A120607 -id:A120607 - cp's OEIS frontend

A120588 G.f. is 1 + x*c(x), where c(x) is the g.f. of the Catalan numbers (A000108).

1, 1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845, 35357670, 129644790, 477638700, 1767263190, 6564120420, 24466267020, 91482563640, 343059613650, 1289904147324, 4861946401452, 18367353072152

Offset: 0

Paul D. Hanna, Jun 16 2006, Jan 24 2008

Previous name was: G.f. satisfies: 3*A(x) = 2 + x + A(x)^2, with A(0) = 1.
This is essentially a duplicate of entry A000108, the Catalan numbers (a(n) = A000108(n-1) for n>0).
In order for the g.f. of an integer sequence to satisfy a functional equation of the form: r*A(x) = c + b*x + A(x)^n, where n > 1, it is necessary that the sequence start with [1, d, m*n*(n-1)/2], where d divides m*n*(n-1)/2 (m>0) and that the coefficients are given by r = n + d^2/m, c = r-1 and b = d^3/m. The remaining terms may then be integer and still satisfy: a_n(k) = r*a(k), where a_n(k) is the k-th term of the n-th self-convolution of the sequence.

			A(x) = 1 + x + x^2 + 2*x^3 + 5*x^4 + 14*x^5 + 42*x^6 + 132*x^7 +...
A(x)^3 = 1 + 2*x + 3*x^2 + 6*x^3 + 15*x^4 + 42*x^5 + 126*x^6 + 396*x^7 +..
More generally, given the functional equation:
r*A(x) = r-1 + b*x + A(x)^n
the series solution is:
A(x) = Sum_{i>=0} C(n*i,i)/(n*i-i+1)*(r-1+bx)^(n*i-i+1)/r^(n*i+1)
which can be expressed as:
A(x) = G( (r-1+bx)^(n-1)/r^n ) * (r-1+bx)/r
where G(x) satisfies: G(x) = 1 + x*G(x)^n .
Also we have:
A(x) = 1 + Series_Reversion[ (1 + r*x - (1+x)^n )/b ].

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

Cf. A000108, A120589 (A(x)^2); A120590 - A120607.

Cf. A001850, A026641.

m:=30; R:=PowerSeriesRing(Rationals(), m); Coefficients(R!( (3 - Sqrt(1-4*x))/2 )); // G. C. Greubel, Feb 18 2019

a[ n_] := SeriesCoefficient[ 1 + (1 - Sqrt[1 - 4 x]) / 2, {x, 0, n}]; (* Michael Somos, May 18 2015 *)

{a(n)=local(A=1+x+x^2+x*O(x^n));for(i=0,n,A=A-3*A+2+x+A^2);polcoeff(A,n)}

{a(n) = my(A); if( n<1, n==0, A = vector(n); A[1] = 1; for( k=2, n, A[k] = sum( j=1, k-1, A[j] * A[k-j])); A[n])} /* Michael Somos, Jul 23 2011 */

((3-sqrt(1-4*x))/2).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Feb 18 2019

G.f.: A(x) = 1 + Series_Reversion(1+3*x - (1+x)^2).
Lagrange Inversion yields g.f.: A(x) = Sum_{n>=0} C(2*n,n)/(n+1)*(2+x)^(n+1)/3^(2*n+1).
G.f.: (3 - sqrt(1-4*x))/2. - Maksym Voznyy (voznyy(AT)mail.ru), Aug 11 2009
a(n) = Sum_{k=1..n-1} a(k) * a(n-k) if n>1. - Michael Somos, Jul 23 2011
G.f.: 2 - G(0), where G(k)= 2*x*(2*k+1) + k +1 - 2*x*(k+1)*(2*k+3)/G(k+1); (continued fraction). - Sergei N. Gladkovskii, Jul 14 2013
G.f.: 2 - G(0), where G(k)= 1 - x/G(k+1) ; (continued fraction). - Sergei N. Gladkovskii, Jul 19 2013
a(n) ~ 2^(2*n-2)/(sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Aug 19 2013
Given g.f. A(x), A001850(n-1) = coefficient of x^n in A(x)^n if n>0, the derivative of log(A(x)) is the g.f. for A026641. - Michael Somos, May 18 2015
A(x) = (1 + 2*Sum_{n >= 1} Catalan(n)*x^n)/(1 + Sum_{n >= 1} Catalan(n)*x^n) = (1 + 3/2*Sum_{n >= 1} binomial(2*n,n)*x^n )/(1 + Sum_{n >= 1} binomial(2*n,n)*x^n). - Peter Bala, Sep 01 2016
D-finite with recurrence n*a(n) +2*(-2*n+3)*a(n-1)=0. - R. J. Mathar, Nov 22 2024

