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A081923
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Expansion of exp(2x)/(1-x)^2.
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6
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1, 4, 18, 92, 536, 3552, 26608, 223456, 2085504, 21450752, 241320704, 2949474816, 38933066752, 552141672448, 8374148696064, 135274709700608, 2318995023429632, 42051109758173184, 804227474125029376
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OFFSET
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0,2
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COMMENTS
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a(n) is the number of rooted labeled forests that satisfy the following conditions:
(i) there are 4 roots labeled 1, 2, 3, and 4;
(ii) there are n non-root vertices labeled 5,..., n+4;
(iii) the trees with roots 1 and 2 have width one;
(iv) the trees with roots 3 and 4 have height at most one.
To construct such a forest, for k=0,...,n, we take the following steps:
(1) choose k non-root vertices for trees with roots 1 and 2;
(2) construct width-one trees on roots 1 and 2 with the k non-root vertices;
(3) with the n-k remaining non-root vertices construct trees of height at most one on roots 3 and 4.
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)
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LINKS
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FORMULA
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E.g.f.: exp(2*x)/(1-x)^2
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
Conjecture: a(n) +(-n-3)*a(n-1) +2*(n-1)*a(n-2)=0. - R. J. Mathar, Nov 24 2012
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
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
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
a(n) = n!*sum(k=0..n, (k+1)*2^(n-k)/(n-k)!). [Dennis P. Walsh, Sep 18 2013]
a(n) ~ n!*n*e^2.
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.
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)
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EXAMPLE
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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
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MAPLE
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seq(n!*add((k+1)*2^(n-k)/(n-k)!, k=0..n), n=0..40); # Dennis P. Walsh, Sep 18 2013
seq(simplify(KummerU(-n, -n - 1, 2)), n = 0..24); # Peter Luschny, May 10 2022
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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