A135148 A binomial recursion: a(n) = q(n) (see formula).

0, 1, 6, 45, 400, 4115, 48146, 631729, 9189972, 146829039, 2556200086, 48167698733, 976792093784, 21211601837803, 491112582793626, 12077021182230057, 314362864408454236, 8635229233659916007, 249631741661080132766, 7575921686807827601701, 240827454421807200901728

Offset

1,3

Links

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

Formula

Let z(1) = x and z(n) = 1 + Sum_{k=1..n-1} ( (2 + binomial(n,k))*z(k)) ), then z(n) = p(n)*x + q(n).

Limit_{n->oo} p(n)/q(n) = (3 - 2*log(2))/(2*log(2) - 1 ) = 4.177398899124179661610768...

a(n) ~ (2*log(2) - 1) * n * n! / (8 * log(2)^(n+2)). - Vaclav Kotesovec, Nov 25 2020

E.g.f.: (1 - exp(x)) * (exp(x) - 2*x - 1) / (2*(2 - exp(x))^2). - Vaclav Kotesovec, Nov 25 2020

Mathematica

z[1] := x; z[n_] := 1 + Sum[(2 + Binomial[n, k])*z[k], {k, 1, n - 1}]; Table[ Coefficient[z[n], x, 0], {n, 1, 20}] (* G. C. Greubel, Sep 28 2016 *)
z[1] := x; z[n_] := z[n] = Expand[1 + Sum[(2 + Binomial[n, k])*z[k], {k, 1, n-1}]]; Table[Coefficient[z[n], x, 0], {n, 1, 30}] (* Vaclav Kotesovec, Nov 25 2020 *)
nmax = 30; Rest[CoefficientList[Series[(1 - E^x)*(E^x - 2*x - 1)/(2*(2 - E^x)^2), {x, 0, nmax}], x] * Range[0, nmax]!] (* Vaclav Kotesovec, Nov 25 2020 *)

Prog

(PARI) r=1; s=2; 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, A135149, A135150, A135074, A135075.

Sequence in context: A243694 A227169 A365184 * A137974 A371541 A291421

Adjacent sequences: A135145 A135146 A135147 * A135149 A135150 A135151

Keyword

nonn

Author

Benoit Cloitre, Nov 20 2007

Status

approved