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A053486
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E.g.f.: exp(3x)/(1-x).
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24
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1, 4, 17, 78, 393, 2208, 13977, 100026, 806769, 7280604, 72865089, 801693126, 9620848953, 125072630712, 1751021612937, 26265338542962, 420245459734113, 7144172944620084, 128595113390582001, 2443307155583319486, 48866143115153174121
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OFFSET
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0,2
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LINKS
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FORMULA
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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
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
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*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
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MAPLE
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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
A053486 := n -> exp(3)*GAMMA(1+n, 3):
seq(simplify(KummerU(-n, -n, 3)), n = 0..20); # Peter Luschny, May 10 2022
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MATHEMATICA
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RecurrenceTable[{a[0]==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 *)
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PROG
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(PARI) x='x+O('x^66); Vec(serlaplace(exp(3*x)/(1-x))) \\ Joerg Arndt, Apr 20 2013
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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