A384266 G.f. A(x) = (3*sqrt(1 - 8*x) - (1 - 4*x)) / (2*(1 - 8*x - 2*x^2)).

1, 4, 22, 136, 892, 6064, 42232, 299296, 2149360, 15596992, 114138592, 841108096, 6234779584, 46448349952, 347541337984, 2610319254016, 19671552622336, 148689857920000, 1126905157115392, 8561360256526336, 65185363066289152, 497307750242234368, 3800975843189291008, 29100188150365757440

Offset: 0

Author

Paul D. Hanna, Jun 06 2025

Keywords

• nonn

Comments

Compare formula (2.a) to C(x) = exp( x*C(x) + Integral C(x) dx ), where C(x) = 1 + x*C(x)^2 is the g.f. of the Catalan numbers (A000108).

Examples

			G.f.: A(x) = 1 + 4*x + 22*x^2 + 136*x^3 + 892*x^4 + 6064*x^5 + 42232*x^6 + 299296*x^7 + 2149360*x^8 + 15596992*x^9 + ...
where log(A(x)) = x*A(x) + Integral A(x) + 2*A(x)^2 dx,
also, A'(x)/A(x) = 2 * A(x) * (1 + A(x)) / (1 - x*A(x)).
SPECIFIC VALUE.
Let z be a shared zero of both (1 - 8*x - 2*x^2) and (3*sqrt(1 - 8*x) - (1 - 4*x)), where z = (3*sqrt(2) - 4)/2 = 0.1213203435..., then A(z) = 2*(3 + 2*sqrt(2))/3 = sqrt(2)/(3*z) = 3.8856180831... (=2*A179587).
		

Links

• Paul D. Hanna, Table of n, a(n) for n = 0..500

Crossrefs

Cf. A179587.

Programs

Mathematica

CoefficientList[Series[(3*Sqrt[1 - 8*x] - (1 - 4*x)) / (2*(1 - 8*x - 2*x^2)), {x, 0, 30}], x] (* Vaclav Kotesovec, Jun 07 2025 *)
CoefficientList[Series[4/(1 + 3*Sqrt[1 - 8*x] - 4*x), {x,0,30}],x] (* Vaclav Kotesovec, Jun 07 2025 *)

PARI

{a(n) = my(A = (3*sqrt(1 - 8*x +x*O(x^n)) - (1 - 4*x)) / (2*(1 - 8*x - 2*x^2)) );
polcoef(A,n)}
for(n=0,30,print1(a(n),", "))

Formula

G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies the following formulas.

(1) A(x) = (3*sqrt(1 - 8*x) - (1 - 4*x)) / (2*(1 - 8*x - 2*x^2)).

(2.a) A(x) = exp( x*A(x) + Integral A(x) + 2*A(x)^2 dx ).

(2.b) A(x) = exp( Integral 2*A(x)*(1 + A(x))/(1 - x*A(x)) dx ).

(2.c) A'(x) = 2 * A(x)^2 * (1 + A(x)) / (1 - x*A(x)).

(3.a) A(x) = (3*sqrt(1 + 8*x^2*A(x)^2) - (1 - 8*x*A(x))) / 2.

(3.b) x/Series_Reversion(x*A(x)) = (3*sqrt(1 + 8*x^2) - (1 - 8*x)) / 2.

(4.a) [x^(2*n+1)] 1/A(x)^(2*n) = 0 for n >= 0.

(4.b) [x^(2*n)] 1/A(x)^(2*n) = [x^(2*n-1)] -1/A(x)^(2*n-1) for n >= 1.

From Vaclav Kotesovec, Jun 07 2025: (Start)

G.f.: 4/(1 + 3*sqrt(1 - 8*x) - 4*x).

Recurrence: n*a(n) = 4*(4*n-3)*a(n-1) - 2*(31*n-48)*a(n-2) - 8*(2*n-3)*a(n-3).

a(n) ~ 3 * 2^(3*n+3) / (sqrt(Pi)*n^(3/2)). (End)