A010844 -id:A010844 - cp's OEIS frontend

A067273 a(n) = n(a(n-1)2+1), a(0) = 0.

0, 1, 6, 39, 316, 3165, 37986, 531811, 8508984, 153161721, 3063234430, 67391157471, 1617387779316, 42052082262229, 1177458303342426, 35323749100272795, 1130359971208729456, 38432239021096801521, 1383560604759484854774, 52575302980860424481431, 2103012119234416979257260, 88326509007845513128804941

Offset: 0

Reinhard Zumkeller, Feb 21 2002

nonn

Harvey P. Dale, Table of n, a(n) for n = 0..403

Cf. A007526.

a := n -> n*hypergeom([1,1-n],[],-2):
seq(simplify(a(n)), n=0..17); # Peter Luschny, May 09 2017

FoldList[2 #1*#2 + #2 &, 0, Range[19]] (* Robert G. Wilson v, Jul 07 2012 *)
a[n_] := 2^(n-1)*Sqrt[E]*n*Gamma[n,1/2];
Table[a[n] // FullSimplify, {n,0,20}] (* Gerry Martens, Jun 28 2015 *)
nxt[{n_,a_}]:={n+1,(n+1)(2*a+1)}; NestList[nxt,{0,0},20][[;;,2]] (* Harvey P. Dale, Jun 26 2023 *)

E.g.f.: x*exp(x)/(1-2*x). a(n) = n!*Sum_{k=1..n} 2^(k-1)/(n-k)! = n*A010844(n-1). - Vladeta Jovovic, Feb 09 2003
a(n) ~ n! * exp(1/2) * 2^(n-1). - Vaclav Kotesovec, Oct 05 2013
a(n) = n*hypergeom([1,1-n], [], -2). - Peter Luschny, May 09 2017
a(n) = Sum_{k=1..n} 2^(k-1)*k!*binomial(n,k). - Ridouane Oudra, Jun 15 2025

More terms from Vladimir Joseph Stephan Orlovsky, Nov 15 2008

A093302 a(n) = (a(n-1)+(2n-1))*(2n) with a(0) = 0.

Original entry on oeis.org

0, 2, 20, 150, 1256, 12650, 151932, 2127230, 34035920, 612646866, 12252937700, 269564629862, 6469551117240, 168208329048890, 4709833213369676, 141294996401091150, 4521439884834917792, 153728956084387206050

Offset: 0

Emrehan Halici (emrehan(AT)halici.com.tr), Apr 24 2004

Altug Alkan, Table of n, a(n) for n = 0..1000

a(n) = A007566(n)-1 = 2*A010844(n)-2n-2. Bisection of A077138.

Cf. A271476.

RecurrenceTable[{a[0]==0,a[n]==(a[n-1]+2n-1)2n},a,{n,20}] (* Harvey P. Dale, May 20 2014 *)

PARI
```
a(n)=2*floor(exp(1/2)*n!*2^n)-2*n-2
```

x='x+O('x^99); concat(0, Vec(serlaplace((2*x+4*x^2)/(1-2*x)*exp(x)))) \\ Altug Alkan, Aug 01 2018

a=vector(99); a[1]=2; for(n=2, #a, a[n] = 2*(a[n-1]+2*n-1)*n); concat(0,a) \\ Altug Alkan, Aug 01 2018

a(n) = 2 * floor(e^(1/2) * n! * 2^n) - 2n - 2.
E.g.f.: (2x+4x^2)/(1-2x) * exp(x).
a(n) = 2*A271476(n) for n >= 1. - Altug Alkan, Aug 01 2018

Edited by Ralf Stephan, Apr 26 2004

A097817 E.g.f. exp(2x)/(1-3x).

Original entry on oeis.org

1, 5, 34, 314, 3784, 56792, 1022320, 21468848, 515252608, 13911820928, 417354628864, 13772702754560, 495817299168256, 19336874667570176, 812148736037963776, 36546693121708402688, 1754241269842003394560

Offset: 0

Paul Barry, Aug 26 2004

Second binomial transform of n!3^n.

Cf. A010844, A097819.

