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

1, 4, 18, 92, 536, 3552, 26608, 223456, 2085504, 21450752, 241320704, 2949474816, 38933066752, 552141672448, 8374148696064, 135274709700608, 2318995023429632, 42051109758173184, 804227474125029376

Offset: 0

Author

Paul Barry, Apr 01 2003

Keywords

• easy
• nonn

Comments

Binomial transform of A001339.

Polynomials in A010027 evaluated at 3. - Ralf Stephan, Dec 15 2004

From Dennis P. Walsh, Sep 18 2013: (Start)

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)

Examples

			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
		

Links

• Harvey P. Dale, Table of n, a(n) for n = 0..448
• Paul Barry, A note on number triangles that are almost their own production matrix, arXiv:1804.06801 [math.CO], 2018.
• Dennis P. Walsh, 18 Forests

Crossrefs

Cf. A081924. A010842.

Cf. A081923(n) (sum(k=0..n, binomial(n,k)*A000522(n-k)*A000522(k))).

Programs

Maple

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

Mathematica

With[{nn=20},CoefficientList[Series[Exp[2x]/(1-x)^2,{x,0,nn}],x] Range[0,nn]!] (* Harvey P. Dale, May 10 2025 *)

Formula

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!*sum(k=0..n, (n-k+1)*2^k/k!). [Dennis P. Walsh, Sep 18 2013]

From Peter Bala, Sep 25 2013: (Start)

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)

a(n) = KummerU(-n, -n - 1, 2). - Peter Luschny, May 10 2022

Extensions

Definition clarified by Harvey P. Dale, May 10 2025