sci.ccny.cuny.edu), Jun 26 2010: (Start)
a(n+1), n >= 0, is column sum for the n-th column of the table R(m,n)=binomial(m+1, n-m)c(m) where c(m) is the m-th Catalan number A000108.
The table entry is nonzero if and only if m <= n <= 2m+1.
R(m,n) gives the number of Rota-Baxter words in one idempotent generator x and one operator of degree m and arity n, or the number of ways of adding m pairs of parentheses to a string of n x's (n necessarily lies between m and 2m+1 inclusive for a nonzero count), such that no two pairs of parentheses are immediately nested and no two x's remain adjacent. (End)
G.f.: A(x) = B(B(x)) where B(x) is the g.f. of A182399. -Paul D. Hanna, Apr 27 2012
G.f.: 1 - x + x*G(0), where G(k) = 1 + 1/(1 - (1+x)/(1 + x/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 01 2013
a(n) ~ (1 + sqrt(2))^(n - 1/2) * 2^(n - 5/4) / (sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Aug 18 2013, simplified Jan 21 2023
O.g.f.: A(x) = x*S(x/(1 + x)), where S(x) = (1 - x - sqrt(1 - 6*x + x^2))/(2*x) is the o.g.f. for the large Schröder numbers A006318. - Peter Bala, Mar 05 2020
G.f.: A(x) satisfies ((A(x) - A(-x))/(2*x))^2 = S(4*x^2), where S(x) is the g.f. for the large Schröder numbers A006318. - Alexander Burstein, May 20 2021
A(x) = (x + x^2)*c(x+x^2), where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the g.f. of the Catalan numbers A000108. Note that (x - x^2)*c(x-x^2) = x. - Peter Bala, Aug 29 2024

A219537 G.f. satisfies A(x) = 1 + x*(A(x)^2 - A(x)^3 + A(x)^4).

Original entry on oeis.org

1, 1, 3, 13, 66, 366, 2148, 13115, 82449, 530095, 3469401, 23037642, 154820262, 1050999343, 7196493255, 49644745965, 344704716018, 2407157839593, 16895247295947, 119121868831235, 843306880720218, 5992060655349521, 42718501097385207, 305476181765843358

Offset: 0

Paul D. Hanna, Nov 21 2012

nonn

a(n) is the number of noncrossing partial matchings on points 1, 2, ... , 3*n where point 1 is unmatched if n>0 and only points congruent modulo 3 can be matched. See Example 57 on p. 47 of the Burstein-Shapiro reference. - Alexander Burstein, Jun 03 2022

			G.f.: A(x) = 1 + x + 3*x^2 + 13*x^3 + 66*x^4 + 366*x^5 + 2148*x^6 +...
Related expansions:
A(x)^2 = 1 + 2*x + 7*x^2 + 32*x^3 + 167*x^4 + 942*x^5 + 5593*x^6 +...
A(x)^3 = 1 + 3*x + 12*x^2 + 58*x^3 + 312*x^4 + 1794*x^5 + 10794*x^6 +...
A(x)^4 = 1 + 4*x + 18*x^2 + 92*x^3 + 511*x^4 + 3000*x^5 + 18316*x^6 +...
A(x)^5 = 1 + 5*x + 25*x^2 + 135*x^3 + 775*x^4 + 4651*x^5 + 28845*x^6 +...
A(x)^6 = 1 + 6*x + 33*x^2 + 188*x^3 + 1116*x^4 + 6852*x^5 + 43204*x^6 +...
where A(x) = 1 + x*(A(x)^2 - A(x)^3 + A(x)^4),
and A(x)^2 = 1 + x*(A(x)^2 + A(x)^5),
and A(x)^3 = 1 + x*(A(x)^2 + A(x)^4 + A(x)^6),
and A(x)^4 = 1 + x*(A(x)^2 + A(x)^4 + A(x)^5 + A(x)^7),
and A(x)^5 = 1 + x*(A(x)^2 + A(x)^4 + A(x)^5 + A(x)^6 + A(x)^8), etc.
The g.f. satisfies A(x) = F(x*A(x)^2) and F(x) = A(x/F(x)^2) where
F(x) = 1 + x + x^2 + 2*x^3 + 4*x^4 + 9*x^5 + 21*x^6 + 51*x^7 +...+ A001006(n-1)*x^n +...
is a g.f. of the Motzkin numbers (A001006, shifted right 1 place).
The g.f. satisfies A(x) = G(x*A(x)) and G(x) = A(x/G(x)) where
G(x) = 1 + x + 2*x^2 + 6*x^3 + 21*x^4 + 80*x^5 + 322*x^6 +...+ A106228(n)*x^n +...
satisfies G(x) = 1 + x*G(x)/(1 - x*G(x)^2).

