A082031 Expansion of e.g.f. exp(2*x)/(1-x)^3.

1, 5, 28, 176, 1240, 9752, 85120, 819296, 8639872, 99209600, 1233416704, 16517058560, 237137769472, 3634932675584, 59263206154240, 1024222802014208, 18706559855656960, 360062627304341504, 7285354765603176448

Offset: 0

Author

Paul Barry, Apr 02 2003

Keywords

• easy
• nonn

Comments

Binomial transform of A082030

Crossrefs

Cf. A052124, A082030.

Programs

Mathematica

With[{nn=20},CoefficientList[Series[Exp[2x]/(1-x)^3,{x,0,nn}],x] Range[0,nn]!] (* Harvey P. Dale, Apr 28 2013 *)

Formula

E.g.f.: exp(2*x)/(1-x)^3.

Conjecture: a(n) +(-n-4)*a(n-1) +2*(n-1)*a(n-2)=0. - R. J. Mathar, Nov 24 2012

From Peter Bala, Sep 20 2013: (Start)

a(n) = (1/2)*( Sum_{k = 0..n} (k+2)!*binomial(n,k)*2^(n-k) ).

Based on this series the ZeilbergerRecurrence command in Maple 17 produces the first-order recurrence (n^2 - 3*n + 4)*a(n) = 2^(n+2) + n*(n^2 - n + 2)*a(n-1). Using this it is easy to verify the second-order recurrence conjectured above by Mathar.

The sequence b(n) := n!*(1 + n*(n-1)/2) = n!*A000124(n-1) also satisfies Mathar's recurrence equation but with starting values b(0) = b(1) = 1. This yields the finite continued fraction expansion a(n)/b(n) = 1/(1 - 4/(5 - 2/(6 - 4/(7 - ... - (2*n - 2)/(n + 4) )))), valid for n >= 2.

Lim_{n -> infinity} a(n)/b(n) = e^2 = 1/(1 - 4/(5 - 2/(6 - 4/(7 - ... - (2*n - 2)/(n + 4 - ...))))).

It can be shown that a(n+1)/b(n+1) = 1 + 16*( Sum_{k = 0..n} 2^k/((k + 1)!*(k^4 + 3*k^2 + 4)) ). Taking the limit gives the series acceleration result e^2 = 1 + 16*( Sum_{k = 0..infinity} 2^k/((k+1)!*(k^4 + 3*k^2 + 4)) ). Cf. A082030 and A052124. (End)