A082031 -id:A082031 - cp's OEIS frontend

A010842 Expansion of e.g.f.: exp(2*x)/(1-x).

1, 3, 10, 38, 168, 872, 5296, 37200, 297856, 2681216, 26813184, 294947072, 3539368960, 46011804672, 644165281792, 9662479259648, 154599668219904, 2628194359869440, 47307498477912064, 898842471080853504, 17976849421618118656, 377513837853982588928

Offset: 0

N. J. A. Sloane, Simon Plouffe

Incomplete Gamma Function at 2, more precisely: a(n) = exp(2)*Gamma(1+n,2).
Let P(A) be the power set of an n-element set A. Then a(n) = the total number of ways to add 0 or more elements of A to each element x of P(A) where the elements to add are not elements of x and order of addition is important. - Ross La Haye, Nov 19 2007
a(n) is the number of ways to split the set {1,2,...,n} into two disjoint subsets S,T with S union T = {1,2,...,n} and linearly order S and then choose a subset of T. - Geoffrey Critzer, Mar 10 2009

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, Tenth Printing, 1972, p. 262.
R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Example 5.1.2.

T. D. Noe, Table of n, a(n) for n = 0..100
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, Tenth Printing, 1972, p. 262.
Ross La Haye, Binary Relations on the Power Set of an n-Element Set, Journal of Integer Sequences, Vol. 12 (2009), Article 09.2.6.
J. W. Layman, The Hankel Transform and Some of its Properties, J. Integer Sequences, 4 (2001), #01.1.5.

Cf. A053484, A053485, A053486, A008290, A137346.

A010843, A000023, A000166, A000142, A000522, A010842, A053486, A053487 have similar properties.

m:=45; R:=PowerSeriesRing(Rationals(), m); b:=Coefficients(R!(Exp(2*x)/(1-x))); [Factorial(n-1)*b[n]: n in [1..m]]; // G. C. Greubel, Oct 16 2018

G(x):=exp(2*x)/(1-x): f[0]:=G(x): for n from 1 to 19 do f[n]:=diff(f[n-1],x) od: x:=0: seq(f[n],n=0..19); # Zerinvary Lajos, Apr 03 2009
seq(simplify(exp(1)^2*GAMMA(n+1, 2)), n=0..19); # Peter Luschny, Apr 28 2016
seq(simplify(KummerU(-n, -n, 2)), n=0..21); # Peter Luschny, May 10 2022

With[{r = Round[n! E^2 - 2^(n + 1)/(n + 1)]}, r - Mod[r, 2^(n - Floor[2/n + Log2[n]])]] (* for n>=4; Stan Wagon, Apr 28 2016 *)
a[n_] := n! Sum[2^i/i!, {i, 0, n}]
Table[a[n], {n, 0, 21}] (* Gerry Martens , May 06 2016 *)
With[{nn=30},CoefficientList[Series[Exp[2x]/(1-x),{x,0,nn}],x] Range[ 0,nn]!] (* Harvey P. Dale, May 27 2019 *)

x='x+O('x^44); Vec(serlaplace(exp(2*x)/(1-x))) \\ Joerg Arndt, Apr 29 2016

a(n) = row sums of A090802. - Ross La Haye, Aug 18 2006
a(n) = n*a(n-1) + 2^n = (n+2)*a(n-1) - (2*n-2)*a(n-2) = n!*Sum_{j=0..n} floor(2^j/j!). - Henry Bottomley, Jul 12 2001
a(n) is the permanent of the n X n matrix with 3's on the diagonal and 1's elsewhere. a(n) = Sum_{k=0..n} A008290(n, k)*3^k. - Philippe Deléham, Dec 12 2003
Binomial transform of A000522. - Ross La Haye, Sep 15 2004
a(n) = Sum_{k=0..n} k!*binomial(n, k)*2^(n-k). - Paul Barry, Apr 22 2005
a(n) = A066534(n) + 2^n. - Ross La Haye, Nov 16 2005
G.f.: hypergeom([1,k],[],x/(1-2*x))/(1-2*x) with k=1,2,3 is the generating function for A010842, A081923, and A082031. - Mark van Hoeij, Nov 08 2011
E.g.f.: 1/E(0), where E(k) = 1 - x/(1-2/(2+(k+1)/E(k+1))); (continued fraction). - Sergei N. Gladkovskii, Nov 21 2011
G.f.: 1/Q(0), where Q(k)= 1 - 2*x - x*(k+1)/(1-x*(k+1)/Q(k+1)); (continued fraction). - Sergei N. Gladkovskii, Apr 18 2013
a(n) ~ n! * exp(2). - Vaclav Kotesovec, Jun 01 2013
From Peter Bala, Sep 25 2013: (Start)
a(n) = n!*e^2 - Sum_{k >= 0} 2^(n + k + 1)/((n + 1)*...*(n + k + 1)).
= n!*e^2 - e^2*( Integral_{t = 0..2} t^n*exp(-t) dt )
= e^2*( Integral_{t >= 2} t^n*exp(-t) dt )
= e^2*( Integral_{t >= 0} t^n*exp(-t)*Heaviside(t-2) dt ),
an integral representation of a(n) as the n-th moment of a nonnegative function on the positive half-axis.
Bottomley's second-order recurrence above a(n) = (n + 2)*a(n-1) - 2*(n - 1)*a(n-2) has n! as a second solution. This yields the finite continued fraction expansion a(n)/n! = 1/(1 - 2/(3 - 2/(4 - 4/(5 - ... - 2*(n - 1)/(n + 2))))) valid for n >= 2. Letting n tend to infinity gives the infinite continued fraction expansion e^2 = 1/(1 - 2/(3 - 2/(4 - 4/(5 - ... - 2*(n - 1)/(n + 2 - ...))))). (End)
a(n) = 2^(n+1)*U(1, n+2, 2), where U is the Bessel U function. - Peter Luschny, Nov 26 2014
For n >= 4, a(n) = r - (r mod 2^(n - floor((2/n) + log_2(n)))) where r = n! * e^2 - 2^(n+1)/(n+1). - Stan Wagon, Apr 28 2016
G.f.: A(x) = 1/(1 - 2*x - x/(1 - x/(1 - 2*x - 2*x/(1 - 2*x/(1 - 2*x - 3*x/(1 - 3*x/(1 - 2*x - 4*x/(1 - 4*x/(1 - 2*x - ... ))))))))). - Peter Bala, May 26 2017
a(n) = Sum_{k=0..n} (-1)^(n-k)*A137346(n, k). - Mélika Tebni, May 10 2022 [This is equivalent to a(n) = KummerU(-n, -n, 2). - Peter Luschny, May 10 2022]
a(n) = F(n), where the function F(x) := 2^(x+1) * Integral_{t >= 0} e^(-2*t)*(1 + t)^x dt smoothly interpolates this sequence to all real values of x. - Peter Bala, Sep 05 2023

