A120590 -id:A120590 - cp's OEIS frontend

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

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)}

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

Original entry on oeis.org

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

A244594 G.f. satisfies: A(x) = (3 + A(x)^3) / (4 - x).

Original entry on oeis.org

1, 1, 4, 29, 263, 2672, 29088, 331749, 3912660, 47329811, 583983656, 7321173872, 92990672635, 1194113490556, 15476763809428, 202197552311829, 2659975668005367, 35205831900984144, 468468683002725372, 6263539340729569047, 84103985900174324256, 1133671250214654009000

Offset: 0

Paul D. Hanna, Jul 01 2014

nonn

			G.f.: A(x) = 1 + x + 4*x^2 + 29*x^3 + 263*x^4 + 2672*x^5 + 29088*x^6 +...
Compare
(4 - x)*A(x) = 4 + 3*x + 15*x^2 + 112*x^3 + 1023*x^4 + 10425*x^5 +...
to:
A(x)^3 = 1 + 3*x + 15*x^2 + 112*x^3 + 1023*x^4 + 10425*x^5 + 113680*x^6 +...

Vaclav Kotesovec, Table of n, a(n) for n = 0..400

Cf. A120590, A001764.

{a(n)=polcoeff(1+serreverse((1+4*x - (1+x)^3)/(1 + x +x*O(x^n))),n)}
for(n=0,30,print1(a(n),", "))

{a(n)=local(A=[1,1],Ax=1+x);for(i=1,n,A=concat(A,0);Ax=Ser(A);A[#A]=Vec( Ax^3 - (4-x)*Ax )[#A]);A[n+1]}
for(n=0,30,print1(a(n),", "))

G.f. satisfies:
(1) A(x) = 1 + Series_Reversion( (1+4*x - (1+x)^3)/(1 + x) ).
(2) A(x) = Sum_{n>=0} C(3*n,n)/(2*n+1) * (3 + x*A(x))^(2*n+1) / 4^(3*n+1).
(3) A(x) = G(x*A(x)) and G(x) = A(x/G(x)) where G(x) = (3+x + G(x)^3)/4  is the g.f. of A120590.
Recurrence: 13*(n-1)*n*a(n) = 96*(n-1)*(2*n-3)*a(n-1) - 8*(6*n^2 - 24*n + 23)*a(n-2) + 2*(n-2)*(2*n-7)*a(n-3). - Vaclav Kotesovec, Jul 03 2014
a(n) ~ sqrt(48-18^(4/3)) * ((24*18^(1/3)+9*18^(2/3)+64)/13)^n / (12*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Jul 03 2014

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

nonn

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

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

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

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A244594 G.f. satisfies: A(x) = (3 + A(x)^3) / (4 - x).

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

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