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%I #42 Sep 08 2022 08:44:58
%S 1,1,9,121,2169,48601,1306809,40994521,1469709369,59277466201,
%T 2656472295609,130952452264921,7042235448544569,410269802967187801,
%U 25740278881968596409,1730295054262416751321,124066865052334175027769
%N Number of 5-level labeled linear rooted trees with n leaves.
%H Michael De Vlieger, <a href="/A050353/b050353.txt">Table of n, a(n) for n = 0..360</a>
%H Norihiro Nakashima, Shuhei Tsujie, <a href="https://arxiv.org/abs/1904.09748">Enumeration of Flats of the Extended Catalan and Shi Arrangements with Species</a>, arXiv:1904.09748 [math.CO], 2019.
%H <a href="/index/Ro#rooted">Index entries for sequences related to rooted trees</a>
%F E.g.f.: (4 - 3*exp(x))/(5 - 4*exp(x)).
%F a(n) is asymptotic to (1/20)*n!/log(5/4)^(n+1). More generally if m>1, the number of m-level labeled linear rooted trees with n leaves is asymptotic to n!/log(m/(m-1))^(n+1)/(m^2-m). - _Benoit Cloitre_, Jan 30 2003
%F For m-level trees (m>1), e.g.f. is (m-1-(m-2)*e^x)/(m-(m-1)*e^x) and number of trees is 1/(m*(m-1))*sum(k>=0, (1-1/m)^k*k^n). Here m=5, so a(n)=(1/20)*sum(k>=0, (4/5)^k*k^n) (for n>0). - _Benoit Cloitre_, Jan 30 2003
%F Let f(x) = (1+x)*(1+2*x). Let D be the operator g(x) -> d/dx(f(x)*g(x)). Then for n>=1, a(n) = D^(n-1)(1) evaluated at x = 3/2. Compare with the result A000670(n) = D^(n-1)(1) at x = 0. See also A194649. - _Peter Bala_, Sep 05 2011
%F E.g.f.: 1 + x/(G(0)-5*x) where G(k)= x + k + 1 - x*(k+1)/G(k+1); (continued fraction Euler's 1st kind, 1-step). - _Sergei N. Gladkovskii_, Jul 11 2012
%F a(n) = (1/20) * Sum_{k>=1} k^n * (4/5)^k for n>0. - _Paul D. Hanna_, Nov 28 2014
%F a(n) = Sum_{k=1..n} Stirling2(n, k) * k! * 4^(k-1). - _Paul D. Hanna_, Nov 28 2014, after _Vladeta Jovovic_ in A050351
%F a(n) = 1 + 4 * Sum_{k=1..n-1} binomial(n,k) * a(k). - _Ilya Gutkovskiy_, Jun 08 2020
%p seq(coeff(series( (4-3*exp(x))/(5-4*exp(x)), x, n+1)*n!, x, n), n = 0..20); # _G. C. Greubel_, Jun 08 2020
%t max = 16; f[x_] := (4-3*E^x) / (5-4*E^x); CoefficientList[ Series[ f[x], {x, 0, max}], x]*Range[0, max]! (* _Jean-François Alcover_, Nov 14 2011, after g.f. *)
%o (PARI) a(n)=n!*if(n<0,0,polcoeff((4-3*exp(x))/(5-4*exp(x))+O(x^(n+1)),n))
%o (PARI) {a(n)=if(n==0,1,(1/20)*round(suminf(k=1, k^n * (4/5)^k *1.)))} \\ _Paul D. Hanna_, Nov 28 2014
%o (Magma) R<x>:=PowerSeriesRing(Rationals(), 20); Coefficients(R!(Laplace( (4-3*Exp(x))/(5-4*Exp(x)) ))); // _G. C. Greubel_, Jun 08 2020
%o (Sage) [1]+[sum( 4^(j-1)*factorial(j)*stirling_number2(n,j) for j in (1..n)) for n in (1..20)] # _G. C. Greubel_, Jun 08 2020
%Y Cf. A000670, A050351 - A050359.
%Y Equals 1/4 * A094417(n) for n>0.
%K nonn
%O 0,3
%A _Christian G. Bower_, Oct 15 1999
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