%I #68 May 10 2022 14:08:23
%S 1,4,17,78,393,2208,13977,100026,806769,7280604,72865089,801693126,
%T 9620848953,125072630712,1751021612937,26265338542962,420245459734113,
%U 7144172944620084,128595113390582001,2443307155583319486,48866143115153174121
%N E.g.f.: exp(3x)/(1-x).
%H Vincenzo Librandi, <a href="/A053486/b053486.txt">Table of n, a(n) for n = 0..200</a>
%H J. W. Layman, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL4/LAYMAN/hankel.html">The Hankel Transform and Some of its Properties</a>, J. Integer Sequences, 4 (2001), #01.1.5.
%F a(n) is the permanent of the n X n matrix with 4's on the diagonal and 1's elsewhere. a(n) = Sum(k=0..n, A008290(n, k)*4^k). - _Philippe Deléham_, Dec 12 2003
%F a(n) = Sum[(n! / k!) * 3^k {k=0...n}]. - _Ross La Haye_, Sep 21 2004
%F a(n) = sum{k=0..n, k!*C(n, k)3^(n-k)}. - _Paul Barry_, Apr 22 2005
%F G.f.: hypergeom([1,1],[],x/(1-3*x))/(1-3*x). - _Mark van Hoeij_, Nov 08 2011
%F D-finite with recurrence -a(n) +(n+3)*a(n-1) +3*(1-n)*a(n-2)=0. - _R. J. Mathar_, Nov 14 2011. This recurrence follows from the Wilf-Zeilberger (WZ) proof technique applied to the sum: Sum[k!* C(n,k)*3^(n-k), {k=0...n}]. - _T. Amdeberhan_, Jul 23 2012
%F E.g.f.: 1/E(0) where E(k)=1-x/(1-3/(3+(k+1)/E(k+1))); (continued fraction). - _Sergei N. Gladkovskii_, Nov 21 2011
%F G.f.: 1/Q(0), where Q(k)= 1 - 3*x - x*(k+1)/(1-x*(k+1)/Q(k+1)); (continued fraction). - _Sergei N. Gladkovskii_, Apr 20 2013
%F a(n) ~ n! * exp(3). - _Vaclav Kotesovec_, Jun 01 2013
%F From _Peter Bala_, Sep 25 2013: (Start)
%F a(n) = n*a(n-1) + 3^n with a(0) = 1.
%F a(n) = n!*e^3 - sum {k >= 0} 3^(n + k + 1)/((n + 1)*...*(n + k + 1))
%F = n!*e^3 - e^3*( int {t = 0..3} t^n*exp(-t) dt )
%F = e^3*( int {t = 3..inf} t^n*exp(-t) dt )
%F = e^3*( int {t = 0..inf} t^n*exp(-t)*Heaviside(t-3) dt ),
%F an integral representation of a(n) as the n-th moment of a nonnegative function on the positive half-axis.
%F Mathar's second-order recurrence above a(n) = (n + 3)*a(n-1) - 3*(n - 1)*a(n-2) has n! as a second solution. This yields the finite continued fraction expansion a(n)/n! = 1/(1 - 3/(4 - 3/(5 - 6/(6 - ...- 3*(n - 1)/(n + 3))))) valid for n >= 2. Letting n tend to infinity gives the infinite continued fraction expansion e^3 = 1/(1 - 3/(4 - 3/(5 - 6/(6 - ...- 3*(n - 1)/(n + 3 - ...))))). (End)
%F G.f.: 1/Q(0), where Q(k) = 1 - 2*x*(k+2) - x^2*(k+1)^2/Q(k+1); (continued fraction). - _Sergei N. Gladkovskii_, Sep 30 2013
%F a(n) = exp(3)*Gamma(1+n,3). - _Peter Luschny_, Dec 18 2017
%F a(n) = KummerU(-n, -n, 3). - _Peter Luschny_, May 10 2022
%p G(x):=exp(3*x)/(1-x): g[0]:=G(x): for n from 1 to 20 do g[n]:=diff(g[n-1],x) od: x:=0: seq(g[n],n=0..20); # _Zerinvary Lajos_, Apr 03 2009
%p A053486 := n -> exp(3)*GAMMA(1+n,3):
%p seq(simplify(A053486(n)), n=0..20); # _Peter Luschny_, Dec 18 2017
%p seq(simplify(KummerU(-n, -n, 3)), n = 0..20); # _Peter Luschny_, May 10 2022
%t RecurrenceTable[{a[0]==1, a[n]== n*a[n-1] + 3^n}, a, {n, 200}] (* _Vincenzo Librandi_, Nov 15 2012 *)
%t With[{nn=20},CoefficientList[Series[Exp[3x]/(1-x),{x,0,nn}],x] Range[ 0,nn]!] (* _Harvey P. Dale_, Aug 03 2017 *)
%o (PARI) x='x+O('x^66); Vec(serlaplace(exp(3*x)/(1-x))) \\ _Joerg Arndt_, Apr 20 2013
%Y Cf. A008290, A000522, A010842.
%K nonn
%O 0,2
%A _N. J. A. Sloane_, Jan 15 2000