Seiichi Manyama, Table of n, a(n) for n = 0..1000
Alexander Burstein and Louis W. Shapiro, Pseudo-Involutions in the Riordan Group, J. Integer Sequences 25 (2022), Article 22.3.6.

Cf. A219534, A219535, A219536, A219538, A363573, A366400.

Cf. A001006, A106228, A271469, A364765.

rec := {(36*n^4+126*n^3+126*n^2+36*n)*a(n)+(-276*n^4-1548*n^3-3198*n^2-2898*n-972)*a(n+1)+(940*n^4+7090*n^3+19916*n^2+24650*n+11316)*a(n+2)+(-845*n^4-9000*n^3-34159*n^2-53004*n-26136)*a(n+3)+(-260*n^4-5200*n^3-37454*n^2-116538*n-133128)*a(n+4)+(459*n^4+9774*n^3+77955*n^2+276012*n+366060)*a(n+5)+(-54*n^4-1242*n^3-10686*n^2-40758*n-58140)*a(n+6), a(0) = 1, a(1) = 1, a(2) = 3, a(3) = 13, a(4) = 66, a(5) = 366}:
f:= gfun:-rectoproc(rec,a(n),remember):
map(f, [$0..50]); # Robert Israel, Feb 25 2018

nmax = 23; sol = {a[0] -> 1};
Do[A[x_] = Sum[a[k] x^k, {k, 0, n}] /. sol; eq = CoefficientList[A[x] - (1 + x (A[x]^2 - A[x]^3 + A[x]^4)) + O[x]^(n + 1), x] == 0 /. sol; sol = sol ~Join~ Solve[eq][[1]], {n, 1, nmax}];
sol /. Rule -> Set;
a /@ Range[0, nmax] (* Jean-François Alcover, Nov 01 2019 *)

/* Formula A(x) = 1 + x*(A(x)^2 - A(x)^3 + A(x)^4): */
{a(n)=local(A=1);for(i=1,n,A=1+x*(A^2-A^3+A^4) +x*O(x^n));polcoeff(A,n)}
for(n=0,25,print1(a(n),", "))

/* Formula using Series Reversion involving Motzkin numbers: */
{a(n)=local(A=1);A=(1+x-sqrt(1-2*x-3*x^2+x^3*O(x^n)))/(2*x); polcoeff(sqrt(1/x*serreverse(x/A^2)), n)}
for(n=0,25,print1(a(n),", "))