With[{nn=20},CoefficientList[Series[Exp[2x]/(1-3x),{x,0,nn}],x] Range[ 0,nn]!] (* Harvey P. Dale, Apr 02 2020 *)

a(n) = 3n*a(n-1)+2^n, n>0, a(0)=1.
a(n) ~ n! * exp(2/3) * 3^n. - Vaclav Kotesovec, Aug 04 2014
a(n) +(-3*n-2)*a(n-1) +6*(n-1)*a(n-2)=0. - R. J. Mathar, Dec 21 2014
From Peter Bala, Jan 30 2015: (Start)
a(n) = int {x = 0..inf} (3*x + 2)^n*exp(-x) dx.
The e.g.f. y = exp(2*x)/(1 - 3*x) satisfies the differential equation (1 - 3*x)*y' = (5 - 6*x)*y. Mathar's recurrence above follows easily from this.
The sequence b(n) = 3^n*n! also satisfies Mathar's recurrence with b(0) = 1, b(1) = 3. This leads to the continued fraction representation a(n) = 3^n*n!*( 1 + 2/(3 - 6/(8 - 12/(11 - ... - (6*n - 6)/(3*n + 2) )))) for n >= 2. Taking the limit gives the continued fraction representation exp(2/3) = 1 + 2/(3 - 6/(8 - 12/(11 - ... - (6*n - 6)/((3*n + 2) - ... )))). (End)
a(n) = 3^n*exp(2/3)*Gamma(n+1,2/3). - Gerry Martens, Jul 24 2015

A097819 E.g.f. exp(3x)/(1-4x).

Original entry on oeis.org

1, 7, 65, 807, 12993, 260103, 6243201, 174811815, 5593984641, 201383466759, 8055338729409, 354434904271143, 17012875405546305, 884669521090002183, 49541493181044905217, 2972489590862708661927, 190239333815213397410049, 12936274699434511153023495

Offset: 0

Paul Barry, Aug 26 2004

Third binomial transform of n!4^n.

Harvey P. Dale, Table of n, a(n) for n = 0..365

Cf. A010844, A097817.

With[{nn=20},CoefficientList[Series[Exp[3x]/(1-4x),{x,0,nn}],x] Range[0,nn]!] (* Harvey P. Dale, Jun 16 2016 *)

a(n) = 4*a(n-1)+3^n, n>0, a(0)=1.
a(n) + (-4*n-3)*a(n-1) + 12*(n-1)*a(n-2) = 0. - R. J. Mathar, Dec 21 2014
From Peter Bala, Jan 30 2015: (Start)
a(n) = int {x = 0..inf} (4*x + 3)^n*exp(-x) dx.
a(n) ~ 4^n*n!*exp(3/4).
The e.g.f. y = exp(3*x)/(1 - 4*x) satisfies the differential equation (1 - 4*x)*y' = (7 - 12*x)*y. Mathar's recurrence above follows easily from this.
The sequence b(n) := 4^n*n! also satisfies Mathar's recurrence with b(0) = 1, b(1) = 4. This leads to the continued fraction representation a(n) = 4^n*n!*( 1 + 3/(4 - 12/(11 - 24/(15 - ... - (12*n - 12)/(4*n + 3) )))) for n >= 2. Taking the limit gives the continued fraction representation exp(3/4) = 1 + 3/(4 - 12/(11 - 24/(15 - ... - (12*n - 12)/((4*n + 3) - ... )))). (End)
a(n) = 4^n*exp(3/4)*Gamma(n+1,3/4). - Gerry Martens, Jul 24 2015

A193229 A double factorial triangle.

Original entry on oeis.org

1, 1, 1, 3, 3, 2, 15, 15, 12, 6, 105, 105, 90, 60, 24, 945, 945, 840, 630, 360, 120, 10395, 10395, 9450, 7560, 5040, 2520, 720, 135135, 135135, 124740, 103950, 75600, 45360, 20160, 5040, 2027025, 2027025, 1891890, 1621620, 1247400, 831600, 453600, 181440, 40320

Offset: 0

Gary W. Adamson, Jul 18 2011

The double factorial triangle coefficients are T(n,k), n >= 0 and 0 <= k <= n.
The T(n,0) equal the double factorial numbers A001147(n) = (2*n-1)!!.
The T(n,n) equal the factorial numbers A000142(n) = n!.
The row sums equal the double factorial numbers A000165(n) = (2*n)!!.
The Kn21(n) sums, see A180662 for the definition of these and other triangle sums, equal A130905(n) while the Kn2p(n) sums equal A130905(n+2*p-2) - (n+2*p-2)!*A010844(p-2)/A000165(p-2), p >= 2. - Johannes W. Meijer, Jul 21 2011

			The first few rows of matrix M[i,j] are:
  1, 1, 0, 0, 0, 0, ...
  2, 2, 2, 0, 0, 0, ...
  3, 3, 3, 3, 0, 0, ...
  4, 4, 4, 4, 4, 0, ...
  5, 5, 5, 5, 5, 5, ...
The first few rows of triangle T(n,k) are:
       1;
       1,      1;
       3,      3,      2;
      15,     15,     12,      6;
     105,    105,     90,     60,    24;
     945,    945,    840,    630,   360,   120;
   10395,  10395,   9450,   7560,  5040,  2520,   720;
  135135, 135135, 124740, 103950, 75600, 45360, 20160, 5040;

