cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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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

Views

Author

Paul Barry, Apr 02 2003

Keywords

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)

Crossrefs

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

A132159 Lower triangular matrix T(n,j) for double application of an iterated mixed order Laguerre transform inverse to A132014. Coefficients of Laguerre polynomials (-1)^n * n! * L(n,-2-n,x).

Original entry on oeis.org

1, 2, 1, 6, 4, 1, 24, 18, 6, 1, 120, 96, 36, 8, 1, 720, 600, 240, 60, 10, 1, 5040, 4320, 1800, 480, 90, 12, 1, 40320, 35280, 15120, 4200, 840, 126, 14, 1, 362880, 322560, 141120, 40320, 8400, 1344, 168, 16, 1, 3628800, 3265920, 1451520, 423360, 90720, 15120
Offset: 0

Views

Author

Tom Copeland, Nov 01 2007

Keywords

Comments

The matrix operation b = T*a can be characterized several ways in terms of the coefficients a(n) and b(n), their o.g.f.'s A(x) and B(x), or their e.g.f.'s EA(x) and EB(x).
1) b(n) = n! Lag[n,(.)!*Lag[.,a1(.),-1],0], umbrally,
where a1(n) = n! Lag[n,(.)!*Lag[.,a(.),-1],0]
2) b(n) = (-1)^n * n! * Lag(n,a(.),-2-n)
3) b(n) = Sum_{j=0..n} (-1)^j * binomial(n,j) * binomial(-2,j) * j! * a(n-j)
4) b(n) = Sum_{j=0..n} binomial(n,j) * (j+1)! * a(n-j)
5) B(x) = (1-xDx))^(-2) A(x), formally
6) B(x) = Sum_{j>=0} (-1)^j * binomial(-2,j) * (xDx)^j A(x)
= Sum_{j>=0} (j+1) * (xDx)^j A(x)
7) B(x) = Sum_{j>=0} (j+1) * x^j * D^j * x^j A(x)
8) B(x) = Sum_{j>=0} (j+1)! * x^j * Lag(j,-:xD:,0) A(x)
9) EB(x) = Sum_{j>=0} x^j * Lag[j,(.)! * Lag[.,a1(.),-1],0]
10) EB(x) = Sum_{j>=0} Lag[j,a1(.),-1] * (-x)^j / (1-x)^(j+1)
11) EB(x) = Sum_{j>=0} x^n * Sum_{j=0..n} (j+1)!/j! * a(n-j) / (n-j)!
12) EB(x) = Sum_{j>=0} (-x)^j * Lag[j,a(.),-2-j]
13) EB(x) = exp(a(.)*x) / (1-x)^2 = (1-x)^(-2) * EA(x)
14) T = A094587^2 = A132013^(-2) = A132014^(-1)
where Lag(n,x,m) are the Laguerre polynomials of order m, D the derivative w.r.t. x and (:xD:)^j = x^j * D^j. Truncating the D operator series at the j = n term gives an o.g.f. for b(0) through b(n).
c = (1!,2!,3!,4!,...) is the sequence associated to T under the list partition transform and associated operations described in A133314. Thus T(n,k) = binomial(n,k)*c(n-k) . c are also the coefficients in formulas 4 and 8.
The reciprocal sequence to c is d = (1,-2,2,0,0,0,...), so the inverse of T is TI(n,k) = binomial(n,k)*d(n-k) = A132014. (A121757 is the reverse of T.)
These formulas are easily generalized for m applications of the basic operator n! Lag[n,(.)!*Lag[.,a(.),-1],0] by replacing 2 by m in formulas 2, 3, 5, 6, 12, 13 and 14, or (j+1)! by (m-1+j)!/(m-1)! in 4, 8 and 11. For further discussion of repeated applications of T, see A132014.
The row sums of T = [formula 4 with a(n) all 1] = [binomial transform of c] = [coefficients of B(x) with A(x) = 1/(1-x)] = A001339. Therefore the e.g.f. of A001339 = [formula 13 with a(n) all 1] = exp(x)*(1-x)^(-2) = exp(x)*exp[c(.)*x)] = exp[(1+c(.))*x].
Note the reciprocal is 1/{exp[(1+c(.))*x]} = exp(-x)*(1-x)^2 = e.g.f. of signed A002061 with leading 1 removed], which makes A001339 and the signed, shifted A002061 reciprocal arrays under the list partition transform of A133314.
The e.g.f. for the row polynomials (see A132382) implies they form an Appell sequence (see Wikipedia). - Tom Copeland, Dec 03 2013
As noted in item 12 above and reiterated in the Bala formula below, the e.g.f. is e^(x*t)/(1-x)^2, and the Poisson-Charlier polynomials P_n(t,y) have the e.g.f. (1+x)^y e^(-xt) (Feinsilver, p. 5), so the row polynomials R_n(t) of this entry are (-1)^n P_n(t,-2). The associated Appell sequence IR_n(t) that is the umbral compositional inverse of this entry's polynomials has the e.g.f. (1-x)^2 e^(xt), i.e., the e.g.f. of A132014 (noted above), and, therefore, the row polynomials (-1)^n PC(t,2). As umbral compositional inverses, R_n(IR.(t)) = t^n = IR_n(R.(t)), where, by definition, P.(t)^n = P_n(t), is the umbral evaluation. - Tom Copeland, Jan 15 2016
T(n,k) is the number of ways to place (n-k) rooks in a 2 x (n-1) Ferrers board (or diagram) under the Goldman-Haglund i-row creation rook mode for i=2. Triangular recurrence relation is given by T(n,k) = T(n-1,k-1) + (n+1-k)*T(n-1,k). - Ken Joffaniel M. Gonzales, Jan 21 2016

Examples

			First few rows of the triangle are
    1;
    2,  1;
    6,  4,  1;
   24, 18,  6, 1;
  120, 96, 36, 8, 1;
		

Crossrefs

Columns: A000142 (k=0), A001563 (k=1), A001286 (k=2), A005990 (k=3), A061206 (k=4), A062199 (k=5), A062148 (k=6).

