A052887 -id:A052887 - cp's OEIS frontend

A288268 Expansion of e.g.f.: exp(Sum_{k>=1} (k-1)*x^k/k).

1, 0, 1, 4, 21, 136, 1045, 9276, 93289, 1047376, 12975561, 175721140, 2581284541, 40864292184, 693347907421, 12548540320876, 241253367679185, 4909234733857696, 105394372192969489, 2380337795595885156, 56410454014314490981, 1399496554158060983080

Offset: 0

Seiichi Manyama, Oct 20 2017

Seiichi Manyama, Table of n, a(n) for n = 0..444
Index entries for sequences related to Laguerre polynomials

Cf. A000296, A002720, A008275, A009940, A052852, A052887, A059114, A288269.

l:= func< n, a, b | Evaluate(LaguerrePolynomial(n, a), b) >;
[1,0]cat[(Factorial(n)/(n-1))*(2*l(n-1,0,-1) - l(n,0,-1)): n in [2..30]]; // G. C. Greubel, Mar 10 2021

a := proc(n) option remember; if n < 3 then [1, 0, 1][n+1] else
-(n^2 - 4*n + 3)*a(n - 2) + (2*n - 2)*a(n - 1) fi end:
seq(a(n), n = 0..21); # Peter Luschny, Feb 20 2022

Table[If[n<2, 1-n, (n!/(n-1))*(2*LaguerreL[n-1, -1] - LaguerreL[n, -1])], {n, 0, 30}] (* G. C. Greubel, Mar 10 2021 *)

{a(n) = n!*polcoeff(exp(sum(k=1, n, (k-1)*x^k/k)+x*O(x^n)), n)}

[1-n if n<2 else (factorial(n)/(n-1))*(2*gen_laguerre(n-1,0,-1) - gen_laguerre(n,0,-1)) for n in (0..30)] # G. C. Greubel, Mar 10 2021

a(0) = 1 and a(n) = (n-1)! * Sum_{k=1..n} (k-1)*a(n-k)/(n-k)! for n > 0.
E.g.f.: (1 - x) * exp(x/(1 - x)). - Ilya Gutkovskiy, Jul 27 2020
a(n) = (n!/(n-1))*( 2*LaguerreL(n-1, -1) - LaguerreL(n, -1) ) with a(0) = 1, a(1) = 0. - G. C. Greubel, Mar 10 2021
a(n) ~ n^(n - 3/4) * exp(-1/2 + 2*sqrt(n) - n) / sqrt(2) * (1 - 65/(48*sqrt(n))). - Vaclav Kotesovec, Mar 10 2021, minor term corrected Dec 01 2021
From Peter Luschny, Feb 20 2022: (Start)
a(n) = n! * Sum_{k=0..n} (-1)^k * LaguerreL(n-k, k-1, -1).
a(n) = 2*(n - 1)*a(n - 1) - (n^2 - 4*n + 3)*a(n - 2) for n >= 3. (End)
From Peter Bala, May 26 2023: (Start)
a(n) = Sum_{k = 0..n} |Stirling1(n,k)|*A000296(k) (follows from the fundamental theorem of Riordan arrays).
Let k be a positive integer. The sequence obtained by reducing a(n) modulo k is purely periodic with the period dividing k. For example, modulo 7 we obtain the purely periodic sequence [1, 0, 1, 4, 0, 3, 2, 1, 0, 1, 4, 0, 3, 2, ...] of period 7. Cf. A047974. (End)
For n>1, a(n) = (2*n*A002720(n-1) - A002720(n))/(n-1). - Vaclav Kotesovec, May 27 2023

A335344 Expansion of e.g.f. exp(x^2/(2*(1 - x)^2)).

Original entry on oeis.org

1, 0, 1, 6, 39, 300, 2715, 28350, 335265, 4422600, 64298745, 1020477150, 17542820295, 324552648420, 6426708843555, 135573281994150, 3034400481137025, 71801374285040400, 1790523094644709425, 46923435009924823350, 1289032229351717425575

Offset: 0

Ilya Gutkovskiy, Jun 02 2020

nonn

For n>0, a(n) is also the number of ways to split n people into nonempty groups, have each group sit around a circular table, and select 2 people from each table (where two seating arrangements are considered identical if each person has the same left neighbors in both of them). See example below. - Enrique Navarrete, Oct 01 2023

			For n = 5, using one table, there are 4! circular seatings and binomial(5,2) ways to select 2 persons, hence 240 ways. Using two tables, the only way we can select 2 persons from each one is seating 3 persons in one table and 2 in the other, which can be done in 20 ways; then choosing 2 persons from each table can be done in 3 ways, hence giving another 60 ways for a total of 300. - _Enrique Navarrete_, Oct 01 2023

Robert Israel, Table of n, a(n) for n = 0..436

Cf. A001286, A052887, A082579, A185369, A335345.