Seiichi Manyama, Rows n = 0..139, flattened

Cf. A000142, A000165, A001147, A102625.
Cf. A180662, A130905, A010844.
T(2*n,n) gives A166334.

nmax:=7: M := Matrix(1..nmax+1,1..nmax+1): for i from 1 to nmax do for j from 1 to i+1 do M[i,j] := i od: od: for n from 0 to nmax do B := M^n: for k from 0 to n do T(n,k) := B[1,k+1] od: od: for n from 0 to nmax do seq(T(n,k),k=0..n) od: seq(seq(T(n,k),k=0..n),n=0..nmax); # Johannes W. Meijer, Jul 21 2011

row(n)=(matrix(n,n,i,j,(i>j-2)*i)^(n-1))[1,]  \\ M. F. Hasler, Jul 24 2011

T(n,k) = the (k+1)-th term in the top row of M^n, where M is an infinite square production matrix; M[i,j] = i, i >= 1 and 1 <= j <= i+1, and M[i,j] = 0, i >= 1 and j >= i+2, see the examples.

It appears that T(n,k) = (2*n-k)!/(2^(n-k)*(n-k)!) with conjectural e.g.f. 1/(x*(1-2*z) + (1-x)*sqrt(1-2*z)) = 1 + (1+x)*z + (3+3*x+2*x^2)*z^2/2! + .... Cf. A102625. - Peter Bala, Jul 09 2012

Corrected, edited and extended by Johannes W. Meijer, Jul 21 2011

More terms from Seiichi Manyama, Apr 06 2019

A295100 a(n) = n! * [x^n] exp(nx)/(1 - 2x).

Original entry on oeis.org

1, 3, 20, 201, 2688, 44815, 894528, 20792205, 551518208, 16438822587, 543934387200, 19783668211153, 784536321392640, 33689132092480839, 1557397919735103488, 77117362592836807125, 4072280214605427376128, 228441851811771488284915, 13566762607790788699226112, 850372121882700252639269337

Offset: 0

Ilya Gutkovskiy, Nov 14 2017

nonn

The n-th term of the n-th binomial transform of A000165.

Robert Israel, Table of n, a(n) for n = 0..375
N. J. A. Sloane, Transforms
Index entries for sequences related to factorial numbers

Cf. A000165, A010844, A063170, A082032, A295098, A295099.

S:= series(exp(n*x)/(1-2*x),x,51):
seq(n!*coeff(S,x,n),n=0..50); # Robert Israel, Nov 14 2017

Table[n! SeriesCoefficient[Exp[n x]/(1 - 2 x), {x, 0, n}], {n, 0, 19}]

a(n) ~ 2^n * exp(n/2) * n!. - Vaclav Kotesovec, Nov 14 2017

a(n) = n! * Sum_{k=0..n} n^k*2^(n-k)/k! = 2^n*Gamma(n+1, n/2)*exp(n/2). - Robert Israel, Nov 14 2017

A331689 E.g.f.: exp(x/(1 - x)) / (1 - 2*x).

Original entry on oeis.org

1, 3, 15, 103, 897, 9471, 117703, 1685475, 27361953, 497111707, 10001175231, 220849928223, 5312868439585, 138337555830423, 3876986580776247, 116375171226474331, 3725295913465848513, 126686907674290095795, 4561317309742758852463, 173343622143918424951767

Offset: 0

Ilya Gutkovskiy, Jan 24 2020

nonn

Alois P. Heinz, Table of n, a(n) for n = 0..403

Cf. A000165, A000262, A000522, A002720, A010842, A010844, A331688.

b:= proc(n) b(n):= `if`(n<0, 0, 1+n*b(n-1)) end:
a:= n-> n!*add(binomial(n, k)*b(k)/k!, k=0..n):
seq(a(n), n=0..23);  # Alois P. Heinz, Jan 24 2020

nmax = 19; CoefficientList[Series[Exp[x/(1 - x)]/(1 - 2 x), {x, 0, nmax}], x] Range[0, nmax]!
A000522[0] = 1; A000522[n_] := Floor[Exp[1] n!]; a[n_] := Sum[Binomial[n, k]^2 k! A000522[n - k], {k, 0, n}]; Table[a[n], {n, 0, 19}]

a(n) = Sum_{k=0..n} binomial(n,k)^2 * k! * A000522(n-k).
a(n) = Sum_{k=0..n} binomial(n,k) * k! * 2^k * A000262(n-k).
a(n) ~ n! * exp(1) * 2^n. - Vaclav Kotesovec, Jan 26 2020

A073474 Triangle T(n,k) read by rows, where o.g.f. for T(n,k) is n!*Sum_{k=0..n} (1+x)^(n-k)/k!.

