A082030 Expansion of e.g.f. exp(x)/(1-x)^3.

1, 4, 19, 106, 685, 5056, 42079, 390454, 4000441, 44881660, 547457611, 7215589954, 102211815589, 1548801969976, 25000879886935, 428332610385166, 7763306399014129, 148412806214119924, 2984692721713278211

Offset: 0

Author

Paul Barry, Apr 02 2003

Keywords

• easy
• nonn

Comments

Binomial transform of A001710 (when preceded by 0).

From Peter Bala, Jul 10 2008: (Start)

a(n) is a difference divisibility sequence, that is, the difference a(n) - a(m) is divisible by n - m for all n and m (provided n is not equal to m). See A000522 for further properties of difference divisibility sequences.

Recurrence relation: a(0) = 1, a(1) = 4, a(n) = (n+3)*a(n-1) - (n-1)*a(n-2) for n >= 2. The sequence b(n) := n!*(n^2+n+1) = A001564(n) satisfies the same recurrence with the initial conditions b(0) = 1, b(1) = 3. This leads to the finite continued fraction expansion a(n)/b(n) = 1/(1-1/(4-1/(5-2/(6-...-(n-1)/(n+3))))).

Lim_{n -> infinity} a(n)/b(n) = e/2 = 1/(1-1/(4-1/(5-2/(6-...-n/((n+4)-...))))).

a(n) = n!*(n^2+n+1)*Sum_{k = 0..n} 1/(k!*(k^4+k^2+1)) since the rhs satisfies the above recurrence with the same initial conditions. Hence e = 2*Sum_{k >= 0} 1/(k!*(k^4+k^2+1)).

For sequences satisfying the more general recurrence a(n) = (n+1+r)*a(n-1) - (n-1)*a(n-2), which yield series acceleration formulas for e/r! that involve the Poisson-Charlier polynomials c_r(-n;-1), refer to A000522 (r = 0), A001339 (r=1), A095000 (r=3) and A095177 (r=4). (End)

Links

• Roland Bacher, Counting Packings of Generic Subsets in Finite Groups, Electr. J. Combinatorics, 19 (2012), #P7. - From _N. J. A. Sloane_, Feb 06 2013
• Eric Weisstein's World of Mathematics, Poisson-Charlier polynomial

Crossrefs

Cf. A082031, A000522, A001339, A095000, A095177, A096307, A096341, A001564.

Programs

Maple

a := n -> hypergeom([3, -n], [], -1); seq(simplify(a(n)), n=0..18); # Peter Luschny, Sep 20 2014
seq(simplify(KummerU(-n, -n - 2, 1)), n = 0..20); # Peter Luschny, May 10 2022

Mathematica

a[n_] := a[n] = If[n == 0, 1, (n (n^2 + n + 1) a[n-1] + 1)/(n^2 - n + 1)];
a /@ Range[0, 18] (* Jean-François Alcover, Oct 16 2019 *)
With[{nn=20},CoefficientList[Series[Exp[x]/(1-x)^3,{x,0,nn}],x] Range[0,nn]!] (* Harvey P. Dale, Aug 07 2022 *)

PARI

{a(n)=n!*polcoeff(exp(x+x*O(x^n))/(1-x)^3,n)} /* Paul D. Hanna, Sep 30 2011 */

PARI

{a(n)=sum(k=0,n,binomial(n,k)*(k+2)!/2)} /* Paul D. Hanna, Sep 30 2011 */

PARI

{a(n)=sum(k=0,n,binomial(n,k)*(k+1)^(k+1)*(-k)^(n-k))} /* Paul D. Hanna, Sep 30 2011 */

PARI

{a(n)=polcoeff(sum(m=0,n,(m+1)^(m+1)*x^m/(1+m*x)^(m+1)+x*O(x^n)),n)} /* Paul D. Hanna, Sep 30 2011 */

Formula

E.g.f.: exp(x)/(1-x)^3.

a(n) = A001340(n)/2.

a(n) = Sum_{k=0..n} A046716(n, k)*3^(n-k). - Philippe Deléham, Jun 12 2004

a(n) = Sum_{k=0..n} binomial(n, k)*(k+2)!/2. - Philippe Deléham, Jun 19 2004

a(n) = Sum_{k=0..n} binomial(n,k)*(k+1)^(k+1)*(-k)^(n-k). - Paul D. Hanna, Sep 30 2011

O.g.f.: Sum_{n>=0} (n+1)^(n+1)*x^n/(1+n*x)^(n+1) = Sum_{n>=0} a(n)*x^n. - Paul D. Hanna, Sep 30 2011

Conjecture: a(n) + (-n-3)*a(n-1) + (n-1)*a(n-2) = 0. - R. J. Mathar, Dec 03 2012

G.f.: (1-x)/(2*x*Q(0)) - 1/2/x, where Q(k) = 1 - x - x*(k+2)/(1 - x*(k+1)/Q(k+1)); (continued fraction). - Sergei N. Gladkovskii, Apr 22 2013

a(n) = hypergeometric([3, -n], [], -1). - Peter Luschny, Sep 20 2014

First-order recurrence: P(n-1)*a(n) = n*P(n)*a(n-1) + 1 with a(0) = 1, where P(n) = n^2 + n + 1 = A001564(n). - Peter Bala, Jul 26 2021

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