G.f. A(x) satisfies [from Paul D. Hanna, Mar 21 2016]: (Start)
(1) A(x)^2 = 1 + x*(A(x)^2 + A(x)^5).
(2) A(x)^3 = 1 + x*(A(x)^2 + A(x)^4 + A(x)^6).
Let F(x) = (1+x - sqrt(1 - 2*x - 3*x^2)) / (2*x), then g.f. A(x) satisfies:
(3) A(x) = sqrt( (1/x)*Series_Reversion(x/F(x)^2) ),
(4) A(x) = F(x*A(x)^2) and F(x) = A(x/F(x)^2),
where F(x) = 1 + x*M(x) such that M(x) = 1 + x*M(x) + x^2*M(x)^2 is the g.f. of the Motzkin numbers (A001006).
Let G(x) = 1 + x*G(x)/(1 - x*G(x)^2), then g.f. A(x) satisfies:
(5) A(x) = (1/x)*Series_Reversion(x/G(x)),
(6) A(x) = G(x*A(x)) and G(x) = A(x/G(x)).
where G(x) is the g.f. of A106228. (End)
Recurrence: 3*n*(3*n-1)*(3*n+1)*(5*n-11)*(5*n-8)*(5*n-6)*a(n) = 6*(5*n-11)*(900*n^5 - 3870*n^4 + 6033*n^3 - 4165*n^2 + 1238*n - 120)*a(n-1) - 2*(n-2)*(5*n-1)*(950*n^4 - 5510*n^3 + 11199*n^2 - 9207*n + 2430)*a(n-2) + 6*(n-3)*(n-2)*(2*n-5)*(5*n-6)*(5*n-3)*(5*n-1)*a(n-3). - Vaclav Kotesovec, Aug 19 2013
a(n) ~ sqrt(300+75*10^(2/3)+30*10^(1/3))/90 * (5/9*10^(2/3)+10/9*10^(1/3)+8/3)^n / (sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Aug 19 2013
Recurrence: 18*n*(2*n+1)*(n+2)*(n+1)*a(n)-(6*(n+1))*(n+2)*(46*n^2+120*n+81)*a(n+1)+(2*(n+2))*(470*n^3+2605*n^2+4748*n+2829)*a(n+2)-(n+3)*(845*n^3+6465*n^2+14764*n+8712)*a(n+3)-(2*(n+4))*(130*n^3+2080*n^2+10407*n+16641)*a(n+4)+(3*(n+5))*(153*n^3+2493*n^2+13520*n+24404)*a(n+5)-(6*(n+5))*(3*n+17)*(3*n+19)*(n+6)*a(n+6) = 0. - Robert Israel, Feb 25 2018
G.f. A(x) satisfies: A(-x*A(x)^5) = 1/A(x). - Alexander Burstein, Jun 03 2022
a(n) = (1/n) * Sum_{k=0..n-1} binomial(n,k) * binomial(2*n+k,n-1-k) for n > 0. - Seiichi Manyama, Aug 05 2023
a(n) = (1/n) * Sum_{k=0..n-1} (-1)^k * binomial(n,k) * binomial(4*n-2*k,n-1-k) for n > 0. - Seiichi Manyama, Aug 06 2023
G.f.: A(x) = sqrt(B(x)) where B(x) is the g.f. of A366400. - Seiichi Manyama, Mar 31 2024
a(n) = (1/n) * Sum_{k=0..floor((n-1)/2)} binomial(n,k) * binomial(3*n-k,n-1-2*k) for n > 0. - Seiichi Manyama, Apr 01 2024
a(n) = Sum_{k=0..n} binomial(n,k) * binomial(n+3*k/2+1/2,n)/(2*n+3*k+1). - Seiichi Manyama, Apr 04 2024
G.f.: Sum_{k>=0} binomial(5*k/2 + 1/2, k)*x^k/((5*k + 1)*(1 - x)^((5*k + 1)/2)). - Miles Wilson, Feb 02 2025

A364395 G.f. satisfies A(x) = 1 + x/A(x)*(1 + 1/A(x)^2).

Original entry on oeis.org

1, 2, -8, 60, -552, 5648, -61712, 705104, -8321696, 100658368, -1241281536, 15546987648, -197234640384, 2529169695232, -32728878054144, 426864306146560, -5605439340018176, 74050470138645504, -983432207024885760, 13122261492710033408, -175836387068096147456

Offset: 0

Seiichi Manyama, Jul 22 2023

sign

Cf. A364393, A364397, A364399.

Cf. A219534.

