A111998 Tenth convolution of Schroeder's (second problem) numbers A001003(n), n>=0.

1, 10, 75, 500, 3135, 18962, 112125, 653200, 3766950, 21571500, 122920642, 697994760, 3953743250, 22357130700, 126273263510, 712639689168, 4019975635855, 22671014908550, 127846248597125, 720994336613980

Offset

0,2

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..300

Formula

G.f.: ((1+x-sqrt(1-6*x+x^2))/(4*x))^10.

a(n) = (10/n)*Sum_{k=1,..,n} binomial(n,k)*binomial(n+k+9,k-1).

a(n) = 10*hypergeom([1-n, n+11], [2], -1), n>=1, a(0)=1.

Contribution from Vaclav Kotesovec, Oct 18 2012: (Start)

Recurrence: n*(n+10)*a(n) = (7*n^2+58*n+45)*a(n-1) - (7*n^2+40*n-18)*a(n-2) + (n-3)*(n+7)*a(n-3)

a(n) ~ 5*sqrt(3*sqrt(2)-4)*(1970-1393*sqrt(2)) * (3+2*sqrt(2))^(n+10)/(64*sqrt(Pi)*n^(3/2))

Generally, G.f. = ((1+x-sqrt(1-6*x+x^2))/(4*x))^k is asymptotic to a(n) ~ sqrt(3*sqrt(2)-4)*k*(1-1/sqrt(2))^(k-1) * (3+2*sqrt(2))^(n+k)/(4*sqrt(Pi)*n^(3/2)).

(End)

Mathematica

CoefficientList[Series[((1+x-Sqrt[1-6*x+x^2])/(4*x))^10, {x, 0, 20}], x] (* Vaclav Kotesovec, Oct 18 2012 *)

Prog

(PARI) x='x+O('x^50); Vec(((1+x-sqrt(1-6*x+x^2))/(4*x))^10) \\ G. C. Greubel, Mar 17 2017

Crossrefs

Cf. Tenth column of convolution triangle A011117.

Sequence in context: A291003 A305784 A053464 * A026935 A110127 A233657

Adjacent sequences: A111995 A111996 A111997 * A111999 A112000 A112001

Keyword

nonn,easy

Author

Wolfdieter Lang, Sep 12 2005

Status

approved