New name by Wolfdieter Lang, Feb 06 2020

A120590 G.f. satisfies: 4*A(x) = 3 + x + A(x)^3, starting with [1,1,3].

Original entry on oeis.org

1, 1, 3, 19, 150, 1326, 12558, 124590, 1278189, 13449205, 144342627, 1573990275, 17389407984, 194228357568, 2189610888840, 24881753664840, 284708154606318, 3277578288381318, 37934510719585350, 441152315040444150

Offset: 0

Paul D. Hanna, Jun 16 2006, Jan 24 2008

nonn

See comments in A120588 for conditions needed for an integer sequence to satisfy a functional equation of the form: r*A(x) = c + b*x + A(x)^n.

			A(x) = 1 + x + 3*x^2 + 19*x^3 + 150*x^4 + 1326*x^5 + 12558*x^6 +...
A(x)^3 = 1 + 3*x + 12*x^2 + 76*x^3 + 600*x^4 + 5304*x^5 + 50232*x^6 +...

Vincenzo Librandi, Table of n, a(n) for n = 0..70
Thomas M. Richardson, The three 'R's and Dual Riordan Arrays, arXiv:1609.01193 [math.CO], 2016.

Cf. A120591 (A(x)^3); A244594, A120588, A120592 - A120607.

FullSimplify[Table[SeriesCoefficient[Sum[Binomial[3*k,k]/(2*k+1)*(3+x)^(2*k+1)/4^(3*k+1),{k,0,Infinity}],{x,0,n}] ,{n,0,20}]] (* Vaclav Kotesovec, Oct 19 2012 *)

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

G.f.: A(x) = 1 + Series_Reversion(1+4*x - (1+x)^3).
G.f.: A(x) = Sum_{n>=0} C(3*n,n)/(2*n+1) * (3+x)^(2*n+1) / 4^(3*n+1), due to Lagrange Inversion.
Recurrence: 13*(n-1)*n*a(n) = 81*(n-1)*(2*n-3)*a(n-1) + 3*(3*n-7)*(3*n-5)*a(n-2). - Vaclav Kotesovec, Oct 19 2012
a(n) ~ sqrt(32-18*sqrt(3))*((81+48*sqrt(3))/13)^n/(12*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 19 2012
G.f.: 4 * sin( arcsin(3 * sqrt(3) * (3 + x) / 16) / 3) / sqrt(3). - Benedict W. J. Irwin, Oct 19 2016

A120592 G.f. satisfies: 5A(x) = 4 + 4x + A(x)^3, starting with [1,2,6].

Original entry on oeis.org

1, 2, 6, 40, 330, 3048, 30156, 312528, 3349170, 36809960, 412651668, 4700098416, 54237852708, 632762593680, 7450815536280, 88435205367456, 1056940049423682, 12708927083800296, 153636691533864900, 1866178021496170800

Offset: 0

Paul D. Hanna, Jun 16 2006, Jan 24 2008

nonn

See comments in A120588 for conditions needed for an integer sequence to satisfy a functional equation of the form: r*A(x) = c + b*x + A(x)^n.

			A(x) = 1 + 2*x + 6*x^2 + 40*x^3 + 330*x^4 + 3048*x^5 + 30156*x^6 +...
A(x)^3 = 1 + 6*x + 30*x^2 + 200*x^3 + 1650*x^4 +15240*x^5 +150780*x^6 +...

Vincenzo Librandi, Table of n, a(n) for n = 0..70

Cf. A120588 - A120591, A120593 - A120607.