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

Original entry on oeis.org

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

Offset: 0

Paul Barry, Apr 02 2003

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)

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

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

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

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 *)

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

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

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

{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 */

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

A052124 Expansion of e.g.f. exp(-2*x)/(1-x)^3.

Original entry on oeis.org

1, 1, 4, 16, 88, 568, 4288, 36832, 354688, 3781504, 44199424, 561823744, 7714272256, 113769309184, 1793341407232, 30085661765632, 535170830467072, 10060645294440448, 199287423535808512, 4148644277780217856, 90545807649965080576, 2067407731760475406336, 49285894020028992323584

Offset: 0

N. J. A. Sloane, Jan 23 2000

nonn

R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Problem 5.64(b).

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Cf. A052127, A082031.
Cf. A000023, A087981.
Cf. A000153, A137775.

A052124 := proc(n) option remember; if n <=1 then 1 else n*A052124(n-1)+2*(n-1)*A052124(n-2); fi; end; # Detlef Pauly

Table[(n+5)*(n+2)*n!*Sum[(-1)^k*2^(k+2)*(k+3)/(k+5)!,{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, Oct 28 2012 *)
With[{nn=20},CoefficientList[Series[Exp[(-2x)]/(1-x)^3,{x,0,nn}],x] Range[ 0,nn]!] (* Harvey P. Dale, Oct 23 2017 *)

my(x='x+O('x^25)); Vec(serlaplace( exp(-2*x)/(1-x)^3)) \\ Michel Marcus, Oct 25 2021

from math import factorial
from fractions import Fraction
def A052124(n): return int((n+5)*(n+2)*factorial(n)*sum(Fraction((-1 if k&1 else 1)*(k+3)<Chai Wah Wu, Apr 20 2023

a(n) = n*a(n-1) + 2*(n-1)*a(n-2). - Detlef Pauly (dettodet(AT)yahoo.de), Sep 22 2003
a(n) = (n+5)*(n+2)*n! * Sum_{k=0..n} (-1)^k*2^(k+2)*(k+3)/(k+5)!. - Vaclav Kotesovec, Oct 28 2012
G.f.: 1/Q(0), where Q(k) = 1 + 2*x - x*(k+3)/(1 - x*(k+1)/Q(k+1)); (continued fraction). - Sergei N. Gladkovskii, Apr 22 2013
a(n) ~ n!*(n+5)*(n+2)/(2*exp(2)). - Vaclav Kotesovec, Jun 15 2013
From Peter Bala, Sep 20 2013: (Start)
a(n) ~ (1/2)*n^2*n!/e^2 for large n.
Letting n -> infinity in the above series for a(n) given by Kotesovec gives the series expansion 1/e^2 = Sum_{k >= 0} (-1)^k*(k+3)*2^(k+3)/(k+5)!.
The sequence b(n) := (1/2)*n!*(n+2)*(n+5) satisfies the recurrence for a(n) given above by Pauly but with the starting values b(0) = 5 and b(1) = 9. This leads to the finite continued fraction expansion a(n) = (1/2)*n!*(n+2)*(n+5)( 1/(5 + 4/(1 + 2/(2 + 4/(3 + ... + 2*(n-1)/n)))) ), valid for n >= 2. Letting n -> infinity in the previous result gives the infinite continued fraction expansion 1/e^2 = 1/(5 + 4/(1 + 2/(2 + 4/(3 + ... + 2*(n-1)/(n + ...))))). Cf. A082031. (End)
a(n) = A087981(n+2)/(2*(n+1)). - Seiichi Manyama, Apr 25 2025

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A010842 Expansion of e.g.f.: exp(2*x)/(1-x).

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

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A052124 Expansion of e.g.f. exp(-2*x)/(1-x)^3.

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