This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A082031 #21 Jul 25 2021 02:35:44
%S A082031 1,5,28,176,1240,9752,85120,819296,8639872,99209600,1233416704,
%T A082031 16517058560,237137769472,3634932675584,59263206154240,
%U A082031 1024222802014208,18706559855656960,360062627304341504,7285354765603176448
%N A082031 Expansion of e.g.f. exp(2*x)/(1-x)^3.
%C A082031 Binomial transform of A082030
%F A082031 E.g.f.: exp(2*x)/(1-x)^3.
%F A082031 Conjecture: a(n) +(-n-4)*a(n-1) +2*(n-1)*a(n-2)=0. - _R. J. Mathar_, Nov 24 2012
%F A082031 From _Peter Bala_, Sep 20 2013: (Start)
%F A082031 a(n) = (1/2)*( Sum_{k = 0..n} (k+2)!*binomial(n,k)*2^(n-k) ).
%F A082031 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.
%F A082031 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.
%F A082031 Lim_{n -> infinity} a(n)/b(n) = e^2 = 1/(1 - 4/(5 - 2/(6 - 4/(7 - ... - (2*n - 2)/(n + 4 - ...))))).
%F A082031 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)
%t A082031 With[{nn=20},CoefficientList[Series[Exp[2x]/(1-x)^3,{x,0,nn}],x] Range[0,nn]!] (* _Harvey P. Dale_, Apr 28 2013 *)
%Y A082031 Cf. A052124, A082030.
%K A082031 easy,nonn
%O A082031 0,2
%A A082031 _Paul Barry_, Apr 02 2003