A001764 -id:A001764 - cp's OEIS frontend

A120920 G.f. satisfies: A(x) = G(x)^3 * A(x^4G(x)^9), where G(x) is the g.f. of the number of ternary trees (A001764): G(x) = 1 + xG(x)^3.

1, 3, 12, 55, 276, 1464, 8058, 45543, 262626, 1538607, 9130446, 54761628, 331403447, 2021021082, 12407102937, 76611488305, 475493441604, 2964664310319, 18560063203353, 116621922800283, 735236268006654

Offset: 0

Paul D. Hanna, Jul 17 2006

nonn

Column 0 of triangle A120919 (cascadence of (1+x)^3).

			A(x) = 1 + 3*x + 12*x^2 + 55*x^3 + 276*x^4 + 1464*x^5 + 8058*x^6 +...
= G(x)^3 * A(x^4*G(x)^9) where
G(x) = 1 + x + 3*x^2 + 12*x^3 + 55*x^4 + 273*x^5 + 1428*x^6 +...
is g.f. of A001764: G(x) = 1 + x*G(x)^3.

Cf. A120919, A120921 (cube-root), A120922, A120923; A001764; variants: A092684, A092687, A120895, A120899, A120920.

{a(n)=local(A=1+x,G=(1/x*serreverse(x/(1+3*x+3*x^2+x^3+x*O(x^n))))^(1/3)); for(i=0,n,A=G^3*subst(A,x,x^4*G^9 +x*O(x^n)));polcoeff(A,n,x)}

A380512 Expansion of e.g.f. exp(xG(x)^3) where G(x) = 1 + xG(x)^3 is the g.f. of A001764.

Original entry on oeis.org

1, 1, 7, 91, 1753, 45001, 1447471, 56041987, 2539200721, 131859347473, 7723214721271, 503787793244011, 36223369111466857, 2846582772323685721, 242741539845295265503, 22325483241906758894611, 2202979676409063904473121, 232158319570869255177386017, 26024052774273208806612761191

Offset: 0

Seiichi Manyama, Jan 26 2025

nonn

Cf. A001764, A251569, A380511.
Cf. A000262, A251568, A380516.
Cf. A382033.

a(n) = if(n==0, 1, (n-1)!*pollaguerre(n-1, 2*n+1, -1));

E.g.f.: exp(G(x)-1), where G(x) is described above.
a(n) = (n-1)! * Sum_{k=0..n-1} binomial(3*n,k)/(n-k-1)! for n > 0.
a(n+1) = n! * LaguerreL(n, 2*n+3, -1).
a(n) = (-1)^(n+1)*U(1-n, 2*(1+n), -1), where U is the Tricomi confluent hypergeometric function. - Stefano Spezia, Jan 26 2025
E.g.f.: exp( Series_Reversion( x/(1+x)^3 ) ). - Seiichi Manyama, Mar 15 2025

A382033 E.g.f. A(x) satisfies A(x) = exp(xB(xA(x))^3), where B(x) = 1 + x*B(x)^3 is the g.f. of A001764.

Original entry on oeis.org

1, 1, 7, 109, 2653, 88261, 3731581, 191571493, 11576241769, 804996352873, 63324553740121, 5559962513556001, 539015912053933645, 57188111522488589293, 6591136171961660099509, 820029701725988751533341, 109537705061927547203868241, 15635869913619342121140932689

Offset: 0

Seiichi Manyama, Mar 12 2025

nonn

Cf. A161630, A382032, A382034.

Cf. A001764, A377554.

a(n) = if(n==0, 1, (n-1)!*sum(k=0, n-1, (k+1)^(n-k-1)*binomial(3*n, k)/(n-k-1)!));

a(n) = (n-1)! * Sum_{k=0..n-1} (k+1)^(n-k-1) * binomial(3*n,k)/(n-k-1)! for n > 0.
Let F(x) be the e.g.f. of A377554. F(x) = log(A(x))/x = B(x*A(x))^3.
E.g.f.: A(x) = exp( Series_Reversion( x/(1 + x*exp(x))^3 ) ).