A364395 := proc(n)
    if n = 0 then
        1;
    else
    (-1)^(n-1)*add( binomial(n,k) * binomial(2*n+2*k-2,n-1),k=0..n)/n ;
    end if;
end proc:
seq(A364395(n),n=0..80); # R. J. Mathar, Jul 25 2023
a := n -> `if`(n=0, 1, (-1)^(n+1)*binomial(2*(n-1), n-1)*hypergeom([n-1/2, -n, n], [(n+1)/2, n/2], -1) / n):
seq(simplify(a(n)), n = 0..20);  # Peter Luschny, Mar 03 2024

nmax = 20; A[_] = 1;
Do[A[x_] = 1 + x/A[x]*(1 + 1/A[x]^2) + O[x]^(nmax+1) // Normal, {nmax+1}];
CoefficientList[A[x], x] (* Jean-François Alcover, Mar 03 2024 *)

a(n) = if(n==0, 1, (-1)^(n-1)*sum(k=0, n, binomial(n, k)*binomial(2*n+2*k-2, n-1))/n);

G.f.: A(x) = 1/B(-x) where B(x) is the g.f. of A219534.
a(n) = (-1)^(n-1) * (1/n) * Sum_{k=0..n} binomial(n,k) * binomial(2*n+2*k-2,n-1) for n > 0.
D-finite with recurrence 9*n*(130549*n-619680) *(3*n-1)*(3*n-2)*a(n) +6*(-15361165*n^4 +161422948*n^3 -662268162*n^2 +955427047*n -435307620)*a(n-1) +4*(-908652649*n^4 +9061174176*n^3 -32838390812*n^2 +51018866685*n -28467674946)*a(n-2) -24*(n-3)*(50425637*n^3 -426659887*n^2 +1128823867*n -890225572)*a(n-3) -16*(n-3)*(n-4) *(4607885*n -6704077)*(2*n-9)*a(n-4)=0. - R. J. Mathar, Jul 25 2023
a(n) ~ c*(-1)^(n+1)*4^n*3F2([n-1/2, -n, n], [(n+1)/2, n/2], -1)*n^(-3/2), with c = 1/(4*sqrt(Pi)). - Stefano Spezia, Oct 21 2023
a(n) = (-1)^(n+1)*binomial(2*(n-1), n-1)*hypergeom([n-1/2, -n, n], [(n+1)/2, n/2], -1) / n. - Peter Luschny, Mar 03 2024

A219538 G.f. satisfies A(x) = 1 + xA(x)^2(1 + A(x))^2/2.

Original entry on oeis.org

1, 2, 12, 98, 924, 9468, 102432, 1151410, 13315692, 157406876, 1893480264, 23103024084, 285233168760, 3556744196000, 44730062281800, 566683825859730, 7225564521956940, 92653105887920556, 1194068058333608136, 15457771628663418748, 200916876963088849992

Offset: 0

Paul D. Hanna, Nov 22 2012

nonn

			G.f.: A(x) = 1 + 2*x + 12*x^2 + 98*x^3 + 924*x^4 + 9468*x^5 + 102432*x^6 +...
Related expansions:
A(x)^2 = 1 + 4*x + 28*x^2 + 244*x^3 + 2384*x^4 + 24984*x^5 +...
A(2)^3 = 1 + 6*x + 48*x^2 + 446*x^3 + 4524*x^4 + 48588*x^5 +...
A(2)^4 = 1 + 8*x + 72*x^2 + 712*x^3 + 7504*x^4 + 82704*x^5 +...
where A(x) = 1 + x*(A(x)^2 + 2*A(x)^3 + A(x)^4)/2.
The g.f. satisfies A(x) = F(x*A(x)^2) and F(x) = A(x/F(x)^2) where
F(x) = 1 + 2*x + 4*x^2 + 10*x^3 + 28*x^4 + 84*x^5 + 264*x^6 +...+ 2*A000108(n)*x^n +...
The g.f. satisfies A(x) = G(x*A(x)) and G(x) = A(x/G(x)) where
G(x) = 1 + 2*x + 8*x^2 + 42*x^3 + 252*x^4 + 1636*x^5 +...+ A100327(n)*x^n +...