FullSimplify[Table[SeriesCoefficient[Sum[Binomial[3*k,k]/(2*k+1)*(4+4*x)^(2*k+1)/5^(3*k+1),{k,0,Infinity}],{x,0,n}],{n,0,20}]] (* Vaclav Kotesovec, Oct 19 2012 *)

{a(n)=local(A=1+2*x+6*x^2+x*O(x^n));for(i=0,n,A=A+(-5*A+4+4*x+A^3)/2);polcoeff(A,n)}

G.f.: A(x) = 1 + Series_Reversion((1+5*x - (1+x)^3)/4).
G.f.: A(x) = Sum_{n>=0} C(3*n,n)/(2*n+1) * (4+4*x)^(2*n+1) / 5^(3*n+1), due to Lagrange Inversion.
Recurrence: 17*(n-1)*n*a(n) = 108*(n-1)*(2*n-3)*a(n-1) + 12*(3*n-7)*(3*n-5)*a(n-2). - Vaclav Kotesovec, Oct 19 2012
a(n) ~ sqrt(250-60*sqrt(15))*((108+30*sqrt(15))/17)^n/(30*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 19 2012

A120593 G.f. satisfies: 5*A(x) = 4 + x + A(x)^4, starting with [1,1,6].

Original entry on oeis.org

1, 1, 6, 76, 1201, 21252, 402892, 8001412, 164321982, 3461110532, 74358814838, 1623152780808, 35897318940028, 802620009567628, 18112759482614328, 412020809942451504, 9437537418826749369, 217486633306640519124

Offset: 0

Paul D. Hanna, Jun 16 2006, Jan 24 2008

nonn

See comments in A120588 for conditions needed for an integer sequence to satisfy a functional equation of the form: r*A(x) = c + b*x + A(x)^n.

			A(x) = 1 + x + 6*x^2 + 76*x^3 + 1201*x^4 + 21252*x^5 +...
A(x)^4 = 1 + 4*x + 30*x^2 + 380*x^3 + 6005*x^4 + 106260*x^5 +...

Cf. A120588 - A120592, A120594 - A120607; A244856.

CoefficientList[1 + InverseSeries[Series[1+5*x - (1+x)^4, {x, 0, 20}], x], x] (* Vaclav Kotesovec, Nov 28 2017 *)

{a(n)=local(A=1+x+6*x^2+x*O(x^n));for(i=0,n,A=A-5*A+4+x+A^4);polcoeff(A,n)}

G.f. satisfies:
(1) A(x) = 1 + Series_Reversion(1+5*x - (1+x)^4).
(2) A(x) = Sum_{n>=0} C(4*n,n)/(3*n+1) * (4+x)^(3*n+1)/5^(4*n+1), by Lagrange Inversion.
(3) A(x) = F(x/A(x)) and F(x) = A(x*F(x)) where F(x) = (4 + F(x)^4)/(5-x)  is the g.f. of A244856. - Paul D. Hanna, Jul 09 2014
a(n) ~ 2^(-7/3 + 3*n) * (-32 + 15*10^(1/3))^(1/2 - n) / (5^(1/3) * n^(3/2) * sqrt(3*Pi)). - Vaclav Kotesovec, Nov 28 2017

A120594 G.f. satisfies: 8A(x) = 7 + 8x + A(x)^4, starting with [1,2,6].

Original entry on oeis.org

1, 2, 6, 44, 394, 3948, 42364, 476120, 5532714, 65935804, 801461012, 9897836520, 123840983812, 1566487308344, 19999112293944, 257365488659376, 3334967582746218, 43477505482249692, 569854228738577572

Offset: 0

Paul D. Hanna, Jun 16 2006

nonn

See comments in A120588 for conditions needed for an integer sequence to satisfy a functional equation of the form: r*A(x) = c + b*x + A(x)^n.

			A(x) = 1 + 2*x + 6*x^2 + 44*x^3 + 394*x^4 + 3948*x^5 + 42364*x^6 +...
A(x)^4 = 1 + 8*x + 48*x^2 + 352*x^3 + 3152*x^4 + 31584*x^5 + 338912*x^6+..

