A082031 - OEIS

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

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OFFSET

0,2

COMMENTS

Binomial transform of A082030

LINKS

Table of n, a(n) for n=0..18.

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)

MATHEMATICA

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

CROSSREFS

Cf. A052124, A082030.

Sequence in context: A143647 A349318 A367889 * A020081 A095676 A324352

Adjacent sequences: A082028 A082029 A082030 * A082032 A082033 A082034

KEYWORD

easy,nonn

AUTHOR

Paul Barry, Apr 02 2003

STATUS

approved