Programs

  • Haskell
    a132159 n k = a132159_tabl !! n !! k
    a132159_row n = a132159_tabl !! n
    a132159_tabl = map reverse a121757_tabl
    -- Reinhard Zumkeller, Mar 06 2014
    
  • Magma
    /* As triangle */ [[Binomial(n,k)*Factorial(n-k+1): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Feb 10 2016
    
  • Maple
    T := proc(n,k) return binomial(n,k)*factorial(n-k+1): end: seq(seq(T(n,k),k=0..n),n=0..10); # Nathaniel Johnston, Sep 28 2011
  • Mathematica
    nn=10;f[list_]:=Select[list,#>0&];Map[f,Range[0,nn]!CoefficientList[Series[Exp[y x]/(1-x)^2,{x,0,nn}],{x,y}]]//Grid  (* Geoffrey Critzer, Feb 15 2013 *)
  • Sage
    flatten([[binomial(n,k)*factorial(n-k+1) for k in (0..n)] for n in (0..15)]) # G. C. Greubel, May 19 2021

Formula

T(n,k) = binomial(n,k)*c(n-k).
From Peter Bala, Jul 10 2008: (Start)
T(n,k) = binomial(n,k)*(n-k+1)!.
T(n,k) = (n-k+1)*T(n-1,k) + T(n-1,k-1).
E.g.f.: exp(x*y)/(1-y)^2 = 1 + (2+x)*y + (6+4*x+x^2)*y^2/2! + ... .
This array is the particular case P(2,1) of the generalized Pascal triangle P(a,b), a lower unit triangular matrix, shown below:
n\k|0....................1...............2.........3.....4
----------------------------------------------------------
0..|1.....................................................
1..|a....................1................................
2..|a(a+b)...............2a..............1................
3..|a(a+b)(a+2b).........3a(a+b).........3a........1......
4..|a(a+b)(a+2b)(a+3b)...4a(a+b)(a+2b)...6a(a+b)...4a....1
...
See A094587 for some general properties of these arrays.
Other cases recorded in the database include: P(1,0) = Pascal's triangle A007318, P(1,1) = A094587, P(2,0) = A038207, P(3,0) = A027465, P(1,3) = A136215 and P(2,3) = A136216. (End)
Let f(x) = (1/x^2)*exp(-x). The n-th row polynomial is R(n,x) = (-x)^n/f(x)*(d/dx)^n(f(x)), and satisfies the recurrence equation R(n+1,x) = (x+n+2)*R(n,x)-x*R'(n,x). Cf. A094587. - Peter Bala, Oct 28 2011
Exponential Riordan array [1/(1 - y)^2, y]. The row polynomials R(n,x) thus form a Sheffer sequence of polynomials with associated delta operator equal to d/dx. Thus d/dx(R(n,x)) = n*R(n-1,x). The Sheffer identity is R(n,x + y) = Sum_{k=0..n} binomial(n,k)*y^(n-k)*R(k,x). Define a polynomial sequence P(n,x) of binomial type by setting P(n,x) = Product_{k = 0..n-1} (2*x + k) with the convention that P(0,x) = 1. Then the present triangle is the triangle of connection constants when expressing the basis polynomials P(n,x + 1) in terms of the basis P(n,x). For example, row 3 is (24, 18, 6, 1) so P(3,x + 1) = (2*x + 2)*(2*x + 3)*(2*x + 4) = 24 + 18*(2*x) + 6*(2*x)*(2*x + 1) + (2*x)*(2*x + 1)*(2*x + 2). Matrix square of triangle A094587. - Peter Bala, Aug 29 2013
From Tom Copeland, Apr 21 2014: (Start)
T = (I-A132440)^(-2) = {2*I - exp[(A238385-I)]}^(-2) = unsigned exp[2*(I-A238385)] = exp[A005649(.)*(A238385-I)], umbrally, where I = identity matrix.
The e.g.f. is exp(x*y)*(1-y)^(-2), so the row polynomials form an Appell sequence with lowering operator D=d/dx and raising operator x+2/(1-D).
With L(n,m,x) = Laguerre polynomials of order m, the row polynomials are (-1)^n * n! * L(n,-2-n,x) = (-1)^n*(-2!/(-2-n)!)*K(-n,-2-n+1,x) where K is Kummer's confluent hypergeometric function (as a limit of n+s as s tends to zero).
Operationally, (-1)^n*n!*L(n,-2-n,-:xD:) = (-1)^n*x^(n+2)*:Dx:^n*x^(-2-n) = (-1)^n*x^2*:xD:^n*x^(-2) = (-1)^n*n!*binomial(xD-2,n) = (-1)^n*n!*binomial(-2,n)*K(-n,-2-n+1,-:xD:) where :AB:^n = A^n*B^n for any two operators. Cf. A235706.
The generalized Pascal triangle Bala mentions is a special case of the fundamental generalized factorial matrices in A133314. (End)
From Peter Bala, Jul 26 2021: (Start)
O.g.f: 1/y * Sum_{k >= 0} k!*( y/(1 - x*y) )^k = 1 + (2 + x)*y + (6 + 4*x + x^2)*y^2 + ....
First-order recurrence for the row polynomials: (n - x)*R(n,x) = n*(n - x + 1)*R(n-1,x) - x^(n+1) with R(0,x) = 1.
R(n,x) = (x + n + 1)*R(n-1,x) - (n - 1)*x*R(n-2,x) with R(0,x) = 1 and R(1,x) = 2 + x.
R(n,x) = A087981 (x = -2), A000255 (x = -1), A000142 (x = 0), A001339 (x = 1), A081923 (x = 2) and A081924 (x = 3). (End)

Extensions

Formula 3) in comments corrected by Tom Copeland, Apr 20 2014
Title modified by Tom Copeland, Apr 23 2014

A095000 E.g.f.: exp(x)/(1-x)^4.