G. C. Greubel, Table of n, a(n) for n = 0..870

Cf. A068875, A100327.

Cf. A219534, A219537, A371658.

CoefficientList[Sqrt[1/x*InverseSeries[Series[x^3/(1-x-Sqrt[1-4*x])^2, {x, 0, 20}], x]],x] (* Vaclav Kotesovec, Dec 28 2013 *)

/* Formula A(x) = 1 + x*A(x)^2*(1 + A(x))^2/2: */
{a(n)=local(A=1);for(i=1,n,A=1+x*A^2*(1+A +x*O(x^n))^2/2);polcoeff(A,n)}
for(n=0,25,print1(a(n),", "))

/* Formula using Series Reversion involving Catalan numbers: */
{a(n)=local(A=1);A=(1-x-sqrt(1-4*x +x^3*O(x^n)))/x; polcoeff(sqrt(1/x*serreverse(x/A^2)), n)}
for(n=0,25,print1(a(n),", "))

Let F(x) = (1-x - sqrt(1 - 4*x)) / x, then g.f. A(x) satisfies:
(1) A(x) = sqrt( (1/x)*Series_Reversion(x/F(x)^2) ),
(2) A(x) = F(x*A(x)^2) and F(x) = A(x/F(x)^2),
where F(x) = 2*C(x) - 1 such that C(x) = 1 + x*C(x)^2 is the g.f. of the Catalan numbers (A000108).
Let G(x) be the g.f. of A100327, then g.f. A(x) satisfies:
(3) A(x) = (1/x)*Series_Reversion(x/G(x)),
(4) A(x) = G(x*A(x)) and G(x) = A(x/G(x)).
Recurrence: 3*n*(3*n-1)*(3*n+1)*(11*n-14)*a(n) = 3*(2*n-1)*(693*n^3 - 1575*n^2 + 1026*n - 176)*a(n-1) + 2*(n-2)*(2*n-3)*(2*n-1)*(11*n-3)*a(n-2). - Vaclav Kotesovec, Dec 28 2013
a(n) ~ sqrt(242+66*sqrt(33)) * (7+11/9*sqrt(33))^n / (66*sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Dec 28 2013
a(n) = (1/n) * Sum_{k=0..floor((n-1)/2)} 2^(n-2*k) * binomial(n,k) * binomial(3*n-k,n-1-2*k) for n > 0. - Seiichi Manyama, Apr 02 2024

A219535 G.f. satisfies A(x) = 1 + x(2A(x)^2 + A(x)^3).

Original entry on oeis.org

1, 3, 21, 192, 2001, 22539, 267276, 3287496, 41556585, 536565225, 7046232285, 93820316412, 1263673602300, 17186898452772, 235709926636296, 3256050894487824, 45263067114496665, 632721425905230213, 8888476706476318047, 125418490224196533096, 1776734673565844413929

Offset: 0

Paul D. Hanna, Nov 21 2012

nonn

			G.f.: A(x) = 1 + 3*x + 21*x^2 + 192*x^3 + 2001*x^4 + 22539*x^5 +...
Related expansions:
A(x)^2 = 1 + 6*x + 51*x^2 + 510*x^3 + 5595*x^4 + 65148*x^5 +...
A(x)^3 = 1 + 9*x + 90*x^2 + 981*x^3 + 11349*x^4 + 136980*x^5 +...
The g.f. satisfies A(x) = G(x*A(x)) and G(x) = A(x/G(x)) where
G(x) = 1 + 3*x + 12*x^2 + 57*x^3 + 300*x^4 + 1686*x^5 +...+ A047891(n+1)*x^n +...

Michael De Vlieger, Table of n, a(n) for n = 0..799
Elżbieta Liszewska, Wojciech Młotkowski, Some relatives of the Catalan sequence, arXiv:1907.10725 [math.CO], 2019.

Column k=2 of A336575.

Cf. A047891, A219534, A219536.