A126042 Expansion of f(x^3)/(1-x*f(x^3)), where f(x) is the g.f. of A001764, whose n-th term is binomial(3n,n)/(2n+1).

Original entry on oeis.org

1, 1, 1, 2, 3, 4, 8, 13, 19, 38, 64, 98, 196, 337, 531, 1062, 1851, 2974, 5948, 10468, 17060, 34120, 60488, 99658, 199316, 355369, 590563, 1181126, 2115577, 3540464, 7080928, 12731141, 21430267, 42860534, 77306428, 130771376, 261542752, 473018396, 803538100

Offset: 0

Paul Barry, Dec 16 2006

Row sums of number triangle A111373.
Interleaves T(3n,2n), T(3n+1,2n+1) and T(3n+2,2n+2) for T(n,k) = A047089(n,k).
One step forward and two steps back: number of nonnegative walks of n steps where the steps are size 1 forwards and size 2 backwards. - David Scambler, Mar 15 2011
Brown's criterion ensures that the sequence is complete (see formulae). - Vladimir M. Zarubin, Aug 05 2019
Number of ordered trees with n+1 edges, having nonroot nodes of outdegree 0 or 3. - Emanuele Munarini, Jun 20 2024

G. C. Greubel, Table of n, a(n) for n = 0..1000
Paul Barry, The Triple Riordan Group, arXiv:2412.05461 [math.CO], 2024. See pp. 7, 10.
Eric Weisstein's World of Mathematics, Brown's Criterion.

Cf. A047089, A111373.

[n lt 3 select 1 else Binomial(n, Floor(n/3)) - (&+[Binomial(n,j): j in [0..Floor(n/3)-1]]): n in [0..40]]; // G. C. Greubel, Jul 30 2022

a:= proc(n) option remember; `if`(n<4, [1$3, 2][n+1], (a(n-1)*
       2*(20*n^4-14*n^3-31*n^2-n+8)-6*(3*n-1)*(5*n-6)*a(n-2)
      +9*(n-2)*(15*n^3-48*n^2+15*n+14)*a(n-3)-54*(n-2)*(n-3)*
      (5*n^2-n-2)*a(n-4))/(2*(2*n+1)*(n+1)*(5*n^2-11*n+4)))
    end:
seq(a(n), n=0..45);  # Alois P. Heinz, Sep 07 2022

Table[Binomial[n, Floor[n/3]] -Sum[Binomial[n,i], {i,0,Floor[n/3] -1}], {n,0,40}] (* David Callan, Oct 26 2017 *)
a[n_] := Binomial[n, Floor[n/3]] (1 + Hypergeometric2F1[1, -n + Floor[n/3], 1 + Floor[n/3], -1]) - 2^n; Table[a[n], {n, 0, 38}] (* Peter Luschny, Jun 20 2024 *)

{a(n)=polcoeff((1/x)*serreverse(x*(1+x)^2/((1+x)^3+x^3+x*O(x^n))),n)}

n=30;
{a0=1;a1=1;a2=1;for(k=1, n/3,print1(a0,", ",a1,", ",a2,", ");
a0=2*a2;a1=2*a0-binomial(3*k,k)/(2*k+1);a2=2*a1-binomial(3*k+1,k)/(k+1))
} \\ Vladimir M. Zarubin, Aug 05 2019

[binomial(n, (n//3)) - sum(binomial(n,j) for j in (0..(n//3)-1)) for n in (0..40)] # G. C. Greubel, Jul 30 2022

a(n) = Sum_{k=0..n} binomial(3*floor((n+2k)/3) - 2k, floor((n+2k)/3)-k)*(k+1)/(2*floor((n+2k)/3) - k + 1)(2*cos(2*Pi*(n-k)/3) + 1)/3.
G.f.: (1/x)*Series_Reversion( x*(1+x)^2/((1+x)^3+x^3) ). - Paul D. Hanna, Mar 15 2011
From Vladimir M. Zarubin, Aug 05 2019: (Start)
a(0) = 1, a(1) = 1, a(2) = 1 and for k>0
a(3*k) = 2*a(3*k-1),
a(3*k+1) = 2*a(3*k) - binomial(3*k,k)/(2*k+1),
a(3*k+2) = 2*a(3*k+1) - binomial(3*k+1,k)/(k+1),
where binomial(3*k,k)/(2*k+1) = A001764(k)
and binomial(3*k+1,k)/(k+1) = A006013(k). (End)
a(n) = (1/(n+1)) * Sum_{k=0..floor(n/3)} (n-3*k+1) * binomial(n+1,k). - Seiichi Manyama, Jan 27 2024