Original entry on oeis.org

1, 2, 1, 5, 6, 2, 16, 33, 24, 6, 65, 196, 228, 120, 24, 326, 1305, 2120, 1740, 720, 120, 1957, 9786, 20550, 23160, 14760, 5040, 720, 13700, 82201, 212352, 305970, 265440, 138600, 40320, 5040, 109601, 767208, 2356424, 4146576, 4571280, 3232320, 1431360, 362880, 40320

Offset: 0

Vladeta Jovovic, Aug 26 2002

Row sums give A010844.

			Triangle begins:
    1;
    2,    1;
    5,    6,    2;
   16,   33,   24,    6;
   65,  196,  228,  120,  24;
  326, 1305, 2120, 1740, 720, 120;
  ...

Alois P. Heinz, Rows n = 0..140, flattened

Cf. A000142, A000522, A073107, A010844 (row sums).

G:=simplify(series(exp(x)/(1-x-x*y),x=0,13)): P[0]:=1: for n from 1 to 11 do P[n]:=sort(n!*coeff(G,x^n)) od: seq(seq(coeff(y*P[n],y^k),k=1..n+1),n=0..9);
# second Maple program:
b:= proc(n, k) option remember; `if`(k>n, 0, `if`(k=0, 1,
      n*(b(n-1, k-1)+b(n-1, k))))
    end:
T:= (n, k)-> b(n+1, k+1)/(n+1):
seq(seq(T(n, k), k=0..n), n=0..9);  # Alois P. Heinz, Sep 12 2019

b[n_, k_] := b[n, k] = If[k>n, 0, If[k==0, 1, n (b[n-1, k-1]+b[n-1, k])]];
T[n_, k_] := b[n+1, k+1]/(n+1);
Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Dec 09 2019, after Alois P. Heinz *)
T[n_, k_] := Sum[Binomial[j, k] FactorialPower[n, j], {j, 0, n}]; (* Peter Luschny, Oct 16 2024 *)

def T(n, k): return sum(binomial(j, k) * falling_factorial(n, j) for j in range(n+1))
for n in range(8): print([T(n, k) for k in range(n+1)])
# Peter Luschny, Oct 16 2024

E.g.f.: exp(x)/(1-x-x*y). - Vladeta Jovovic, Oct 17 2003

T(n, k) = Sum_{j=0..n} binomial(j, k)*FallingFactorial(n, j). - Peter Luschny, Oct 16 2024

Edited by Emeric Deutsch, Jun 10 2004

A111139 a(n) = n!*Sum_{k=0..n} Fibonacci(k)/k!.

Original entry on oeis.org

0, 1, 3, 11, 47, 240, 1448, 10149, 81213, 730951, 7309565, 80405304, 964863792, 12543229529, 175605213783, 2634078207355, 42145251318667, 716469272418936, 12896446903543432, 245032491167329389, 4900649823346594545

Offset: 0

Vladeta Jovovic, Oct 17 2005

Eigensequence of a triangle with the Fibonacci series as the left border, the natural numbers (1, 2, 3, ...) as the right border; and the rest zeros. - Gary W. Adamson, Aug 01 2016

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Eric Weisstein's MathWorld, Incomplete Gamma Function.
Eric Weisstein's MathWorld, Fibonacci Number.
Eric Weisstein's MathWorld, Golden Ratio.

Cf. A000045, A263823.

Cf. A009102, A009551, A000142, A000166, A000522, A000023, A053486, A010844 (incomplete Gamma function values at other points).

a:=n->sum(fibonacci (j)*n!/j!,j=0..n):seq(a(n),n=0..20); # Zerinvary Lajos, Mar 19 2007

f[n_] := n!*Sum[Fibonacci[k]/k!, {k, 0, n}]; Table[ f[n], {n, 0, 20}] (* or *)
Simplify[ Range[0, 20]!CoefficientList[ Series[2/Sqrt[5]*Exp[x/2]*Sinh[Sqrt[5]*x/2]/(1 - x), {x, 0, 20}], x]] (* Robert G. Wilson v, Oct 21 2005 *)
Module[{nn=20,fibs,fct},fct=Range[0,nn]!;fibs=Accumulate[ Fibonacci[ Range[ 0,nn]]/fct];Times@@@Thread[{fct,fibs}]] (* Harvey P. Dale, Feb 19 2014 *)
Round@Table[(E^GoldenRatio Gamma[n+1, GoldenRatio] - E^(1-GoldenRatio) Gamma[n+1, 1-GoldenRatio])/Sqrt[5], {n, 0, 20}] (* Vladimir Reshetnikov, Oct 27 2015 *)

