A397088 E.g.f. A(x) satisfies A'(x) = exp( A(x)*A'(x)^4 ), with A(0)=0.

1, 1, 10, 206, 6520, 279736, 15181296, 997487440, 76992086336, 6829150642240, 684529750280576, 76524539192084096, 9440085602432363520, 1273845857661955460096, 186655067329617703720960, 29514780183694348261738496, 5009445771827759781276491776, 908381310192753648470120243200

Offset

1,3

Comments

In general, for k > 0, if e.g.f. A(x) satisfies A'(x) = exp(A(x)*A'(x)^k), then a(n) ~ (k+1)^(2*n-3) * exp(n/k - 1) * n^(n-2) / (sqrt(k) * (k + exp(1 + 1/k))^(n - 3/2)). - Vaclav Kotesovec, Jun 18 2026

Links

Vaclav Kotesovec, Table of n, a(n) for n = 1..320

Formula

For k >= 0, e.g.f. A(x) satisfies A'(x) = exp( A(x)*A'(x)^k ), with A(0)=0.

Let R(x) = Series_Reversion( (1-(1-k*(k+1)*x) * exp(-(k+1)*x))/(k+1)^2 ).

A(x) = R(x) * exp(-k*R(x)).

For k > 0, with V(x) = LambertW(e^(1/k) * (1-(k+1)^2*x)/k), we have R(x) = (1-k*V(x))/(k*(k+1)) and A(x) = (1-k*V(x))/(k*(k+1)) * exp((k*V(x)-1)/(k+1)).

a(n) ~ 5^(2*n-3) * exp(n/4 - 1) * n^(n-2) / (2 * (4 + exp(5/4))^(n - 3/2)). - Vaclav Kotesovec, Jun 18 2026

Prog

(PARI) my(k=4, N=20, x='x+O('x^N), R=serreverse((1-(1-k*(k+1)*x)*exp(-(k+1)*x))/(k+1)^2)); Vec(serlaplace(R*exp(-k*R)))

Crossrefs

Cf. A104150, A214654, A392206, A392208.

Cf. A376174.

Sequence in context: A027014 A088746 A293970 * A356973 A356960 A364989

Adjacent sequences: A397082 A397083 A397084 * A397089 A397090 A397091

Keyword

nonn,changed

Author

Seiichi Manyama, Jun 16 2026

Status

approved