A132436 A binomial recursion: a(n) = p(n) (see comment).

1, 1, 4, 20, 129, 1020, 9542, 103063, 1262134, 17279744, 261531315, 4335950346, 78146040374, 1521220672933, 31808447321848, 711019048106744, 16919695824732249, 427046133330613512, 11394750238551713066, 320486422239301377007, 9476411014096567341034

Offset

1,3

Comments

Let z(1) = x and z(n) = 1 + Sum_{k=1..n-1} (-1 + binomial(n,k))*z(k), then z(n) = p(n)*x + q(n).

Links

Vaclav Kotesovec, Table of n, a(n) for n = 1..400

Formula

Limit_{n->oo} p(n)/q(n) = (Pi-2)/(4-Pi) = 1.329896183162743847239353...

From Vaclav Kotesovec, Nov 25 2020: (Start)

a(n) ~ (Pi - 2) * n! / (2*sqrt(Pi*n) * log(2)^(n + 1/2)).

a(n) ~ (Pi - 2) * n^n / (sqrt(2) * exp(n) * log(2)^(n + 1/2)).

E.g.f.: 1 + x + exp(x/2)*(2*arcsin(exp(x/2)/sqrt(2)) - 1 - Pi/2) / sqrt(2 - exp(x)).

(End)

Mathematica

z[1] := x; z[n_] := z[n] = Expand[1 + Sum[(-1 + Binomial[n, k])*z[k], {k, 1, n-1}]]; Table[Coefficient[z[n], x], {n, 1, 30}] (* Vaclav Kotesovec, Nov 25 2020 *)
nmax = 30; Rest[Simplify[CoefficientList[Series[1 + x + E^(x/2)*(2*ArcSin[E^(x/2)/Sqrt[2]] - 1 - Pi/2)/Sqrt[2 - E^x], {x, 0, nmax}], x] * Range[0, nmax]!]] (* Vaclav Kotesovec, Nov 25 2020 *)

Prog

(PARI) r=1; s=-1; v=vector(120, j, x); for(n=2, 120, g=r+sum(k=1, n-1, (s+binomial(n, k))*v[k]); v[n]=g); z(n)=v[n]; p(n)=polcoeff(z(n), 1); q(n)=polcoeff(z(n), 0); a(n)=p(n);

Crossrefs

Cf. A135147, A135148, A135149, A135150, A135074, A135075.

Sequence in context: A196557 A082032 A140585 * A360467 A361533 A307006

Adjacent sequences: A132433 A132434 A132435 * A132437 A132438 A132439

Keyword

nonn

Author

Benoit Cloitre, Nov 20 2007

Status

approved