A376174 E.g.f. A(x) satisfies: A'(x) = 1 + A(x)*A'(x)^4.

1, 1, 9, 165, 4629, 175689, 8424801, 488756205, 33292495341, 2605108910481, 230300167685049, 22701604019859765, 2468971586334241989, 293687751861227612889, 37930838135497768301841, 5286141242516127169100925, 790686768925097978507354781, 126349187477795223746580576801

Offset: 1

Author

Paul D. Hanna, Sep 14 2024

Keywords

• nonn

Comments

In general, if k > 1 and e.g.f. A(x) satisfies: A'(x) = 1 + A(x)*A'(x)^k, then a(n) ~ k^(k/2 + 1) * (k^(k+1)*(k+1))^(n - 3/2) * n^(n-2) / ((k-1)^((k+3)/2) * ((k-1)^(k+1) + k^k)^(n - 3/2) * exp(n)). - Vaclav Kotesovec, Sep 17 2024

Examples

			E.g.f.: A(x) = x + x^2/2! + 9*x^3/3! + 165*x^4/4! + 4629*x^5/5! + 175689*x^6/6! + 8424801*x^7/7! + 488756205*x^8/8! + 33292495341*x^9/9! + ...
where A'(x) = 1 + A(x)*A'(x)^4.
Also,
A'(x) = 1 + A(x) + 4*A(x)^2 + 22*A(x)^3 + 140*A(x)^4 + 969*A(x)^5 + 7084*A(x)^6 + 53820*A(x)^7 + ... + A002293(n)*A(x)^n + ...
RELATED SERIES.
Series D(x) = 1 + x*D(x)^4 begins
D(x) = 1 + x + 4*x^2 + 22*x^3 + 140*x^4 + 969*x^5 + 7084*x^6 + 53820*x^7 + ... + A002293(n)*x^n + ...
where Integral( dx/D(x) ) = x - x^2/2! - 6*x^3/3! - 90*x^4/4! - 2184*x^5/5! - 73440*x^6/6! - 3160080*x^7/7! - 165765600*x^8/8! + ...
and A(x) = Series_Reversion( Integral( dx/D(x) ) ).
A'(x)^4 = 1 + 4*x + 48*x^2/2! + 1008*x^3/3! + 30672*x^4/4! + 1229616*x^5/5! + 61348752*x^6/6! + 3668121072*x^7/7! + ...
		

Links

• Paul D. Hanna, Table of n, a(n) for n = 1..300

Crossrefs

Cf. A180254, A376175, A002293.

Programs

Mathematica

nmax = 20; B[] = 1; Do[B[x] = 1 + Integrate[B[x], x]*B[x]^4 + O[x]^nmax // Normal, nmax]; A[x_] = Integrate[B[x], x]; CoefficientList[A[x]/x, x] * Range[nmax]!  (* Vaclav Kotesovec, Sep 17 2024 *)

PARI

/* A'(x) = 1 + A(x)*A'(x)^4 */
{a(n) = my(A=x); for(i=1, n, A = intformal(1 + A*(A')^4 + x*O(x^n))); n!*polcoeff(A, n)}
for(n=1, 20, print1(a(n)", "))

PARI

/* A(x) = Series_Reversion( Integral( dx/D ) ) where D = 1 + x*D^4 */
{a(n) = my(D = ((1/x)*serreverse(x/(1 + x + x*O(x^n))^4))^(1/4), A); A = serreverse(intformal(1/D)); n!*polcoeff(A, n)}
for(n=1, 20, print1(a(n)", "))

Formula

E.g.f. A(x) = Sum_{n>=1} a(n)*x^n/n! satisfies:

(1) A(x) = Series_Reversion( Integral( dx/D(x) ) ) where D(x) = 1 + x*D(x)^4 is the g.f. of A002293.

(2) A'(x) = Sum_{n>=0} A002293(n)*A(x)^n where A002293(n) = binomial(4*n,n)/(3*n+1).

(3) A'(x) = 1 + A(x)*A'(x)^4.

(4) A'(x) = 1/(1 - A(x)*A'(x)^3).

(5) A(x) = x + Sum_{n>=0} d^n/dx^n G(x)^(n+1)/(n+1)!, where G(x) = Integral( x*D(x)^3 dx ) and D(x) = 1 + x*D(x)^4.

(6) A(x - B(x)) = x where B(x) = Sum_{n>=0} binomial(4*n+2,n) * x^(n+2) / ((n+1)*(n+2)).

a(n) ~ 2^(10*n-9) * 5^(n - 3/2) * n^(n-2) / (3^(7/2) * exp(n) * 499^(n - 3/2)). - Vaclav Kotesovec, Sep 17 2024