Original entry on oeis.org

1, 5, 29, 193, 1457, 12341, 116125, 1203329, 13627073, 167525317, 2222710781, 31665408545, 482196718129, 7817359305653, 134443910166077, 2444991262876321, 46883166605035265, 945426638499719429, 20002372214708227933, 443036881445294292737, 10252840082607606694961
Offset: 0

Views

Author

Philippe Deléham, Jun 19 2004

Keywords

Comments

Sum_{k=0..n} A094816(n,k)*x^k give A000522(n), A001339(n), A082030(n) for x = 1, 2, 3 respectively.
From Peter Bala, Jul 10 2008: (Start)
Recurrence relation: a(0) = 1, a(1) = 5, a(n) = (n+4)*a(n-1) - (n-1)*a(n-2) for n >= 2. Let p_3(n) = n^3+2*n-1 = n^(3)-3*n^(2)+3*n^(1)-1, where n^(k) denotes the rising factorial n*(n+1)*...*(n+k-1). The polynomial p_3(n) is an example of a Poisson-Charlier polynomial c_k(x;a) at k = 3, x = -n and a = -1.
The sequence b(n) := n!*p_3(n+1) = A001565(n) satisfies the same recurrence as a(n) but with the initial conditions b(0) = 2, b(1) = 11. This leads to the finite continued fraction expansion a(n)/b(n) = 1/(2+1/(5-1/(6-2/(7-...-(n-1)/(n+4))))).
Lim_{n -> infinity} a(n)/b(n) = e/6 = 1/(2+1/(5-1/(6-2/(7-...-n/((n+5)-...))))).
a(n) = -b(n) * Sum_{k = 0..n} 1/(k!*p_3(k)*p_3(k+1)) - since the rhs satisfies the above recurrence with the same initial conditions. Hence e = -6 * Sum_{k>=0} 1/(k!*p_3(k)*p_3(k+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), A082030 (r=2) and A095177 (r=4).
{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. (End)

Crossrefs

Programs

  • Maple
    a := n -> hypergeom([4, -n], [], -1); seq(round(evalf(a(n), 100)), n=0..18); # Peter Luschny, Sep 20 2014
  • Mathematica
    Table[n!*SeriesCoefficient[E^(x)/(1-x)^4,{x,0,n}],{n,0,20}] (* Vaclav Kotesovec, Oct 14 2012 *)
  • PARI
    x='x+O('x^66); Vec(serlaplace(exp(x)/(1-x)^4)) \\ Joerg Arndt, May 11 2013

Formula

a(n) = Sum_{k=0..n} A094816(n, k)*4^k.
a(n) = Sum_{k=0..n} binomial(n, k)*(k+3)!/6.
a(n) ~ n!*n^3*e/6. - Vaclav Kotesovec, Oct 14 2012
a(n) = hypergeom([4, -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^3 + 3*n^2 + 5*n + 2 = A001565(n). - Peter Bala, Jul 26 2021
D-finite with recurrence a(n) +(-n-4)*a(n-1) +(n-1)*a(n-2)=0. - R. J. Mathar, Aug 01 2022
a(n) = A001341(n)/6. - Alois P. Heinz, Jan 17 2025

A095177 E.g.f.: exp(x)/(1-x)^5.

Original entry on oeis.org

1, 6, 41, 316, 2721, 25946, 271801, 3105936, 38474561, 513796366, 7360674441, 112632827396, 1833790646881, 31656637715106, 577636838177561, 11109543835539736, 224635867973671041, 4764236394052127126
Offset: 0

Views

Author

Philippe Deléham, Jun 20 2004

Keywords

Comments

Sum_{k = 0..n} A094816(n,k)*x^k give A000522(n), A001339(n), A082030(n), A095000(n) for x = 1, 2, 3, 4 respectively.
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) = 6, a(n) = (n+5)*a(n-1) - (n-1)*a(n-2) for n >= 2. Let p_4(n) = n^4+2*n^3+5*n^2+1 = n^(4)-4*n^(3)+6*n^(2)-4*n^(1)+1, where n^(k) denotes the rising factorial n*(n+1)*...*(n+k-1). The polynomial p_4(n) is an example of a Poisson-Charlier polynomial c_k(x;a) at k = 4, x = -n and a = -1.
The sequence b(n) := n!*p_4(n+1) = A001688(n) satisfies the same recurrence as a(n) but with the initial conditions b(0) = 9, b(1) = 53. This leads to the finite continued fraction expansion expansion a(n)/b(n) = 1/(9-1/(6-1/(7-2/(8-...-(n-1)/(n+5))))).
Lim n -> infinity a(n)/b(n) = e/24 = 1/(9-1/(6-1/(7-2/(8-...-n/((n+6)-...))))).
a(n) = b(n) * sum {k = 0..n} 1/(k!*p_4(k)*p_4(k+1)) - since the rhs satisfies the above recurrence with the same initial conditions. Hence e = 24 * sum {k = 0..inf} 1/(k!*p_4(k)p_4(k+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), A082030 (r=2), A095000 (r=3). (End)