A251569 E.g.f.: exp(xG(x)) where G(x) = 1 + xG(x)^3 is the g.f. of A001764.

Original entry on oeis.org

1, 1, 3, 25, 385, 8661, 255211, 9280573, 401106945, 20075281705, 1141518933811, 72671265032961, 5119905952974913, 395447744211899965, 33224120086567957275, 3016468531370564888101, 294296638636407727046401, 30704676897459478866984273, 3411268107193733242307499235

Offset: 0

Paul D. Hanna, Dec 05 2014

nonn

It appears that a(n) - 1 is divisible by n*(n - 1) for n >= 2. Cf. A251568. - Peter Bala, Feb 15 2015

			E.g.f.: A(x) = 1 + x + 3*x^2/2! + 25*x^3/3! + 385*x^4/4! + 8661*x^5/5! +...
such that A(x) = exp(x*G(x)) where G(x) = 1 + x*G(x)^3:
G(x) = 1 + x + 3*x^2 + 12*x^3 + 55*x^4 + 273*x^5 + 1428*x^6 +...

Cf. A001764, A251568.

Flatten[{1,Table[Sum[n!/k! * Binomial[3*n-2*k-1, n-k] * k/(2*n-k),{k,0,n}],{n,1,20}]}] (* Vaclav Kotesovec, Feb 15 2015 *)

{a(n)=local(G=1);for(i=1,n,G=1+x*G^3 +x*O(x^n));n!*polcoeff(exp(x*G),n)}
for(n=0,20,print1(a(n),", "))

{a(n) = if(n==0,1,sum(k=1,n, n!/k! * binomial(3*n-2*k-1, n-k) * k/(2*n-k) ))}
for(n=0,20,print1(a(n),", "))

a(n) = Sum_{k=0..n} n!/k! * binomial(3*n-2*k-1, n-k) * k/(2*n-k) for n>0 with a(0)=1.

Recurrence: 2*(2*n-1)*(54*n^2 - 171*n + 116)*a(n) = (1458*n^4 - 7533*n^3 + 12474*n^2 - 6624*n - 7)*a(n-1) - (324*n^3 - 1080*n^2 + 759*n + 95)*a(n-2) + 8*(n-2)*(54*n^2 - 63*n - 1)*a(n-3). - Vaclav Kotesovec, Feb 15 2015

A363309 Expansion of g.f. A(x) = F(xF(x)^5), where F(x) = 1 + xF(x)^3 is the g.f. of A001764.

Original entry on oeis.org

1, 1, 8, 67, 590, 5403, 51034, 494268, 4886794, 49153835, 501631980, 5182767291, 54115252508, 570206217940, 6055948422280, 64765311313944, 696876526961130, 7539151412082315, 81957518070961472, 894826829565106185, 9808173152466891270, 107888887505651377475

Offset: 0

Paul D. Hanna, May 29 2023

nonn

Compare the g.f. A(x) = F(x*F(x)^5) to F(-x*F(x)^5) = 1/F(x), where F(x) = 1 + x*F(x)^3 is the g.f. of A001764.

Conjecture: given A(x) = F(x*F(x)^(2*n-1)) where F(x) = 1 + x*F(x)^n, let B(x) = A(x*B(x)^(n-1)), then ((B(x) - 1)/x)^(1/(2*n-1)) is an integer series for n >= 1. Incidentally, the function A(x) = F(x*F(x)^(2*n-1)) is interesting because F(-x*F(x)^(2*n-1)) = 1/F(x) when F(x) = 1 + x*F(x)^n. This sequence illustrates the case for n = 3; for n = 2, see A363308.

