A214553 -id:A214553 - cp's OEIS frontend

A214377 G.f. A(x) satisfies A(x) = 1 + 4xA(x)^(3/2).

1, 4, 24, 168, 1280, 10296, 86016, 739024, 6488064, 57946200, 524812288, 4808643120, 44493176832, 415146189360, 3901709352960, 36902658748320, 350980432461824, 3354743017001880, 32207616155320320, 310446853795570800, 3003167577200394240, 29146910264615460240

Offset: 0

Paul D. Hanna, Jul 14 2012

nonn

Radius of convergence of g.f. A(x) is r = sqrt(3)/18 where A(r) = 3.

			G.f.: A(x) = 1 + 4*x + 24*x^2 + 168*x^3 + 1280*x^4 + 10296*x^5 + 86016*x^6 + ... where A(x) = 1 + 4*x*A(x)^(3/2).
Radius of convergence: r = 1/(2*3^(3/2)) = 0.09622504486...
Related expansions:
A(x)^(3/2) = 1 + 6*x + 42*x^2 + 320*x^3 + 2574*x^4 + 21504*x^5 + 184756*x^6 + ...
A(x)^(1/2) = 1 + 2*x + 10*x^2 + 64*x^3 + 462*x^4 + 3584*x^5 + 29172*x^6 + ... + A078531(n)*x^n + ...

Bruce C. Berndt, Ramanujan's Notebooks Part I, Springer Verlag, 1985, p. 305.

G. C. Greubel, Table of n, a(n) for n = 0..975
Alois Panholzer, Parking function varieties for combinatorial tree models, arXiv:2007.14676 [math.CO], 2020.

Cf. A078531, A135863, A214553, A098616.

Table[4^n*Binomial[3*n/2, n]*2/(n+2), {n, 0, 20}] (* Vaclav Kotesovec, Oct 20 2015 *)

{a(n)=4^n*binomial(3/2*n,n)/(n/2+1)}
for(n=0,30,print1(a(n),", "))

a(n) = 2^(2*n+1) * binomial(3*n/2, n) / (n+2).
Self-convolution of A078531.
A(-x) = 1/x * series reversion( x*(2*x + sqrt(1 + 4*x^2))^2 ) follows from the Lagrange inversion formula and equation 1.13, p. 305 in Berndt. Cf. A098616. - Peter Bala, Oct 19 2015
a(n) ~ 2^(n + 1/2) * 3^(3*n/2 + 1/2) / (sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Oct 20 2015
G.f.: 4*x*(sin(asin(216*x^2-1)/3)/(6*x)+1/(12*x))^3+1. - Vladimir Kruchinin, Sep 30 2022
From Paul D. Hanna, Feb 03 2023: (Start)
G.f. A(x) satisfies: A(x) = 1/(1 - 4*x*A(x)^(1/2)).
G.f. A(x) satisfies: A(x) = B(x*A(x)) where B(x) = A(x/B(x)) = 1 + 8*x^2 + 4*x*sqrt(1 + 4*x^2) is the g.f. of A135863. (End)
From Karol A. Penson, Mar 23 2024: (Start)
G.f. = 1/(48*z^2) - 2F1([-2/3, -1/3], [-1/2], 108*z^2)/(48*z^2) + 4*z*2F1([5/6, 7/6],[5/2],108*z^2);                                     a(n) = Integral_{x=0..sqrt(108)} x^n*W(x), with W(x) = ((72*(g1(x) - g2(x)) + x^2*(-g1(x) + g2(x)) + 4*sqrt(-3*x^2 + 324)*(g1(x) + g2(x)))*3^(1/6))/(96*Pi*(x^2)^(5/6)),
  where g1(x) = (18 - sqrt(324 - 3*x^2))^(2/3) and
  g2(x) = (18 + sqrt(324 - 3*x^2))^(2/3).
This integral representation is unique as W(x) is the solution of the Hausdorff power moment problem on x = (0, sqrt(108)). Using only the definition of a(n), W(x) can be proven to be positive. W(x) is singular at x = 0, with singularity x^(-1/3), and for x > 0 is monotonically decreasing to zero at x = sqrt(108). For x -> sqrt(108), W'(x) tends to -infinity. (End)
G.f. A(x) satisfies A(x) = 1/A(-x*A(x)^2). - Seiichi Manyama, Jun 17 2025
D-finite with recurrence -(n+2)*(n-1)*a(n) +12*(3*n-2)*(3*n-4)*a(n-2)=0. - R. J. Mathar, Jul 30 2025

A245112 G.f. A(x) satisfies A(x)^2 = 1 + 4xA(x)^5.