CoefficientList[1/x*InverseSeries[Series[2*x^2/(1-2*x-Sqrt[1-8*x+4*x^2]), {x, 0, 21}], x],x] (* Vaclav Kotesovec, Dec 28 2013 *)

/* Formula A(x) = 1 + x*(2*A(x)^2 + A(x)^3): */
{a(n)=my(A=1);for(i=1,n,A=1+x*(2*A^2+A^3) +x*O(x^n));polcoeff(A,n)}
for(n=0,25,print1(a(n),", "))

/* Formula using Series Reversion: */
{a(n)=my(A=1,G=(1-2*x-sqrt(1-8*x+4*x^2+x^3*O(x^n)))/(2*x));A=(1/x)*serreverse(x/G);polcoeff(A,n)}
for(n=0,25,print1(a(n),", "))

a(n) = sum(k=0, n, 2^(n-k)*binomial(n, k)*binomial(2*n+k+1, n)/(2*n+k+1)); \\ Seiichi Manyama, Jul 28 2020

a(n) = sum(k=0, n, 2^k*binomial(2*n+1, k)*binomial(3*n-k, n-k))/(2*n+1); \\ Seiichi Manyama, Jul 28 2020

Let G(x) = (1-2*x - sqrt(1 - 8*x + 4*x^2)) / (2*x), then g.f. A(x) satisfies:
(1) A(x) = (1/x)*Series_Reversion(x/G(x)),
(2) A(x) = G(x*A(x)) and G(x) = A(x/G(x)),
where x*G(x) is the g.f. of A047891.
Recurrence: 2*n*(2*n+1)*(11*n - 16)*a(n) = (649*n^3 - 1593*n^2 + 1130*n - 240)*a(n-1) + 16*(n-2)*(2*n-3)*(11*n-5)*a(n-2). - Vaclav Kotesovec, Dec 28 2013
a(n) ~ sqrt((33+17*sqrt(33))/11) * ((59+11*sqrt(33))/8)^n / (4 * sqrt(2*Pi) * n^(3/2)). - Vaclav Kotesovec, Dec 28 2013
From Seiichi Manyama, Jul 28 2020: (Start)
a(n) = Sum_{k=0..n} 2^(n-k) * binomial(n,k) * binomial(2*n+k+1,n)/(2*n+k+1).
a(n) = (1/(2*n+1)) * Sum_{k=0..n} 2^k * binomial(2*n+1,k) * binomial(3*n-k,n-k). (End)
From Seiichi Manyama, Aug 10 2023: (Start)
a(n) = (1/n) * Sum_{k=0..n-1} (-2)^k * 3^(n-k) * binomial(n,k) * binomial(3*n-k,n-1-k) for n > 0.
a(n) = (1/n) * Sum_{k=1..n} 3^k * binomial(n,k) * binomial(2*n,k-1) for n > 0. (End)

A219536 G.f. satisfies A(x) = 1 + x(A(x)^2 + 2A(x)^3).

Original entry on oeis.org

1, 3, 24, 255, 3102, 40854, 566934, 8164263, 120864390, 1827982362, 28122626760, 438720097638, 6923868098820, 110346550539780, 1773394661610258, 28707809007278775, 467677404522668742, 7661583171651546786, 126137791939032756960, 2085923447593966281378

Offset: 0

Paul D. Hanna, Nov 21 2012

			G.f.: A(x) = 1 + 3*x + 24*x^2 + 255*x^3 + 3102*x^4 + 40854*x^5 +...
Related expansions:
A(x)^2 = 1 + 6*x + 57*x^2 + 654*x^3 + 8310*x^4 + 112560*x^5 +...
A(x)^3 = 1 + 9*x + 99*x^2 + 1224*x^3 + 16272*x^4 + 227187*x^5 +...
The g.f. satisfies A(x) = G(x*A(x)) and G(x) = A(x/G(x)) where
G(x) = 1 + 3*x + 15*x^2 + 93*x^3 + 645*x^4 + 4791*x^5 +...+ A103210(n)*x^n +...