			G.f.: A(x) = 1 + x + 8*x^2 + 67*x^3 + 590*x^4 + 5403*x^5 + 51034*x^6 + 494268*x^7 + 4886794*x^8 + 49153835*x^9 + 501631980*x^10 + ...
such that A(x) = F(x*F(x)^5) where F(x) = 1 + x*F(x)^3 begins
F(x) = 1 + x + 3*x^2 + 12*x^3 + 55*x^4 + 273*x^5 + 1428*x^6 + 7752*x^7 + ... + A001764(n)*x^n + ...
RELATED SERIES.
Let B(x) = A(x*B(x)^2) which begins
B(x) = 1 + x + 10*x^2 + 120*x^3 + 1620*x^4 + 23560*x^5 + 360352*x^6 + 5714800*x^7 + 93129840*x^8 + ... + A363310(n)*x^n + ...
then
( (B(x) - 1)/x )^(1/5) = 1 + 2*x + 16*x^2 + 180*x^3 + 2360*x^4 + 33760*x^5 + 510928*x^6 + 8043440*x^7 + ... + A363311(n)*x^n + ...
is an integer series.

Paul D. Hanna, Table of n, a(n) for n = 0..300

Cf. A363308, A363309, A363310, A363311, A363111, A001764.

{a(n) = if(n==0, 1, sum(k=1, n, 5*k* binomial(3*k+1, k) * binomial(3*n+2*k, n-k) / ((3*k+1)*(3*n+2*k)) ) )}
for(n=0, 30, print1(a(n), ", "))

/* G.f. A(x) = F(x*F(x)^5), where F(x) = 1 + x*F(x)^3 */
{a(n) = my(F = 1); for(i=1,n, F = 1 + x*F^3 + x*O(x^n));
polcoeff( subst(F, x, x*F^5), n)}
for(n=0, 30, print1(a(n), ", "))

G.f. A(x) = Sum_{n>=0} a(n)*x^n may be defined as follows; here, F(x) is the g.f. of A001764.
(1) A(x) = F(x*F(x)^5), where F(x) = 1 + x*F(x)^3.
(2) A(x) = B(x/A(x)^2) where B(x) = A(x*B(x)^2) = F( x*B(x)^2 * F(x*B(x)^2)^5 ) is the g.f. of A363310.
(3) a(n) = Sum_{k=1..n} 5*k* binomial(3*k+1, k) * binomial(3*n+2*k, n-k) / ((3*k+1)*(3*n+2*k)) for n > 0, with a(0) = 1.

A380511 Expansion of e.g.f. exp(xG(x)^2) where G(x) = 1 + xG(x)^3 is the g.f. of A001764.

Original entry on oeis.org

1, 1, 5, 55, 961, 23141, 711421, 26631235, 1175535425, 59786520841, 3442729157461, 221413508687471, 15730688410899265, 1223574846548300845, 103417508018836074701, 9437941200860641295611, 924934291227615821904001, 96881241931552168636182545, 10801002623361396194857667365

Offset: 0

Seiichi Manyama, Jan 26 2025

nonn

Cf. A251569, A380512.
Cf. A080893, A380515.
Cf. A082579, A251568, A380514.
Cf. A001764, A006013, A370054, A382058.

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

a(n) = 2 * n! * Sum_{k=0..n-1} binomial(2*n+k,k)/((2*n+k) * (n-k-1)!) for n > 0.
a(n) = U(1-n, 2-3*n, 1), where U is the Tricomi confluent hypergeometric function. - Stefano Spezia, Jan 26 2025
E.g.f.: exp( Series_Reversion( x*(1-x)^2 ) ). - Seiichi Manyama, Mar 15 2025

A143603 Triangle, read by rows, such that the g.f. of column k = G(x)^(2k+1) where G(x) = 1 + x*G(x)^3 is the g.f. of A001764 (ternary trees).

