A132437 A binomial recursion: a(n) = q(n) (see comment).

0, 1, 3, 15, 97, 767, 7175, 77497, 949047, 12993303, 196655437, 3260367539, 58761008087, 1143864229549, 23917992791139, 534642521054391, 12722568903456817, 321112383611040455, 8568150193087139231, 240986045600284560553, 7125677277725450247087

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)

E.g.f.: -2-x + exp(x/2)*((4+Pi)/2 - 2*arcsin(exp(x/2)/sqrt(2))) / sqrt(2-exp(x)).

a(n) ~ (4 - Pi) * n! / (2*sqrt(Pi*n) * log(2)^(n + 1/2)).

a(n) ~ (4 - Pi) * n^n / (sqrt(2) * exp(n) * log(2)^(n + 1/2)). (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, 0], {n, 1, 30}] (* Vaclav Kotesovec, Nov 25 2020 *)
Rest[CoefficientList[Series[-2 - x + E^(x/2)*((4 + Pi)/2 - 2*ArcSin[E^(x/2) / Sqrt[2]]) / Sqrt[2 - E^x], {x, 0, 20}], x] * Range[0, 20]!] (* 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: A060148 A143435 A331325 * A331610 A331618 A128081

Adjacent sequences: A132434 A132435 A132436 * A132438 A132439 A132440

Keyword

nonn

Author

Benoit Cloitre, Nov 20 2007

Status

approved