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[Exp[x]/(1-x)^5, {x, 0, 20}], x]* Range[0, 20]! (* Vaclav Kotesovec, Jun 21 2013 *)
    Table[HypergeometricPFQ[{5, -n}, {}, -1], {n, 0, 20}] (* Benedict W. J. Irwin, May 27 2016 *)
  • PARI
    a(n) = sum(k=0,n, binomial(n, k)*(k+4)!/4! ); \\ Joerg Arndt, Apr 22 2013

Formula

a(n) = Sum_{k = 0..n} A094816(n, k)*5^k.
a(n) = Sum_{k=0..n} binomial(n, k)*(k+4)!/4!.
G.f.: 1/Q(0), where Q(k) = 1 - x - x*(k+5)/(1 - x*(k+1)/Q(k+1)); (continued fraction). - Sergei N. Gladkovskii, Apr 22 2013
a(n) ~ n! *exp(1)*n^4/24. - Vaclav Kotesovec, Jun 21 2013
a(n) = 2F0(5,-n;;-1). - Benedict W. J. Irwin, May 27 2016
First-order recurrence: P(n-1)*a(n) = n*P(n)*a(n-1) + 1 with a(0) = 1, where P(n) = n^4 + 6*n^3 + 17*n^2 + 20*n + 9 = A094793(n). - Peter Bala, Jul 26 2021

A061206 a(n) = total number of occurrences of the consecutive pattern 1324 in all permutations of [n+3].

Original entry on oeis.org

1, 10, 90, 840, 8400, 90720, 1058400, 13305600, 179625600, 2594592000, 39956716800, 653837184000, 11333177856000, 207484333056000, 4001483566080000, 81096733605888000, 1723305589125120000, 38318206628782080000, 889833909490606080000, 21543347282404147200000
Offset: 1

Views

Author

Melvin J. Knight (knightmj(AT)juno.com), May 30 2001

Keywords

Comments

a(n) is the number of sequences of n+3 balls colored with at most n colors such that exactly four balls are the same color as some other ball in the sequence. - Jeremy Dover, Sep 27 2017

Examples

			a(4)=840 because 4*(7!)/24 = 4*7*6*5 = 840.
		

Crossrefs

Programs

  • Magma
    [n*Factorial(n+3)/24: n in [1..20]]; // Vincenzo Librandi, Oct 11 2011
    
  • Maple
    a := n -> n!*binomial(-n,4): seq(a(n),n=1..20); # Peter Luschny, Apr 29 2016
  • Mathematica
    Array[# (# + 3)!/24 &, 20] (* or *) Array[#!*Binomial[-#, 4] &, 20] (* Michael De Vlieger, Sep 30 2017 *)
  • PARI
    a(n) = n*(n+3)!/24; \\ Altug Alkan, Oct 08 2017
  • Sage
    [binomial(n,4)*factorial (n-3) for n in range(4, 21)] # Zerinvary Lajos, Jul 07 2009
    

Formula

a(n) = n*(n+3)!/24.
If we define f(n,i,x) = Sum_{k=i..n} Sum_{j=i..k} binomial(k,j)*Stirling1(n,k)*Stirling2(j,i) * x^(k-j), then a(n-3) = (-1)^n*f(n,4,-2), (n >= 4). - Milan Janjic, Mar 01 2009
E.g.f.: x/(1-x)^5. (This was initiated by e-mail exchange with Gary Detlefs.) - Wolfdieter Lang, May 28 2010
a(n) = ((n+4)!/6) * Sum_{k=1..n} (k+2)!/(k+4)!. - Gary Detlefs, Aug 05 2010
a(n) = Sum_{k>0} k * A264173(n+3,k). - Alois P. Heinz, Nov 06 2015
a(n) = n!*binomial(-n,4). - Peter Luschny, Apr 29 2016
From Amiram Eldar, Sep 24 2022: (Start)
Sum_{n>=1} 1/a(n) = 118/3 - 16*e - 4*gamma + 4*Ei(1), where gamma is Euler's constant (A001620) and Ei(1) is the exponential integral at 1 (A091725).
Sum_{n>=1} (-1)^(n+1)/a(n) = 2/3 - 8/e + 4*gamma - 4*Ei(-1), where -Ei(-1) is the negated exponential integral at -1 (A099285). (End)

Extensions

More terms from Jason Earls, Jun 12 2001
Corrected by Zerinvary Lajos, Jul 07 2009
More precise definition from Alois P. Heinz, Nov 06 2015

A064618 Stirling transform of (n!)^2.

Original entry on oeis.org

1, 1, 5, 49, 821, 21121, 775205, 38516689, 2490976661, 203419086241, 20474978755205, 2490729330118129, 360263844701062901, 61114158974786823361, 12017074366801186956005, 2711409826920884006692369, 695820350706240448128979541, 201526362605605903609254528481
Offset: 0