Cf. A120588 - A120593, A120595 - A120607.

CoefficientList[1 + InverseSeries[Series[(1+8*x - (1+x)^4)/8, {x, 0, 20}], x], x] (* Vaclav Kotesovec, Nov 28 2017 *)

{a(n)=local(A=1+2*x+6*x^2+x*O(x^n));for(i=0,n,A=A+(-8*A+7+8*x+A^4)/4);polcoeff(A,n)}

G.f.: A(x) = 1 + Series_Reversion((1+8*x - (1+x)^4)/8). Lagrange Inversion yields: G.f.: A(x) = Sum_{n>=0} C(4*n,n)/(3*n+1) * (7+8*x)^(3*n+1)/8^(4*n+1). - Paul D. Hanna, Jan 24 2008

a(n) ~ 2^(-11/6 + 3*n) * (-7 + 6*2^(1/3))^(1/2 - n) / (n^(3/2) * sqrt(3*Pi)). - Vaclav Kotesovec, Nov 28 2017

A120595 G.f. satisfies: 13A(x) = 12 + 27x + A(x)^4, starting with [1,3,6].

Original entry on oeis.org

1, 3, 6, 36, 249, 1932, 16044, 139500, 1253934, 11558316, 108658902, 1037800920, 10041891132, 98230257636, 969814634424, 9651213968784, 96710160474513, 974967422602428, 9881687141571732, 100632995795535588

Offset: 0

Paul D. Hanna, Jun 16 2006

nonn

See comments in A120588 for conditions needed for an integer sequence to satisfy a functional equation of the form: r*A(x) = c + b*x + A(x)^n.

			A(x) = 1 + 3*x + 6*x^2 + 36*x^3 + 249*x^4 + 1932*x^5 +...
A(x)^4 = 1 + 12*x + 78*x^2 + 468*x^3 + 3237*x^4 + 25116*x^5 +...

Cf. A120588 - A120594, A120596 - A120607; A245043.

CoefficientList[1 + InverseSeries[Series[(1+13*x - (1+x)^4)/27, {x, 0, 20}], x], x] (* Vaclav Kotesovec, Nov 27 2017 *)

{a(n)=local(A=1+3*x+6*x^2+x*O(x^n));for(i=0,n,A=A+(-13*A+12+27*x+A^4)/9);polcoeff(A,n)}

G.f.: A(x) = 1 + Series_Reversion((1+13*x - (1+x)^4)/27).
G.f.: A(x) = Sum_{n>=0} C(4*n,n)/(3*n+1) * (12 + 27*x)^(3*n+1) / 13^(4*n+1). - Paul D. Hanna, Jan 24 2008
a(n) ~ 2^(3*n - 7/3) * 3^(2*n) / (13^(1/3) * sqrt(Pi) * n^(3/2) * (2^(1/3)*13^(4/3) - 32)^(n - 1/2)). - Vaclav Kotesovec, Nov 27 2017

A120589 Self-convolution of A120588, such that a(n) = 3*A120588(n) for n >= 2.

Original entry on oeis.org

1, 2, 3, 6, 15, 42, 126, 396, 1287, 4290, 14586, 50388, 176358, 624036, 2228700, 8023320, 29084535, 106073010, 388934370, 1432916100, 5301789570, 19692361260, 73398801060, 274447690920, 1029178840950, 3869712441972, 14585839204356

Offset: 0

Paul D. Hanna, Jun 16 2006

For n >= 2, a(n) equals 2^(2n+1) times the coefficient of Pi in 2F1([3/2, n+1], [5/2], -1). - John M. Campbell, Jul 17 2011

			A(x) = 1 + 2*x + 3*x^2 + 6*x^3 + 15*x^4 + 42*x^5 + 126*x^6 + 396*x^7 + ...
A(x)^(1/2) = 1 + x + x^2 + 2*x^3 + 5*x^4 + 14*x^5 + 42*x^6 + 132*x^7 + ...

