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A081923 Expansion of e.g.f.: exp(2x)/(1-x)^2.

%I A081923 #52 May 10 2025 17:21:12
%S A081923 1,4,18,92,536,3552,26608,223456,2085504,21450752,241320704,
%T A081923 2949474816,38933066752,552141672448,8374148696064,135274709700608,
%U A081923 2318995023429632,42051109758173184,804227474125029376
%N A081923 Expansion of e.g.f.: exp(2x)/(1-x)^2.
%C A081923 Binomial transform of A001339.
%C A081923 Polynomials in A010027 evaluated at 3. - _Ralf Stephan_, Dec 15 2004
%C A081923 From _Dennis P. Walsh_, Sep 18 2013: (Start)
%C A081923 a(n) is the number of rooted labeled forests that satisfy the following conditions:
%C A081923 (i) there are 4 roots labeled 1, 2, 3, and 4;
%C A081923 (ii) there are n non-root vertices labeled 5,..., n+4;
%C A081923 (iii) the trees with roots 1 and 2 have width one;
%C A081923 (iv) the trees with roots 3 and 4 have height at most one.
%C A081923 To construct such a forest, for k=0,...,n, we take the following steps:
%C A081923 (1) choose k non-root vertices for trees with roots 1 and 2;
%C A081923 (2) construct width-one trees on roots 1 and 2 with the k non-root vertices;
%C A081923 (3) with the n-k remaining non-root vertices construct trees of height at most one on roots 3 and 4.
%C A081923 Thus a(n) is the sum (over k) of the product of the number of ways to do each step: a(n)=sum(k=0..n, binomial(n,k)*(k+1)!*2^(n-k)).  (End)
%H A081923 Harvey P. Dale, <a href="/A081923/b081923.txt">Table of n, a(n) for n = 0..448</a>
%H A081923 Paul Barry, <a href="https://arxiv.org/abs/1804.06801">A note on number triangles that are almost their own production matrix</a>, arXiv:1804.06801 [math.CO], 2018.
%H A081923 Dennis P. Walsh, <a href="http://capone.mtsu.edu/dwalsh/FORESTEX.pdf">18 Forests</a>
%F A081923 E.g.f.: exp(2*x)/(1-x)^2
%F A081923 E.g.f.: 1/U(0) where U(k)=   1 - 2*x/( 1 + x/(2 - x - 4/( 2 - x*(k+1)/U(k+1)))) ; (continued fraction, 3rd kind, 4-step). - _Sergei N. Gladkovskii_, Oct 28 2012
%F A081923 Conjecture: a(n) +(-n-3)*a(n-1) +2*(n-1)*a(n-2)=0. - _R. J. Mathar_, Nov 24 2012
%F A081923 G.f.: 2/x/G(0) - 1/x, where G(k)= 1 + 1/(1 - x*(2*k+2)/(x*(2*k+4) - 1 + x*(2*k+2)/G(k+1))); (continued fraction). - _Sergei N. Gladkovskii_, May 31 2013
%F A081923 G.f.: (sum(k>=0, k!*(x/(1-2*x))^k ) - 1)/x = Q(0)/(2*x) - 1/x, where Q(k)= 1 + 1/(1 - x*(k+1)/(x*(k+1) + (1-2*x)/Q(k+1) )); (continued fraction). - _Sergei N. Gladkovskii_, Aug 09 2013
%F A081923 G.f.: W(0)/x - 1/x, where W(k) = 1 - x*(k+1)/( x*(k+3) - 1/(1 - x*(k+1)/( x*(k+1) - 1/W(k+1) ))); (continued fraction). - _Sergei N. Gladkovskii_, Aug 26 2013
%F A081923 a(n) = n!*sum(k=0..n, (k+1)*2^(n-k)/(n-k)!). [_Dennis P. Walsh_, Sep 18 2013]
%F A081923 a(n) = n!*sum(k=0..n, (n-k+1)*2^k/k!). [_Dennis P. Walsh_, Sep 18 2013]
%F A081923 From _Peter Bala_, Sep 25 2013: (Start)
%F A081923 a(n) ~ n!*n*e^2.
%F A081923 Applying Maple's ZeilbergerRecurrence command to the above series of Walsh for a(n) results in the first-order recurrence equation (n - 1)*a(n+1) = n*(n + 1)*a(n) - 2^(n+2) with a(0) = 1 and a(2) = 18. Using this it is easy to verify that a(n) satisfies the second-order recurrence a(n) = (n + 3)*a(n-1) - 2*(n - 1)*a(n-2) conjectured above by Mathar.
%F A081923 The sequence b(n) = n!*(n - 1) satisfies the same second-order recurrence but with the initial conditions b(0) = -1 and b(1) = 0. This leads to the finite continued fraction expansion a(n)/b(n) = 9 - 2*( 4/(6 - 6/(7 - 8/(9 - ... - 2*n/(n + 4)))) ) valid for n >= 2. Letting n tend to infinity produces the infinite continued fraction expansion e^2 = 9 - 2*( 4/(6 - 6/(7 - 8/(9 - ... - 2*n/(n + 4 - ...)))) ). (End)
%F A081923 a(n) = KummerU(-n, -n - 1, 2). - _Peter Luschny_, May 10 2022
%e A081923 For n=2, the a(2)=18 forests that satisfy the specified conditions are given in the link above. - _Dennis P. Walsh_, Sep 20 2013
%p A081923 seq(n!*add((k+1)*2^(n-k)/(n-k)!,k=0..n),n=0..40); # _Dennis P. Walsh_, Sep 18 2013
%p A081923 seq(simplify(KummerU(-n, -n - 1, 2)), n = 0..24); # _Peter Luschny_, May 10 2022
%t A081923 With[{nn=20},CoefficientList[Series[Exp[2x]/(1-x)^2,{x,0,nn}],x] Range[0,nn]!] (* _Harvey P. Dale_, May 10 2025 *)
%Y A081923 Cf. A081924. A010842.
%Y A081923 Cf. A081923(n) (sum(k=0..n, binomial(n,k)*A000522(n-k)*A000522(k))).
%K A081923 easy,nonn
%O A081923 0,2
%A A081923 _Paul Barry_, Apr 01 2003
%E A081923 Definition clarified by _Harvey P. Dale_, May 10 2025