Views

Author

Karol A. Penson, Sep 26 2001

Keywords

Comments

From Thomas Wieder, Oct 21 2004: (Start)
"Also the number of hierarchies with labeled elements and labeled levels where the levels are permuted. Let l_x denote level x, e.g. l_2 is level 2. Let 1 denote an element and 2 a second element and so on. Then l_1:123 means elements 1,2 and 3 are on level 1.
"Let | indicate separation between levels. Then l_1:1|l_2:346|l_3:5 denotes a hierarchy of n=6 unlabeled elements with element 1 on level 1, elements 3,4 and 6 on level 2 and element 5 on level 3.
"E.g. for n=3 one has a(3) = 49 possible hierarchies:
"l_1:123,
"l_1:12|l_2:3, l_1:13|l_2:2, l_1:23|l_2:1,
"l_2:12|l_1:3, l_2:13|l_1:2, l_2:23|l_1:1,
"l_1:1|l_2:23, l_1:2|l_2:13, l_1:3|l_2:12,
"l_2:1|l_1:23, l_2:2|l_1:13, l_2:3|l_1:12,
"l_1:1|l_2:2|l_3:3 and further five permutations of the elements with levels fixed,
"l_3:1|l_1:2|l_2:3 and further five permutations of the elements with levels fixed,. etc., up to
"l_3:1|l_2:2|l_1:3 and further five permutations of the elements with levels fixed. this gives 1 + 6 +6 + 6*6 = 49 = a(3) possible hierarchies.
"See A001339 for the number of hierarchies with unlabeled elements and labeled levels."
(End)
Conjecture: for fixed k = 1,2,..., the sequence a(n) (mod k) is eventually periodic with the exact period dividing phi(k), where phi(k) is the Euler totient function A000010. For example, modulo 10 the sequence becomes (1, 1, 5, 9, 1, 1, 5, 9, ...), with an apparent period 1, 1, 5, 9 of length 4 = phi(10) beginning at a(0). - Peter Bala, Jan 15 2018

Crossrefs

Programs

  • Maple
    a:= n-> add(Stirling2(n, k)*(k!^2), k=0..n):
    seq(a(n), n=0..20);  # Alois P. Heinz, Apr 21 2012
  • Mathematica
    Table[Sum[(k!)^2*StirlingS2[n, k], {k, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, May 10 2014 *)
  • PARI
    /* By Vladeta Jovovic's formula: */
    {a(n) = my(X=x+x*O(x^n)); n!*polcoeff( sum(m=0,n, m!*(exp(X)-1)^m), n)} /* Paul D. Hanna, Feb 15 2012 */

Formula

a(n) = Sum_{k=0..n} Stirling2(n, k)*(k!)^2.
E.g.f: hypergeom([1, 1], [], exp(x)-1). - Vladeta Jovovic, Sep 14 2003
O.g.f.: Sum_{n>=0} n!^2 * Product_{k=1..n} x/(1 - k*x). - Paul D. Hanna, Nov 25 2012
a(n) ~ exp(1/2) * (n!)^2. - Vaclav Kotesovec, May 10 2014

A096307 E.g.f.: exp(x)/(1-x)^6.

Original entry on oeis.org

1, 7, 55, 481, 4645, 49171, 566827, 7073725, 95064361, 1369375615, 21054430591, 344231563897, 5964569413645, 109196040092491, 2106381399472435, 42705264827626261, 907920105215691217, 20198878182718877815
Offset: 0

Views

Author

Philippe Deléham, Jun 26 2004

Keywords

Comments

Sum_{k = 0..n} A094816(n,k)*x^k give A000522(n), A001339(n), A082030(n), A095000(n), A095177(n) for x = 1, 2, 3, 4, 5 respectively.

Crossrefs

Programs

  • Mathematica
    Table[HypergeometricPFQ[{6, -n}, {}, -1], {n, 0, 20}] (* Benedict W. J. Irwin, May 27 2016 *)
    With[{nn = 250}, CoefficientList[Series[Exp[x]/(1 - x)^6, {x, 0, nn}], x] Range[0, nn]!] (* G. C. Greubel, May 27 2016 *)

Formula

a(n) = Sum_{k = 0..n} A094916(n, k)*6^k.
a(n) = Sum_{k = 0..n} binomial(n, k)*(k+5)!/5!.
a(n) = 2F0(6,-n;;-1). - Benedict W. J. Irwin, May 27 2016
From Peter Bala, Jul 25 2021: (Start)
a(n) = (n+6)*a(n-1) - (n-1)*a(n-2) with a(0) = 1 and a(1) = 7. Cf. A001689.
First-order recurrence: P(n-1)*a(n) = n*P(n)*a(n-1) - 1 with a(0) = 1, where P(n) = n^5 + 10*n^4 + 45*n^3 + 100*n^2 + 109*n + 44 = A094794(n).
(End)

A248727 A046802(x,y) --> A046802(x,y+1), transform of e.g.f. for the graded number of positroids of the totally nonnegative Grassmannians G+(k,n); enumerates faces of the stellahedra.

Original entry on oeis.org

1, 2, 1, 5, 5, 1, 16, 24, 10, 1, 65, 130, 84, 19, 1, 326, 815, 720, 265, 36, 1, 1957, 5871, 6605, 3425, 803, 69, 1, 13700, 47950, 65646, 44240, 15106, 2394, 134, 1, 109601, 438404, 707840, 589106, 267134, 63896, 7094, 263, 1
Offset: 0