G. C. Greubel, Table of n, a(n) for n = 0..790
Elżbieta Liszewska, Wojciech Młotkowski, Some relatives of the Catalan sequence, arXiv:1907.10725 [math.CO], 2019.

Column k=2 of A336574.

Cf. A003168, A103210, A219534, A219535.

CoefficientList[1/x*InverseSeries[Series[4*x^2/(1-x-Sqrt[1-10*x+x^2]), {x, 0, 20}], x],x] (* Vaclav Kotesovec, Dec 28 2013 *)

/* Formula A(x) = 1 + x*(A(x)^2 + 2*A(x)^3): */
{a(n)=my(A=1);for(i=1,n,A=1+x*(A^2+2*A^3) +x*O(x^n));polcoeff(A,n)}
for(n=0,25,print1(a(n),", "))

/* Formula using Series Reversion: */
{a(n)=my(A=1,G=(1-x-sqrt(1-10*x+x^2+x^3*O(x^n)))/(4*x));A=(1/x)*serreverse(x/G);polcoeff(A,n)}
for(n=0,25,print1(a(n),", "))

a(n) = sum(k=0, n, 2^k*binomial(n, k)*binomial(2*n+k+1, n)/(2*n+k+1)); \\ Seiichi Manyama, Jul 26 2020

a(n) = sum(k=0, n, 2^(n-k)*binomial(2*n+1, k)*binomial(3*n-k, n-k))/(2*n+1); \\ Seiichi Manyama, Jul 26 2020

Let G(x) = (1-x - sqrt(1 - 10*x + x^2)) / (4*x), then g.f. A(x) satisfies:
(1) A(x) = (1/x)*Series_Reversion(x/G(x)),
(2) A(x) = G(x*A(x)) and G(x) = A(x/G(x)),
where G(x) is the g.f. of A103210.
Recurrence: 4*n*(2*n+1)*(19*n-26)*a(n) = (2717*n^3 - 6435*n^2 + 4342*n - 840)*a(n-1) + 2*(n-2)*(2*n-3)*(19*n-7)*a(n-2). - Vaclav Kotesovec, Dec 28 2013
a(n) ~ (3/19)^(1/4) * (5+sqrt(57)) * ((143 + 19*sqrt(57))/16)^n / (16*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Dec 28 2013
From Seiichi Manyama, Jul 26 2020: (Start)
a(n) = Sum_{k=0..n} 2^k * binomial(n,k) * binomial(2*n+k+1,n)/(2*n+k+1).
a(n) = (1/(2*n+1)) * Sum_{k=0..n} 2^(n-k) * binomial(2*n+1,k) * binomial(3*n-k,n-k). (End)
From Seiichi Manyama, Aug 10 2023: (Start)
a(n) = (1/n) * Sum_{k=0..n-1} (-1)^k * 3^(n-k) * binomial(n,k) * binomial(3*n-k,n-1-k) for n > 0.
a(n) = (1/n) * Sum_{k=1..n} 3^k * 2^(n-k) * binomial(n,k) * binomial(2*n,k-1) for n > 0. (End)
a(n) = (-1)^(n+1) * (3/n) * Jacobi_P(n-1, 1, n+1, -5) for n >= 1. - Peter Bala, Sep 08 2024

A371693 G.f. satisfies A(x) = ( 1 + x * A(x) * (1 + A(x)) )^2.

Original entry on oeis.org

1, 4, 28, 248, 2480, 26688, 301648, 3531424, 42449088, 520858496, 6497190528, 82146802944, 1050370074624, 13559126110720, 176469550681344, 2313050095245824, 30506619439926272, 404558181197010944, 5391161355764205568, 72156618656648237056, 969557980700415827968

Offset: 0

Seiichi Manyama, Apr 03 2024

nonn

Column k=2 of A378239.
Cf. A006318, A371681.
Cf. A219534.

a(n, r=2, t=2, u=2) = r*sum(k=0, n, binomial(n, k)*binomial(t*n+u*k+r, n)/(t*n+u*k+r));

G.f.: B(x)^2 where B(x) is the g.f. of A219534.

a(n) = Sum_{k=0..n} binomial(n,k) * binomial(2*n+2*k+2,n)/(n+k+1).

