A135147 A binomial recursion : a(n) = p(n) (see formula).

1, 4, 25, 188, 1671, 17190, 201125, 2638984, 38390179, 613363466, 10678267425, 201215691660, 4080450217247, 88609322165902, 2051573162708125, 50450534991347216, 1313219083705400475, 36072797094375866898, 1042811362801447763225, 31647646914322017237652, 1006032342980535954429463

Offset

1,2

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).

Lim n-->oo p(n)/q(n) = (3 - 2*log(2))/(2*log(2) - 1) = 4.17739889912417966161076...

a(n) ~ (3 - 2*log(2)) * n * n! / (8 * log(2)^(n+2)). - Vaclav Kotesovec, Nov 25 2020

E.g.f.: (1 - exp(x)) * (2*x - 1 - exp(x)) / (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], {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], {n, 1, 30}] (* Vaclav Kotesovec, Nov 25 2020 *)
nmax = 30; Rest[CoefficientList[Series[(1 - E^x)*(-1 - E^x + 2*x)/(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. A135148, A135149, A135150, A135074, A135075.

Sequence in context: A166697 A054368 A365764 * A221382 A215791 A175913

Adjacent sequences: A135144 A135145 A135146 * A135148 A135149 A135150

Keyword

nonn

Author

Benoit Cloitre, Nov 20 2007

Extensions

More terms from Amiram Eldar, Nov 25 2020

Status

approved