Original entry on oeis.org

1, 1, 1, 3, 3, 1, 12, 12, 5, 1, 55, 55, 25, 7, 1, 273, 273, 130, 42, 9, 1, 1428, 1428, 700, 245, 63, 11, 1, 7752, 7752, 3876, 1428, 408, 88, 13, 1, 43263, 43263, 21945, 8379, 2565, 627, 117, 15, 1, 246675, 246675, 126500, 49588, 15939, 4235, 910, 150, 17, 1

Offset: 1

Paul D. Hanna, Aug 29 2008

From Peter Bala, Aug 07 2014: (Start)
Riordan array (G(x), x*G(x)). Let C(x) = 1 + x + 2*x^2 + 5*x^3 + 14*x^4 + ... be the o.g.f. of the Catalan numbers A000108. Then C(x*G(x)) = G(x).
This leads to a factorization of this array in the group of Riordan matrices as (1, x*G(x))*(C(x), x*C(x)) = (1 + A110616)*A033184 (here, in the final product, 1 refers to the 1 X 1 identity matrix and + means direct sum - see the Example section). (End)

			Triangle begins:
1;
1, 1;
3, 3, 1;
12, 12, 5, 1;
55, 55, 25, 7, 1;
273, 273, 130, 42, 9, 1;
1428, 1428, 700, 245, 63, 11, 1;
7752, 7752, 3876, 1428, 408, 88, 13, 1; ...
where g.f. of column k = G(x)^(2k+1) where G(x) = 1 + x*G(x)^3.
Matrix inverse begins:
1;
-1, 1;
0, -3, 1;
0, 3, -5, 1;
0, -1, 10, -7, 1;
0, 0, -10, 21, -9, 1;
0, 0, 5, -35, 36, -11, 1;
0, 0, -1, 35, -84, 55, -13, 1; ...
where g.f. of column k = (1-x)^(2k+1) for k>=0.
From _Peter Bala_, Aug 07 2014: (Start)
Matrix factorization as (1 + A110616)*A033184 begins
/1           \/ 1         \    / 1           \
|0  1        || 1  1       |   | 1  1        |
|0  1 1      || 2  2 1     | = | 3  3  1     |
|0  3 2 1    || 5  5 3 1   |   |12 12  5 1   |
|0 12 7 3 1  ||14 14 9 4 1 |   |55 55 25 7 1 |
(End)

Yuxuan Zhang and Yan Zhuang, A subfamily of skew Dyck paths related to k-ary trees, arXiv:2306.15778 [math.CO], 2023.

Cf. columns: A001764, A102893, A102594; row sums: A006013. A033184, A110616.

{T(n,k)=binomial(3*n-k,n-k)*(2*k+1)/(2*n+1)}

T(n,k) = C(3n-k,n-k)*(2k+1)/(2n+1) for 0<=k<=n.
Let M = the production matrix:
1, 1
2, 2, 1
3, 3, 2, 1
4, 4, 3, 2, 1
5, 5, 4, 3, 2, 1
...
Top row of M^(n-1) gives n-th row. - Gary W. Adamson, Jul 07 2011

A168478 G.f. satisfies: A(x/A(x)^3) = G(x) where G(x) = 1 + x*G(x)^3 is the g.f. of A001764.

Original entry on oeis.org

1, 1, 6, 60, 803, 13071, 244917, 5101603, 115451307, 2794682082, 71579132742, 1924722618873, 54022011952266, 1575777019075715, 47606721776494443, 1485688929610479498, 47790055655273649449, 1581727833458617151379

Offset: 0

Paul D. Hanna, Dec 06 2009

nonn

			G.f.: A(x) = 1 + x + 6*x^2 + 60*x^3 + 803*x^4 + 13071*x^5 +...
A(x/A(x)^3) = 1 + x + 3*x^2 + 12*x^3 + 55*x^4 + 273*x^5 + 1428*x^6 +...+ A001764(n)*x^n +...

Cf. A168479 (cube), A168448 (variant), A001764.