vector(100, n, n--; n!*sum(k=0, n, fibonacci(k)/k!)) \\ Altug Alkan, Oct 28 2015

E.g.f.: (2/sqrt(5))*exp(x/2)*sinh(sqrt(5)*x/2)/(1-x).
Recurrence: a(n) = (n+1)*a(n-1) - (n-2)*a(n-2) - (n-2)*a(n-3). - Vaclav Kotesovec, Oct 18 2012
a(n) ~ 2*sqrt(e/5)*sinh(sqrt(5)/2)*n!. - Vaclav Kotesovec, Oct 18 2012
From Vladimir Reshetnikov, Oct 27 2015: (Start)
Let phi=(1+sqrt(5))/2.
a(n) = (phi^n*hypergeom([1,-n], [], 1-phi)-(1-phi)^n*hypergeom([1,-n], [], phi))/sqrt(5).
a(n) = (exp(phi)*Gamma(n+1, phi)-exp^(1-phi)*Gamma(n+1, 1-phi))/sqrt(5), where Gamma(a, x) is the upper incomplete Gamma function.
Gamma(n+1, phi)*exp(phi) = a(n)*phi + A263823(n).
a(n) ~ exp(phi-n)*n^(n+1/2)*sqrt(2*Pi/5)*(1-exp(-sqrt(5))).
(End)

A136807 Hankel transform of double factorial numbers n!*2^n=A000165(n).

Original entry on oeis.org

1, 4, 256, 589824, 86973087744, 1282470362637926400, 2723154477021188283432960000, 1133321924829207204666583887642624000000, 120746421332702772771144114237731253721340313600000000

Offset: 0

Paul Barry, Jan 23 2008

By the properties of the Hankel transform, a(n)=2^(n(n+1))*A055209(n).

Also Hankel transform of A000354, A010844, A082032. - Philippe Deléham, Jan 23 2008

G. C. Greubel, Table of n, a(n) for n = 0..28
Michael Z. Spivey and Laura L. Steil, The k-Binomial Transforms and the Hankel Transform, Journal of Integer Sequences, Vol. 9 (2006), Article 06.1.1.

Cf. A000354, A010844, A082032, A055209.

[1] cat [(&*[(2*k)^(2*(n-k+1)): k in [1..n]]): n in [1..10]]; // G. C. Greubel, Oct 14 2018

Table[Product[(2k)^(2(n-k+1)),{k,n}],{n,0,10}] (* Harvey P. Dale, Apr 11 2013 *)

for(n=0,10, print1(prod(k=1,n,(2*k)^(2*(n-k+1))), ", ")) \\ G. C. Greubel, Oct 14 2018

a(n) = Product_{k=1..n} (2k)^(2(n-k+1)).

a(n) ~ 2^((n+1)^2) * Pi^(n+1) * n^(n^2 + 2*n + 5/6) / (A^2 * exp(3*n^2/2 + 2*n - 1/6)), where A is the Glaisher-Kinkelin constant A074962. - Vaclav Kotesovec, Feb 24 2019

cp's OEIS Frontend

A067273 a(n) = n*(a(n-1)*2+1), a(0) = 0.

Views

Author

Keywords

Links

Crossrefs

Programs

Maple

Mathematica

Formula

Extensions

A093302 a(n) = (a(n-1)+(2n-1))*(2n) with a(0) = 0.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

PARI

PARI

PARI

Formula

Extensions

A097817 E.g.f. exp(2x)/(1-3x).

Views

Author

Keywords

Comments

Crossrefs

Programs

Mathematica

Formula

A097819 E.g.f. exp(3x)/(1-4x).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Formula

A193229 A double factorial triangle.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

PARI

Formula

Extensions

A295100 a(n) = n! * [x^n] exp(n*x)/(1 - 2*x).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

Formula

A331689 E.g.f.: exp(x/(1 - x)) / (1 - 2*x).

Views

Author

Keywords

Links

Crossrefs

Programs

Maple

Mathematica

Formula

A073474 Triangle T(n,k) read by rows, where o.g.f. for T(n,k) is n!*Sum_{k=0..n} (1+x)^(n-k)/k!.

Views

A067273 a(n) = n(a(n-1)2+1), a(0) = 0.

A295100 a(n) = n! * [x^n] exp(nx)/(1 - 2x).