Views

Author

Tom Copeland, Oct 12 2014

Keywords

Comments

This is a transform of A046802 treating it as an array of h-vectors, so y is replaced by (y+1) in the e.g.f. for A046802.
An e.g.f. for the reversed row polynomials with signs is given by exp(a.(0;t)x) = [e^{(1+t)x} [1+t(1-e^(-x))]]^(-1) = 1 - (1+2t)x + (1+5t+5t^2)x^2/2! + ... . The reciprocal is an e.g.f. for the reversed face polynomials of the simplices A074909, i.e., exp(b.(0;t)x) = e^{(1+t)x} [1+t(1-e^(-x))] = 1 + (1+2t)x +(1+3t+3t^2) x^2/2! + ... , so the relations of A133314 apply between the two sets of polynomials. In particular, umbrally [a.(0;t)+b.(0;t)]^n vanishes except for n=0 for which it's unity, implying the two sets of Appell polynomials formed from the two bases, a_n(z;t) = (a.(0;t)+z)^n and b_n(z;t) = (b.(0;t) + z)^n, are an umbral compositional inverse pair, i.e., b_n(a.(x;t);t)= x^n = a_n(b.(x;t);t). Raising operators for these Appell polynomials are related to the polynomials of A028246, whose reverse polynomials are given by A123125 * A007318. Compare: A248727 = A007318 * A123125 * A007318 and A046802 = A007318 * A123125. See A074909 for definitions and related links. - Tom Copeland, Jan 21 2015
The o.g.f. for the umbral inverses is Og(x) = x / (1 - x b.(0;t)) = x / [(1-tx)(1-(1+t)x)] = x + (1+2t) x^2 + (1+3t+3t^2) x^3 + ... . Its compositional inverse is an o.g.f for signed A033282, the reverse f-polynomials for the simplicial duals of the Stasheff polytopes, or associahedra of type A, Oginv(x) =[1+(1+2t)x-sqrt[1+2(1+2t)x+x^2]] / (2t(1+t)x) = x - (1+2t) x^2 + (1+5t+5t^2) x^3 + ... . Contrast this with the o.g.f.s related to the corresponding h-polynomials in A046802. - Tom Copeland, Jan 24 2015
Face vectors, or coefficients of the face polynomials, of the stellahedra, or stellohedra. See p. 59 of Buchstaber and Panov. - Tom Copeland, Nov 08 2016
See A008279 for a relation between the e.g.f.s enumerating the faces of permutahedra and stellahedra. - Tom Copeland, Nov 14 2016

Examples

			The triangle T(n, k) starts:
n\k    0     1     2     3     4    5   6  7 ...
1:     1
2:     2     1
3:     5     5     1
4:    16    24    10     1
5:    65   130    84    19     1
6:   326   815   720   265    36    1
7:  1957  5871  6605  3425   803   69   1
8: 13700 47950 65646 44240 15106 2394 134  1
... reformatted, _Wolfdieter Lang_, Mar 27 2015
		

Crossrefs

Programs

  • Mathematica
    (* t = A046802 *) t[, 1] = 1; t[n, n_] = 1; t[n_, 2] = 2^(n - 1) - 1; t[n_, k_] = Sum[((i - k + 1)^i*(k - i)^(n - i - 1) - (i - k + 2)^i*(k - i - 1)^(n - i - 1))*Binomial[n - 1, i], {i, 0, k - 1}]; T[n_, j_] := Sum[Binomial[k, j]*t[n + 1, k + 1], {k, j, n}]; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jan 23 2015, after Tom Copeland *)

Formula

Let M(n,k)= sum{i=0,..,k-1, C(n,i)[(i-k)^i*(k-i+1)^(n-i)- (i-k+1)^i*(k-i)^(n-i)]} with M(n,0)=1. Then M(n,k)= A046802(n,k), and T(n,j)= sum(k=j,..,n, C(k,j)*M(n,k)) for j>0 with T(n,0)= 1 + sum(k=1,..,n, M(n,k)) for n>0 and T(0,0)=1.
E.g.f: y * exp[x*(y+1)]/[y+1-exp(x*y)].
Row sums are A007047. Row polynomials evaluated at -1 are unity. Row polynomials evaluated at -2 are A122045.
First column is A000522. Second column appears to be A036918/2 = (A001339-1)/2 = n*A000522(n)/2.
Second diagonal is A052944. (Changed from conjecture to fact on Nov 08 2016.)
The raising operator for the reverse row polynomials with row signs is R = x - (1+t) - t e^(-D) / [1 + t(1-e^(-D))] evaluated at x = 0, with D = d/dx. Also R = x - d/dD log[exp(a.(0;t)D], or R = - d/dz log[e^(-xz) exp(a.(0;t)z)] = - d/dz log[exp(a.(-x;t)z)] with the e.g.f. defined in the comments and z replaced by D. Note that t e ^(-D) / [1+t(1-e^(-D))] = t - (t+t^2) D + (t+3t^2+2t^3) D^2/2! - ... is an e.g.f. for the signed reverse row polynomials of A028246. - Tom Copeland, Jan 23 2015
Equals A007318*(padded A090582)*A007318*A097808 = A007318*(padded (A008292*A007318))*A007318*A097808 = A007318*A130850 = A007318*(mirror of A028246). Padded means in the same way that A097805 is padded A007318. - Tom Copeland, Nov 14 2016
Umbrally, the row polynomials are p_n(x) = (1 + q.(x))^n, where (q.(x))^k = q_k(x) are the row polynomials of A130850. - Tom Copeland, Nov 16 2016
From the previous umbral statement, OP(x,d/dy) y^n = (y + q.(x))^n, where OP(x,y) = exp[y * q.(x)] = x/((1+x)*exp(-x*y) - 1), the e.g.f. of A130850, so OP(x,d/dy) y^n evaluated at y = 1 is p_n(x), the n-th row polynomial of this entry, with offset 0. - Tom Copeland, Jun 25 2018
Consolidating some formulas in this entry and A046082, in umbral notation for concision, with all offsets 0: Let A_n(x;y) = (y + E.(x))^n, an Appell sequence in y where E.(x)^k = E_k(x) are the Eulerian polynomials of A123125. Then the row polynomials of A046802 (the h-polynomials of the stellahedra) are given by h_n(x) = A_n(x;1); the row polynomials of this entry (A248727, the face polynomials of the stellahedra), by f_n(x) = A_n(1 + x;1); the Swiss-knife polynomials of A119879, by Sw_n(x) = A_n(-1;1 + x); and the row polynomials of the Worpitsky triangle (A130850), by w_n(x) = A(1 + x;0). Other specializations of A_n(x;y) give A090582 (the f-polynomials of the permutohedra, cf. also A019538) and A028246 (another version of the Worpitsky triangle). - Tom Copeland, Jan 24 2020

Extensions

Title expanded by Tom Copeland, Nov 08 2016

A143409 Square array read by antidiagonals: form the Euler-Seidel matrix for the sequence {k!} and then divide column k by k!.