A378239 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals downwards, where T(n,0) = 0^n and T(n,k) = k * Sum_{r=0..n} binomial(n,r) * binomial(2n+2r+k,n)/(2n+2r+k) for k > 0.

Original entry on oeis.org

1, 1, 0, 1, 2, 0, 1, 4, 12, 0, 1, 6, 28, 100, 0, 1, 8, 48, 248, 968, 0, 1, 10, 72, 452, 2480, 10208, 0, 1, 12, 100, 720, 4680, 26688, 113792, 0, 1, 14, 132, 1060, 7728, 51504, 301648, 1318832, 0, 1, 16, 168, 1480, 11800, 87104, 591312, 3531424, 15732064, 0

Offset: 0

Seiichi Manyama, Nov 20 2024 based on suggestions from Mikhail Kurkov

			Square array begins:
  1,      1,      1,      1,       1,       1,       1, ...
  0,      2,      4,      6,       8,      10,      12, ...
  0,     12,     28,     48,      72,     100,     132, ...
  0,    100,    248,    452,     720,    1060,    1480, ...
  0,    968,   2480,   4680,    7728,   11800,   17088, ...
  0,  10208,  26688,  51504,   87104,  136352,  202560, ...
  0, 113792, 301648, 591312, 1017184, 1621280, 2454256, ...

Columns k=0..4 give A000007, A219534, A371693, A378155, A378156.

Cf. A033877, A071949, A378236, A378237, A378238, A378240.

T(n, k, t=2, u=2) = if(k==0, 0^n, k*sum(r=0, n, binomial(n, r)*binomial(t*n+u*r+k, n)/(t*n+u*r+k)));
matrix(7, 7, n, k, T(n-1, k-1))

G.f. A_k(x) of column k satisfies A_k(x) = ( 1 + x * A_k(x)^(2/k) * (1 + A_k(x)^(2/k)) )^k for k > 0.
G.f. of column k: B(x)^k where B(x) is the g.f. of A219534.
B(x)^k = B(x)^(k-1) + x * B(x)^(k+1) + x * B(x)^(k+3). So T(n,k) = T(n,k-1) + T(n-1,k+1) + T(n-1,k+3) for n > 0.

cp's OEIS Frontend

A025227 a(n) = a(1)*a(n-1) + a(2)*a(n-2) + ... + a(n-1)*a(1) for n >= 3.

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A219537 G.f. satisfies A(x) = 1 + x*(A(x)^2 - A(x)^3 + A(x)^4).

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A364395 G.f. satisfies A(x) = 1 + x/A(x)*(1 + 1/A(x)^2).

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A219538 G.f. satisfies A(x) = 1 + x*A(x)^2*(1 + A(x))^2/2.

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A219535 G.f. satisfies A(x) = 1 + x*(2*A(x)^2 + A(x)^3).

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A219536 G.f. satisfies A(x) = 1 + x*(A(x)^2 + 2*A(x)^3).

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A371693 G.f. satisfies A(x) = ( 1 + x * A(x) * (1 + A(x)) )^2.

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A025227 a(n) = a(1)a(n-1) + a(2)a(n-2) + ... + a(n-1)*a(1) for n >= 3.

A219538 G.f. satisfies A(x) = 1 + xA(x)^2(1 + A(x))^2/2.

A219535 G.f. satisfies A(x) = 1 + x(2A(x)^2 + A(x)^3).

A219536 G.f. satisfies A(x) = 1 + x(A(x)^2 + 2A(x)^3).

A378239 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals downwards, where T(n,0) = 0^n and T(n,k) = k * Sum_{r=0..n} binomial(n,r) * binomial(2n+2r+k,n)/(2n+2r+k) for k > 0.