G. C. Greubel, Table of n, a(n) for n = 0..1000
Guo-Niu Han, Enumeration of Standard Puzzles, 2011. [Cached copy]
Guo-Niu Han, Enumeration of Standard Puzzles, arXiv:2006.14070 [math.CO], 2020.
S. Kitaev, J. Remmel and M. Tiefenbruck, Marked mesh patterns in 132-avoiding permutations I, arXiv:1201.6243 [math.CO], 2012. - From _N. J. A. Sloane_, May 09 2012
Sergey Kitaev, Jeffrey Remmel, and Mark Tiefenbruck, Quadrant Marked Mesh Patterns in 132-Avoiding Permutations II, arXiv:1302.2274 [math.CO], 2013.
Sergey Kitaev, Jeffrey Remmel, Mark Tiefenbruck, Quadrant Marked Mesh Patterns in 132-Avoiding Permutations II, Electronic Journal of Combinatorial Number Theory, 15 (2015), #A16.

Cf. A120588 (A(x)^(1/2)); A120590-A120607.

m:=30; R:=PowerSeriesRing(Rationals(), m); Coefficients(R!( (5-2*x-3*Sqrt(1-4*x))/2 )); // G. C. Greubel, Feb 18 2019

A120589List := proc(m) local A, P, n; A := [1,2,3]; P := [3];
for n from 1 to m - 2 do P := ListTools:-PartialSums([op(P), P[-1]]);
A := [op(A), P[-1]] od; A end: A120589List(26); # Peter Luschny, Mar 26 2022

Join[{1,2,3}, Table[3*(2*n)!/n!/(n+1)!, {n,2,40}]]
CoefficientList[Series[(5-2x -3Sqrt[1-4x])/2, {x, 0, 30}], x] (* G. C. Greubel, Feb 18 2019 *)

{a(n)=local(A=1+x+x^2+x*O(x^n));for(i=0,n,A=A-3*A+2+x+A^2);polcoeff(A^2,n)}

my(x='x+O('x^30)); Vec((5-2*x-3*sqrt(1-4*x))/2) \\ G. C. Greubel, Feb 18 2019

((5-2*x-3*sqrt(1-4*x))/2).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Feb 18 2019

a(n) = 3*A000108(n-1) for n >= 2, where A000108 are the Catalan numbers.
G.f.: (5 - 2*x - 3*sqrt(1-4*x))/2. - G. C. Greubel, Feb 18 2019
a(n) ~ 3 * 2^(2*n-2) / (sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Aug 20 2025

A120591 Self-convolution cube of A120590, such that a(n) = 4*A120590(n) for n>=2.

Original entry on oeis.org

1, 3, 12, 76, 600, 5304, 50232, 498360, 5112756, 53796820, 577370508, 6295961100, 69557631936, 776913430272, 8758443555360, 99527014659360, 1138832618425272, 13110313153525272, 151738042878341400, 1764609260161776600

Offset: 0

Paul D. Hanna, Jun 16 2006

nonn

			A(x) = 1 + 3*x + 12*x^2 + 76*x^3 + 600*x^4 + 5304*x^5 + 50232*x^6 +...
A(x)^(1/3) = 1 + x + 3*x^2 + 19*x^3 + 150*x^4 + 1326*x^5 + 12558*x^6 +...

Cf. A120590 (A(x)^(1/3)); A120588, A120592 - A120607.

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

A120596 G.f. satisfies: 6*A(x) = 5 + x + A(x)^5, starting with [1,1,10].

Original entry on oeis.org

1, 1, 10, 210, 5505, 161601, 5082420, 167451780, 5705082795, 199354509755, 7105393162010, 257312347583330, 9440808323869455, 350189693739455535, 13110655796699158800, 494772468434359266960, 18801468275832345890970

Offset: 0

Paul D. Hanna, Jun 16 2006

nonn

See comments in A120588 for conditions needed for an integer sequence to satisfy a functional equation of the form: r*A(x) = c + b*x + A(x)^n.

			A(x) = 1 + x + 10*x^2 + 210*x^3 + 5505*x^4 + 161601*x^5 +...
A(x)^5 = 1 + 5*x + 60*x^2 + 1260*x^3 + 33030*x^4 + 969606*x^5 +...

Cf. A120588 - A120595, A120597 - A120607.

