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 A053486 E.g.f.: exp(3x)/(1-x). 20
 1, 4, 17, 78, 393, 2208, 13977, 100026, 806769, 7280604, 72865089, 801693126, 9620848953, 125072630712, 1751021612937, 26265338542962, 420245459734113, 7144172944620084, 128595113390582001, 2443307155583319486, 48866143115153174121 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 LINKS Vincenzo Librandi, Table of n, a(n) for n = 0..200 J. W. Layman, The Hankel Transform and Some of its Properties, J. Integer Sequences, 4 (2001), #01.1.5. FORMULA 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 a(n) = Sum[(n! / k!) * 3^k {k=0...n}]. - Ross La Haye, Sep 21 2004 a(n) = sum{k=0..n, k!*C(n, k)3^(n-k)}. - Paul Barry, Apr 22 2005 G.f.: hypergeom([1,1],[],x/(1-3*x))/(1-3*x). - Mark van Hoeij, Nov 08 2011 Conjecture: -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 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 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 a(n) ~ n! * exp(3). - Vaclav Kotesovec, Jun 01 2013 From Peter Bala, Sep 25 2013: (Start) a(n) = n*a(n-1) + 3^n with a(0) = 1. a(n) = n!*e^3 - sum {k >= 0} 3^(n + k + 1)/((n + 1)*...*(n + k + 1)) = n!*e^3 - e^3*( int {t = 0..3} t^n*exp(-t) dt ) = e^3*( int {t = 3..inf} t^n*exp(-t) dt ) = e^3*( int {t = 0..inf} t^n*exp(-t)*Heaviside(t-3) dt ), an integral representation of a(n) as the n-th moment of a nonnegative function on the positive half-axis. 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) 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 a(n) = exp(3)*Gamma(1+n,3). - Peter Luschny, Dec 18 2017 MAPLE G(x):=exp(3*x)/(1-x): g:=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 A053486 := n -> exp(3)*GAMMA(1+n, 3): seq(simplify(A053486(n)), n=0..20); # Peter Luschny, Dec 18 2017 MATHEMATICA RecurrenceTable[{a==1, a[n]== n*a[n-1] + 3^n}, a, {n, 200}] (* Vincenzo Librandi, Nov 15 2012 *) With[{nn=20}, CoefficientList[Series[Exp[3x]/(1-x), {x, 0, nn}], x] Range[ 0, nn]!] (* Harvey P. Dale, Aug 03 2017 *) PROG (PARI) x='x+O('x^66); Vec(serlaplace(exp(3*x)/(1-x))) \\ Joerg Arndt, Apr 20 2013 CROSSREFS Cf. A008290, A000522, A010842. Sequence in context: A005494 A257072 A193782 * A151249 A289924 A218134 Adjacent sequences:  A053483 A053484 A053485 * A053487 A053488 A053489 KEYWORD nonn AUTHOR N. J. A. Sloane, Jan 15 2000 STATUS approved

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Last modified May 30 08:43 EDT 2020. Contains 334712 sequences. (Running on oeis4.)