{a(n)=local(A=1+x, F=sum(k=0, n, binomial(3*k+1, k)/(3*k+1)*x^k)+x*O(x^n)); for(i=0, n, A=subst(F, x, serreverse(x/(A+x*O(x^n))^3))); polcoeff(A, n)}

{a(n)=local(A=1+x);for(i=1,n,A=1+A^3*serreverse(x/(A+x*O(x^n))^3)); polcoeff(A,n)}

G.f. satisfies: A(x) = 1 + A(x)^3*Series_Reversion[x/A(x)^3].
G.f. satisfies: A( (x*(1-x)^2)/A(x*(1-x)^2)^3 ) = 1/(1-x).
G.f. satisfies: A( (x/(1+x)^3)/A(x/(1+x)^3)^3 ) = 1 + x.

A251690 G.f. A(x) satisfies the condition that G(A(x)) is a power series in x consisting entirely of positive integer coefficients such that G(A(x) - x^k) has negative coefficients for k>0, where G(x) = 1 + x*G(x)^3 is the g.f. of A001764.

Original entry on oeis.org

1, -1, -2, -2, 0, -1, 0, -3, 0, -3, -3, 0, -3, -2, -3, -1, -2, 0, -1, -2, 0, 0, -2, 0, 0, 0, -2, 0, -3, 0, -2, 0, -1, 0, -3, -2, -1, -1, -3, -1, 0, -2, -2, -3, -1, -3, -1, -1, 0, 0, -1, -1, -3, -3, -1, 0, -1, 0, -2, 0, -3, -3, -3, -2, -1, -2, -1, -2, -2, -2, -3, -1, -3, -1, -3, -1, 0, -2, -2, -2, -1, -1, -2, -2, 0, -3, -3, -2, -3, -1, -3, -2, 0, 0, 0, -2, -2, -2, -2, -3, -3, 0, -2, 0, -3, -1, 0, -2, -3, -1, -3, 0, -1, 0, -2, -1, -1, -3, -3, -1, -3, -3, 0, -3, -2, -3, -2

Offset: 1

Paul D. Hanna, Dec 31 2014

sign

Compare to the similar series F(x) for the Catalan function C(x) = 1 + x*C(x)^2, where C(F(x)) consists entirely of positive integer coefficients such that C(F(x) - x^k) has negative coefficients for k>0; in which case F(x) = (x+x^2) - (x+x^2)^2, and C(F(x)) = 1/(1-x-x^2).

It seems that a(n) lies in [-3,0] for n>1; does this continue to hold?

			G.f.: A(x) = x - x^2 - 2*x^3 - 2*x^4 - x^6 - 3*x^8 - 3*x^10 - 3*x^11 - 3*x^13 - 2*x^14 - 3*x^15 - x^16 - 2*x^17 - x^19 - 2*x^20 - 2*x^23 - 2*x^27 - 3*x^29 +...
Given G(x) = 1 + x*G(x)^3, which begins
G(x) = 1 + x + 3*x^2 + 12*x^3 + 55*x^4 + 273*x^5 + 1428*x^6 + 7752*x^7 +...
then
G(A(x)) = 1 + x + 2*x^2 + 4*x^3 + 8*x^4 + 17*x^5 + 36*x^6 + 78*x^7 + 169*x^8 + 370*x^9 + 813*x^10 + 1793*x^11 + 3971*x^12 +...+ A251691(n)*x^n +...
consists entirely of positive integer coefficients such that
G(A(x) - x^k) has negative coefficients for k>0, as illustrated by:
k=1: G(A(x) - x) = 1 - x^2 - 2*x^3 + x^4 + 12*x^5 + 11*x^6 - 48*x^7 +...
k=2: G(A(x) - x^2) = 1 + x + x^2 - 2*x^3 - 19*x^4 - 83*x^5 - 267*x^6 +...
k=3: G(A(x) - x^3) = 1 + x + 2*x^2 + 3*x^3 + 2*x^4 - 13*x^5 - 97*x^6 - 471*x^7 +...
k=4: G(A(x) - x^4) = 1 + x + 2*x^2 + 4*x^3 + 7*x^4 + 11*x^5 + 6*x^6 - 58*x^7 - 413*x^8 +...
k=5: G(A(x) - x^5) = 1 + x + 2*x^2 + 4*x^3 + 8*x^4 + 16*x^5 + 30*x^6 + 48*x^7 + 33*x^8 - 215*x^9 - 1632*x^10 +...
k=6: G(A(x) - x^6) = 1 + x + 2*x^2 + 4*x^3 + 8*x^4 + 17*x^5 + 35*x^6 + 72*x^7 + 139*x^8 + 234*x^9 + 228*x^10 - 655*x^11 - 6102*x^12 +...
etc.
The position of the first negative term in G(A(x) - x^n), n>=1, begins:
[2, 3, 5, 7, 9, 11, 13, 16, 18, 20, 23, 25, 28, 30, 32, 35, 37, 40, 42, 45, 47, 50, 52, 55, 57, 59, 62, 64, 67, 69, 72, 74, 77, 79, 82, 84, 87, 89, 92, 94, 97, 99, 102, 104, 106, 109, 111, 114, 116, 119, ...];
does the ratio of the position of the first negative term in G(A(x)-x^n) divided by n approach some constant?