CoefficientList[1 + InverseSeries[Series[1+6*x - (1+x)^5, {x, 0, 20}], x], x] (* Vaclav Kotesovec, Nov 28 2017 *)

{a(n)=local(A=1+x+10*x^2+x*O(x^n));for(i=0,n,A=A-6*A+5+x+A^5);polcoeff(A,n)}

G.f.: A(x) = 1 + Series_Reversion(1+6*x - (1+x)^5). Lagrange Inversion yields: G.f.: A(x) = Sum_{n>=0} C(5*n,n)/(4*n+1) * (5+x)^(4*n+1)/6^(5*n+1). - Paul D. Hanna, Jan 24 2008

a(n) ~ (-5 + 4*(6/5)^(5/4))^(1/2 - n) / (2^(15/8) * 3^(3/8) * 5^(1/8) * n^(3/2) * sqrt(Pi)). - Vaclav Kotesovec, Nov 28 2017

A120597 G.f. satisfies: 9A(x) = 8 + 8x + A(x)^5, starting with [1,2,10].

Original entry on oeis.org

1, 2, 10, 120, 1770, 29208, 516180, 9554640, 182867970, 3589443160, 71861735660, 1461730482160, 30123451315620, 627598216410480, 13197173403868200, 279728425129963680, 5970277970921643570, 128199003794219752920

Offset: 0

Paul D. Hanna, Jun 16 2006

nonn

See comments in A120588 for conditions needed for an integer sequence to satisfy a functional equation of the form: r*A(x) = c + b*x + A(x)^n.

			A(x) = 1 + 2*x + 10*x^2 + 120*x^3 + 1770*x^4 + 29208*x^5 +...
A(x)^5 = 1 + 10*x + 90*x^2 + 1080*x^3 + 15930*x^4 + 262872*x^5 +...

Cf. A120588 - A120596, A120598 - A120607.

CoefficientList[1 + InverseSeries[Series[(1+9*x - (1+x)^5)/8, {x, 0, 20}], x], x] (* Vaclav Kotesovec, Nov 28 2017 *)

{a(n)=local(A=1+2*x+10*x^2+x*O(x^n));for(i=0,n,A=A+(-9*A+8+8*x+A^5)/4);polcoeff(A,n)}

G.f.: A(x) = 1 + Series_Reversion((1+9*x - (1+x)^5)/8). Lagrange Inversion yields: G.f.: A(x) = Sum_{n>=0} C(5*n,n)/(4*n+1) * (8+8*x)^(4*n+1)/9^(5*n+1). - Paul D. Hanna, Jan 24 2008

a(n) ~ (-1 + 9*sqrt(3)/(10*5^(1/4)))^(1/2 - n) / (3^(3/4) * 5^(1/8) * n^(3/2) * sqrt(Pi)). - Vaclav Kotesovec, Nov 28 2017

cp's OEIS Frontend

A120588 G.f. is 1 + x*c(x), where c(x) is the g.f. of the Catalan numbers (A000108).

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A120590 G.f. satisfies: 4*A(x) = 3 + x + A(x)^3, starting with [1,1,3].

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A120592 G.f. satisfies: 5*A(x) = 4 + 4*x + A(x)^3, starting with [1,2,6].

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A120593 G.f. satisfies: 5*A(x) = 4 + x + A(x)^4, starting with [1,1,6].

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A120594 G.f. satisfies: 8*A(x) = 7 + 8*x + A(x)^4, starting with [1,2,6].

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A120595 G.f. satisfies: 13*A(x) = 12 + 27*x + A(x)^4, starting with [1,3,6].

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A120589 Self-convolution of A120588, such that a(n) = 3*A120588(n) for n >= 2.

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A120592 G.f. satisfies: 5A(x) = 4 + 4x + A(x)^3, starting with [1,2,6].

A120594 G.f. satisfies: 8A(x) = 7 + 8x + A(x)^4, starting with [1,2,6].

A120595 G.f. satisfies: 13A(x) = 12 + 27x + A(x)^4, starting with [1,3,6].

A120597 G.f. satisfies: 9A(x) = 8 + 8x + A(x)^5, starting with [1,2,10].