Original entry on oeis.org

1, 2, 1, 5, 3, 1, 16, 11, 4, 1, 65, 49, 19, 5, 1, 326, 261, 106, 29, 6, 1, 1957, 1631, 685, 193, 41, 7, 1, 13700, 11743, 5056, 1457, 316, 55, 8, 1, 109601, 95901, 42079, 12341, 2721, 481, 71, 9, 1, 986410, 876809, 390454, 116125, 25946, 4645, 694, 89, 10, 1
Offset: 0

Views

Author

Peter Bala, Aug 14 2008

Keywords

Comments

The Euler-Seidel matrix for the sequence {k!} is array A076571 read as a square, whose k-th column entries have a common factor of k!. Removing these common factors gives the current table.
This table is closely connected to the constant 1/e. The row, column and diagonal entries of this table occur in series acceleration formulas for 1/e.
For a similar table based on the differences of the sequence {k!} and related to the constant e, see A086764. For other arrays similarly related to constants see A143410 (for sqrt(e)), A143411 (for 1/sqrt(e)), A008288 (for log(2)), A108625 (for zeta(2)) and A143007 (for zeta(3)).

Examples

			The Euler-Seidel matrix for the sequence {k!} begins
==============================================
n\k|.....0.....1.....2.....3.....4.....5.....6
==============================================
0..|.....1.....1.....2.....6....24...120...720
1..|.....2.....3.....8....30...144...840
2..|.....5....11....38...174...984
3..|....16....49...212..1158
4..|....65...261..1370
5..|...326..1631
6..|..1957
...
Dividing the k-th column by k! gives
==============================================
n\k|.....0.....1.....2.....3.....4.....5.....6
==============================================
0..|.....1.....1.....1.....1.....1.....1.....1
1..|.....2.....3.....4.....5.....6.....7
2..|.....5....11....19....29....41
3..|....16....49...106...193
4..|....65...261...685
5..|...326..1631
6..|..1957
...
Examples of series formula for 1/e:
Row 2: 1/e = 2*(1/5 - 1/(1!*5*11) + 1/(2!*11*19) - 1/(3!*19*29) + ...).
Column 4: 24/e = 9 - (0!/(1*6) + 1!/(6*41) + 2!/(41*316) + ...).
...
Displayed as a triangle:
0 |     1
1 |     2,     1
2 |     5,     3,    1
3 |    16,    11,    4,    1
4 |    65,    49,   19,    5,   1
5 |   326,   261,  106,   29,   6,  1
6 |  1957,  1631,  685,  193,  41,  7, 1
7 | 13700, 11743, 5056, 1457, 316, 55, 8, 1
		

Crossrefs

Cf. A008288, A076571, A086764, A108625, A143007, A143410, A143411, A143413, A001517 (main diagonal), A028387 (row 2), A000522 (column 0), A001339 (column 1), A082030 (column 2), A095000 (column 3), A095177 (column 4).

Programs

  • Maple
    T := (n, k) -> 1/k!*add(binomial(n,j)*(k+j)!, j = 0..n):
    for n from 0 to 9 do seq(T(n, k), k = 0..9) end do;
    # Alternate:
    T:= proc(n,k) option remember;
      if n = 0 then return 1 fi;
      (n+k)*procname(n-1,k) + procname(n-1,k-1);
    end proc:
    seq(seq(T(s-n,n),n=0..s),s=0..10); # Robert Israel, Jul 07 2017
    # Or:
    A143409 := (n,k) -> hypergeom([k+1, k-n], [], -1):
    seq(seq(simplify(A143409(n,k)),k=0..n),n=0..9); # Peter Luschny, Oct 05 2017
  • Mathematica
    T[n_, k_] := HypergeometricPFQ[{k+1,k-n}, {}, -1];
    Table[T[n,k], {n,0,9}, {k,0,n}] // Flatten (* Peter Luschny, Oct 05 2017 *)

Formula

T(n,k) = (1/k!)*Sum_{j = 0..n} binomial(n,j)*(k+j)!.
T(n,k) = ((n+k)!/k!)*Num_Pade(n,k), where Num_Pade(n,k) denotes the numerator of the Padé approximation for the function exp(x) of degree (n,k) evaluated at x = 1.
Recurrence relations:
T(n,k) = T(n-1,k) + (k+1)*T(n-1,k+1);
T(n,k) = (n+k)*T(n-1,k) + T(n-1,k-1).
E.g.f. for column k: exp(y)/(1-y)^(k+1).
E.g.f. for array: exp(y)/(1-x-y) = (1 + x + x^2 + ...) + (2 + 3*x + 4*x^2 + ...)*y + (5 + 11*x + 19*x^2 + ...)*y^2/2! + ... .
Row n lists the values of the Poisson-Charlier polynomial x^(n) + C(n,1)*x^(n-1) + C(n,2)*x^(n-2) + ... + C(n,n) for x = 1,2,3,..., where x^(m) denotes the rising factorial x*(x+1)*...*(x+m-1).
Main diagonal is A001517.
Series formulas for 1/e:
Row n: 1/e = n!*[1/T(n,0) - 1/(1!*T(n,0)*T(n,1)) + 1/(2!*T(n,1)*T(n,2)) - 1/(3!*T(n,2)*T(n,3)) + ...].
Column k: k!/e = A000166(k) + (-1)^(k+1)*[0!/(T(0,k)*T(1,k)) + 1!/(T(1,k)*T(2,k)) + 2!/(T(2,k)*T(3,k)) + ...].
Main diagonal: 1/e = 1 - 2*Sum_{n>=0} (-1)^n/(T(n,n)*T(n+1,n+1)) = 1 - 2*[1/(1*3) - 1/(3*19) + 1/(19*193) - ...].
Second subdiagonal: 1/e = 2*(1^2/(1*5) - 2^2/(5*49) + 3^2/(49*685) - ...).
Compare with A143413.
From Peter Luschny, Oct 05 2017: (Start)
T(n, k) = hypergeom([k+1, k-n], [], -1).
When seen as a triangular array then the row sums are A273596 and the alternating row sums are A003470. (End)