Paul D. Hanna, Table of n, a(n) for n = 1..200

Cf. A251691, A001764.

/* Prints initial N terms: */
N=100;
/* G(x) = 1 + x*G(x)^3 is the g.f. of A001764: */
{G=1+serreverse(x/(1+x +x*O(x^(3*N+10)))^3);}
/* Print terms as you build vector A, then print A at the end: */
{A=[1,-1];print1("1, -1, ");
for(l=1,N,A=concat(A,-4);
for(i=1,4,A[#A]=A[#A]+1;
V=Vec(subst(G,x,x*truncate(Ser(A)) +O(x^floor(3*#A+1)) ));
if((sign(V[3*#A])+1)/2==1, print1(A[#A],", "); break));); A}

cp's OEIS Frontend

A120920 G.f. satisfies: A(x) = G(x)^3 * A(x^4*G(x)^9), where G(x) is the g.f. of the number of ternary trees (A001764): G(x) = 1 + x*G(x)^3.

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A380512 Expansion of e.g.f. exp(x*G(x)^3) where G(x) = 1 + x*G(x)^3 is the g.f. of A001764.

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A382033 E.g.f. A(x) satisfies A(x) = exp(x*B(x*A(x))^3), where B(x) = 1 + x*B(x)^3 is the g.f. of A001764.

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A126042 Expansion of f(x^3)/(1-x*f(x^3)), where f(x) is the g.f. of A001764, whose n-th term is binomial(3n,n)/(2n+1).

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A251569 E.g.f.: exp(x*G(x)) where G(x) = 1 + x*G(x)^3 is the g.f. of A001764.

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A363309 Expansion of g.f. A(x) = F(x*F(x)^5), where F(x) = 1 + x*F(x)^3 is the g.f. of A001764.

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A380511 Expansion of e.g.f. exp(x*G(x)^2) where G(x) = 1 + x*G(x)^3 is the g.f. of A001764.

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A143603 Triangle, read by rows, such that the g.f. of column k = G(x)^(2k+1) where G(x) = 1 + x*G(x)^3 is the g.f. of A001764 (ternary trees).

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A120920 G.f. satisfies: A(x) = G(x)^3 * A(x^4G(x)^9), where G(x) is the g.f. of the number of ternary trees (A001764): G(x) = 1 + xG(x)^3.

A380512 Expansion of e.g.f. exp(xG(x)^3) where G(x) = 1 + xG(x)^3 is the g.f. of A001764.

A382033 E.g.f. A(x) satisfies A(x) = exp(xB(xA(x))^3), where B(x) = 1 + x*B(x)^3 is the g.f. of A001764.

A251569 E.g.f.: exp(xG(x)) where G(x) = 1 + xG(x)^3 is the g.f. of A001764.

A363309 Expansion of g.f. A(x) = F(xF(x)^5), where F(x) = 1 + xF(x)^3 is the g.f. of A001764.

A380511 Expansion of e.g.f. exp(xG(x)^2) where G(x) = 1 + xG(x)^3 is the g.f. of A001764.