Original entry on oeis.org

1, 2, 18, 224, 3230, 50688, 840420, 14483456, 256856886, 4656988160, 85929839996, 1608379269120, 30463651429484, 582796191989760, 11245047027447240, 218581150665277440, 4276257634911525670, 84135742205488791552, 1663738200769421021580, 33047906167191995678720

Offset: 0

Paul D. Hanna, Jul 31 2014

Radius of convergence of g.f. A(x) is r = (3/5)^(5/2) / 6 where A(r) = sqrt(5/3).

			G.f.: A(x) =  = 1 + 2*x + 18*x^2 + 224*x^3 + 3230*x^4 + 50688*x^5 +...
where A(x)^2 = 1 + 4*x*A(x)^5:
A(x)^2 = 1 + 4*x + 40*x^2 + 520*x^3 + 7680*x^4 + 122360*x^5 +...
A(x)^5 = 1 + 10*x + 130*x^2 + 1920*x^3 + 30590*x^4 + 512512*x^5 +...
Related series:
A(x)^4 = 1 + 8*x + 96*x^2 + 1360*x^3 + 21120*x^4 + 347760*x^5 +...
A(x)^8 = 1 + 16*x + 256*x^2 + 4256*x^3 + 73216*x^4 + 1294560*x^5 +...
where A(x) = sqrt(1 + 4*x^2*A(x)^8) + 2*x*A(x)^4.

Gi-Sang Cheon, Sung-Tae Jin, and Louis W. Shapiro, A combinatorial equivalence relation for formal power series, Linear Algebra and its Applications, Volume 491, 15 February 2016, Pages 123-137.

Cf. A000108, A024491, A214553, A245113.

a[n_] := 4^n * Binomial[(5*n - 1)/2, n] / (3*n + 1); Array[a, 20, 0] (* Amiram Eldar, Sep 02 2025 *)

/* From A(x)^2 = 1 + 4*x*A(x)^5 : */
{a(n) = local(A=1+x);for(i=1,n,A=sqrt(1 + 4*x*A^5 +x*O(x^n)));polcoeff(A,n)}
for(n=0,20,print1(a(n),", "))

{a(n) = 4^n * binomial((5*n - 1)/2, n) / (3*n + 1)}
for(n=0,20,print1(a(n),", "))

/* From A(x) = sqrt(1 + 4*x^2*A(x)^8) + 2*x*A(x)^4 : */
{a(n) = local(A=1+x);for(i=1,n,A = sqrt(1 + 4*x^2*A^8 +x*O(x^n)) + 2*x*A^4);polcoeff(A,n)}
for(n=0,20,print1(a(n),", "))

a(n) = 4^n * binomial((5*n - 1)/2, n) / (3*n + 1).
G.f. A(x) satisfies: A(x) = sqrt(1 + 4*x^2*A(x)^8) + 2*x*A(x)^4.
Self convolution yields A214553.
G.f. A(x) = 1/x * series reversion of x*sqrt(1 - 4*x*C(4*x)), where C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the generating function of the Catalan numbers A000108. See A024491. - Peter Bala, Mar 27 2023
D-finite with recurrence 3*n*(n-1)*(3*n-1)*(3*n+1)*a(n) - 20*(5*n-9)*(5*n-3)*(5*n-7)*(5*n-1)*a(n-2) = 0. - R. J. Mathar, Nov 22 2024
G.f. A(x) satisfies A(x) = 1/A(-x*A(x)^8). - Seiichi Manyama, Jun 20 2025
a(n) ~ 2^(n-1/2) * 5^(5*n/2) / (3^(3*n/2+1) * n^(3/2) * sqrt(Pi)). - Amiram Eldar, Sep 02 2025

A214668 G.f. A(x) satisfies A(x) = 1 + 9xA(x)^(4/3).

Original entry on oeis.org

1, 9, 108, 1458, 21060, 318087, 4960116, 79227720, 1289516436, 21308126895, 356506456680, 6027199821864, 102804351279084, 1766931074710515, 30570993847594800, 532022685332573016, 9306598678048938420, 163549467160708850910, 2886019647490699098588

Offset: 0

Paul D. Hanna, Jul 24 2012

Radius of convergence of g.f. A(x) is r = 1/(3*4^(4/3)) where A(r) = 4.

Self-convolution cube of A078532.