A052582 a(n) = 2*n*n!.

Original entry on oeis.org

0, 2, 8, 36, 192, 1200, 8640, 70560, 645120, 6531840, 72576000, 878169600, 11496038400, 161902540800, 2440992153600, 39230231040000, 669529276416000, 12093372555264000, 230485453406208000, 4622513815535616000, 97316080327065600000, 2145819571211796480000
Offset: 0

Views

Author

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

Keywords

Comments

Total number of pairs (a_i,a_(i+1)) in all permutations on [n] such that a_i,a_(i+1) are consecutive integers. - David Callan, Nov 04 2003
Number of permutations of {1,2,...,n+2} such that there is exactly one entry between the entries 1 and 2. Example: a(2)=8 because we have 1324, 1423, 2314, 2413, 3142, 4132, 3241 and 4231. - Emeric Deutsch, Apr 06 2008
Number of permutations of 0 to n distinct letters (ABC...) 1 times ("-" (0), A (1), AB (1-1), ABC (1-1-1), ABCD (1-1-1-1 )etc...) and one after the other to resemble motif:( "-",... BB (0-2), ABB (1-2-0), AABB (2-2-0-0), AAABB (3-2-0-0-0) AAAABB (4-2-0-0-0-0), AAAAABB (5-2-0-0-0-0-0), AAAAAABB (6-2-0-0-0-0-0-0), etc... 0 fixed point (or free fixed point). Example: if ABC (1-1-1) and motif ABB (1-2-0) then 2 * 0 (free) fixed point, if ABCD (1-1-1-1), and motif AABB (2-2-0-0) then 8 * 0 (free) fixed point, if ABCDE (1-1-1-1-1), and motif AAABB (3-2-0-0-0), then 36 * 0 (free) fixed point, if ABCDEF (1-1-1-1-1-1), and motif AAAABB (4-2-0-0-0-0), then 192 * 0 (free) fixed point, if ABCDEFG (1-1-1-1-1-1-1), and motif AAAAABB (5-2-0-0-0-0-0), then 1200 * 0 (free) fixed point, etc... - Zerinvary Lajos, Dec 07 2009

Crossrefs

Programs

  • Haskell
    a052582 n = a052582_list !! n
    a052582_list =  0 : 2 : zipWith
       div (zipWith (*) (tail a052582_list) (drop 2 a000290_list)) [1..]
    -- Reinhard Zumkeller, Nov 12 2011
  • Maple
    spec := [S,{S=Prod(Sequence(Z),Sequence(Z),Union(Z,Z))},labeled]: seq(combstruct[count](spec,size=n), n=0..20);
  • Mathematica
    a[ n_] := If[ n<0, 0, n! SeriesCoefficient[ 2 x / (1 - x)^2, {x, 0, n}]]; (* Michael Somos, Oct 20 2011 *)
    a[ n_] := If[ n<0, 0, 2 n n!]; (* Michael Somos, Oct 20 2011 *)
  • PARI
    {a(n) = if( n<0, 0, 2 * n * n!)}; /* Michael Somos, Oct 20 2011 */
    

Formula

E.g.f.: 2*x / (1 - x)^2.
Recurrence: {a(0)=0, a(1)=2, (-n^2-2*n-1)*a(n)+a(n+1)*n=0.}.
a(n) = A138770(n+2,1). - Emeric Deutsch, Apr 06 2008
a(n) = A001339(n) - A007808(n). - Michael Somos, Oct 20 2011
a(n) = (a(n-1)^2 - 2 * a(n-2)^2 + a(n-2) * a(n-3) - 4 * a(n-1) * a(n-3)) / (a(n-2) - a(n-3)) if n>2. - Michael Somos, Oct 20 2011
a(n) = 2*n*n!. - Gary Detlefs, Sep 16 2010
a(n+1) = a(n) * (n+1)^2 / n. - Reinhard Zumkeller, Nov 12 2011
0 = a(n)*(+a(n+1) -4*a(n+2) +a(n+3)) +a(n+1)*(+2*a(n+1) -a(n+3)) + a(n+2)*(+a(n+2)) if n>=0. - Michael Somos, Jun 26 2017
From Amiram Eldar, Feb 14 2021: (Start)
Sum_{n>=1} 1/a(n) = (Ei(1) - gamma)/2 = (A091725 - A001620)/2, where Ei(x) is the exponential integral.
Sum_{n>=1} (-1)^(n+1)/a(n) = (gamma - Ei(-1))/2 = (A001620 + A099285)/2. (End)
a(n) = 2 * A001563(n). - Alois P. Heinz, Sep 03 2024
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