			G.f.: A(x) = 1 + 9*x + 108*x^2 + 1458*x^3 + 21060*x^4 + 318087*x^5 + ...
where A(x) = 1 + 9*x*A(x)^(4/3).
Radius of convergence: r = 1/(3*4^(4/3)) = 0.052496710...
Related expansions:
A(x)^(4/3) = 1 + 12*x + 162*x^2 + 2340*x^3 + 35343*x^4 + 551124*x^5 + ... + a(n+1)/9*x^n + ...
A(x)^(1/3) = 1 + 3*x + 27*x^2 + 315*x^3 + 4158*x^4 + 59049*x^5 + 880308*x^6 + 13586859*x^7 + 215233605*x^8 + ... + A078532(n)*x^n + ...

Cf. A078532, A214377, A214553.

a[n_] := 9^n * Binomial[4*n/3, n] / (n/3 + 1); Array[a, 20, 0] (* Amiram Eldar, Sep 02 2025 *)

PARI
```
{a(n)=9^n*binomial(4*n/3, n)/(n/3+1)}
```

{a(n)=local(A=1+x);for(i=1,n,A =1+9*x*(A+x*O(x^n))^(4/3));polcoeff(A,n)}
for(n=0,30,print1(a(n),", "))

a(n) = 9^n * binomial(4*n/3, n) / (n/3 + 1).
From Karol A. Penson, Mar 24 2024: (Start)
G.f. = 4F3([1/4, 1/2, 3/4, 1], [1/3, 2/3, 2], 6912*z^3) + 9*z*3F2([7/12, 5/6, 13/12], [2/3, 7/3], 6912*z^3) + 108*z^2*3F2([11/12, 7/6, 17/12], [4/3, 8/3], 6912*z^3);
a(n) = Integral_{x=0..6912^(1/3)} x^n*W(x), where
 W(x) = h1(x) + h2(x) + h3(x), with
 h1(x) = sqrt(6)*3F2([-3/4, 7/12, 11/12], [1/2, 3/4], x^3/6912)/(18*Pi*x^(1/4)),
 h2(x) = sqrt(x)*3F2([-1/2, 5/6, 7/6], [3/4, 5/4], x^3/6912)/(36*Pi),
 h3(x) = (5*sqrt(6)*x^(5/4)*3F2([-1/4, 13/12, 17/12], [5/4, 3/2], x^3/6912))/(5184*Pi).
This integral representation is unique as W(x) is the solution of the Hausdorff power moment problem on x = (0, 6912^(1/3)). Using only the definition of a(n), W(x) can be proven to be positive. W(x) is singular at x = 0, with singularity x^(-1/4), and for x > 0 is first monotonically decreasing up to a local minimum at x around x = 2, then it is monotonically increasing up to a local maximum at x around x = 10.8, and then finally is monotonically decreasing up to zero at x = 6912^(1/3). For x -> 6912^(1/3), W'(x) tends to -infinity. (End)
G.f. A(x) satisfies A(x) = 1/A(-x*A(x)^(5/3)). - Seiichi Manyama, Jun 18 2025
D-finite with recurrence (n-1)*(n-2)*(n+3)*a(n) - 216*(4*n-9)*(4*n-3)*(2*n-3)*a(n-3) = 0. - R. J. Mathar, Jul 30 2025
a(n) ~ 2^(8*n/3 + 1/2) * 3^(n+1) / (sqrt(Pi) * n^(3/2)). - Amiram Eldar, Sep 02 2025

cp's OEIS Frontend

A214377 G.f. A(x) satisfies A(x) = 1 + 4xA(x)^(3/2).

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A245112 G.f. A(x) satisfies A(x)^2 = 1 + 4xA(x)^5.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

PARI

PARI

Formula

A214668 G.f. A(x) satisfies A(x) = 1 + 9xA(x)^(4/3).

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

PARI

PARI

Formula

cp's OEIS Frontend

A214377 G.f. A(x) satisfies A(x) = 1 + 4*x*A(x)^(3/2).

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A245112 G.f. A(x) satisfies A(x)^2 = 1 + 4*x*A(x)^5.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

PARI

PARI

Formula

A214668 G.f. A(x) satisfies A(x) = 1 + 9*x*A(x)^(4/3).

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

PARI

PARI

Formula

A214377 G.f. A(x) satisfies A(x) = 1 + 4xA(x)^(3/2).

A245112 G.f. A(x) satisfies A(x)^2 = 1 + 4xA(x)^5.

A214668 G.f. A(x) satisfies A(x) = 1 + 9xA(x)^(4/3).