f:= gfun:-rectoproc({a(n) = -(n-1)*(3*n-7)*a(n-2) + 3*(n-1)*a(n-1) + (n - 1)*(n - 2)*(n - 3)*a(n-3),a(0)=1,a(1)=0,a(2)=1},a(n),remember):
map(f, [$0..30]); # Robert Israel, Jun 04 2020

nmax = 20; CoefficientList[Series[Exp[x^2/(2 (1 - x)^2)], {x, 0, nmax}], x] Range[0, nmax]!
a[0] = 1; a[n_] := a[n] = (1/2) Sum[Binomial[n - 1, k - 1] (k - 1) k! a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 20}]

seq(n)=Vec(serlaplace(exp(x^2/(2*(1 - x)^2) + O(x*x^n)))) \\ Andrew Howroyd, Jun 02 2020

a(0) = 1; a(n) = Sum_{k=1..n} binomial(n-1,k-1) * A001286(k) * a(n-k).
D-finite with recurrence a(n) = -(n - 1)*(3*n - 7)*a(n - 2) + 3*(n - 1)*a(n - 1) + (n - 1)*(n - 2)*(n - 3)*a(n - 3). - Robert Israel, Jun 04 2020
a(n) ~ n^(n - 1/6) * exp(1/6 - n^(1/3)/2 + 3*n^(2/3)/2 - n) / sqrt(3). - Vaclav Kotesovec, Jun 11 2020
a(n) = n! * Sum_{k=0..floor(n/2)} binomial(n-1,n-2*k)/(2^k * k!). - Seiichi Manyama, Mar 16 2023

A361572 Expansion of e.g.f. exp( (x / (1-x))^3 ).

Original entry on oeis.org

1, 0, 0, 6, 72, 720, 7560, 90720, 1270080, 20381760, 364694400, 7125148800, 150186960000, 3393726336000, 81882210009600, 2102315389574400, 57244753133568000, 1647544166940672000, 49957730917981286400, 1591303422125646028800

Offset: 0

Seiichi Manyama, Mar 16 2023

nonn

Cf. A000262, A052887, A361576.

Cf. A091695.

Table[n! * Sum[Binomial[n-1,n-3*k]/k!, {k,0,n/3}], {n,0,20}] (* Vaclav Kotesovec, Mar 17 2023 *)
With[{nn=20},CoefficientList[Series[Exp[(x/(1-x))^3],{x,0,nn}],x] Range[0,nn]!] (* Harvey P. Dale, Sep 07 2024 *)

my(N=20, x='x+O('x^N)); Vec(serlaplace(exp((x/(1-x))^3)))

a_vector(n) = my(v=vector(n+1)); v[1]=1; for(i=1, n, v[i+1]=(i-1)!*sum(j=3, i, (-1)^(j-3)*j*binomial(-3, j-3)*v[i-j+1]/(i-j)!)); v;

a(n) = n! * Sum_{k=0..floor(n/3)} binomial(n-1,n-3*k)/k!.
a(0) = 1; a(n) = (n-1)! * Sum_{k=3..n} (-1)^(k-3) * k * binomial(-3,k-3) * a(n-k)/(n-k)!.
From Vaclav Kotesovec, Mar 17 2023: (Start)
a(n) = 4*(n-1)*a(n-1) - 6*(n-2)*(n-1)*a(n-2) + (n-2)*(n-1)*(4*n - 9)*a(n-3) - (n-4)*(n-3)*(n-2)*(n-1)*a(n-4).
a(n) ~ 3^(1/8) * exp(-1/4 + 5*3^(-1/4)*n^(1/4)/8 - sqrt(3*n)/2 + 4*3^(-3/4)*n^(3/4) - n) * n^(n - 1/8) / 2 * (1 - (409/2560)*3^(1/4)/n^(1/4)). (End)

A387244 Expansion of e.g.f. exp(x^2/(1-x)^4).

Original entry on oeis.org

1, 0, 2, 24, 252, 2880, 38280, 594720, 10565520, 209502720, 4558407840, 107702179200, 2744400415680, 75016089308160, 2189152249764480, 67906418407027200, 2230160988344889600, 77271779968704921600, 2815893910009609228800, 107629691727791474841600, 4304364116456244429388800

Offset: 0

Vaclav Kotesovec, Aug 24 2025

In general, if s >= 1, 1 <= r <= s and e.g.f. = exp(x^r/(1-x)^s) then for n > 0, a(n) = n! * Sum_{k=1..n} binomial(n + (s-r)*k - 1, s*k - 1)/k!.

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

Cf. A293012, A000262, A082579, A091695, A361283.

Cf. A052845, A052887, A386514.

m:=25; R:=PowerSeriesRing(Rationals(), m); b:=Coefficients(R!(Exp(x^2/(1-x)^4))); [Factorial(n-1)*b[n]: n in [1..m]]; // Vincenzo Librandi, Aug 25 2025

nmax=20; CoefficientList[Series[E^(x^2/(1-x)^4), {x, 0, nmax}], x] * Range[0, nmax]!
nmax=20; Join[{1}, Table[n!*Sum[Binomial[n+2*k-1, 4*k-1]/k!, {k, 1, n}], {n, 1, nmax}]]
Join[{1}, Table[n!*n*(n - 1)*(n + 1)/6 * HypergeometricPFQ[{1 - n/2, 3/2 - n/2, 1 + n/2, 3/2 + n/2}, {5/4, 3/2, 7/4, 2}, 1/16], {n, 1, 20}]]

For n > 0, a(n) = n! * Sum_{k=1..n} binomial(n + 2*k - 1, 4*k - 1)/k!.
a(n) = 5*(n-1)*a(n-1) - 2*(n-1)*(5*n-11)*a(n-2) + 2*(n-2)*(n-1)*(5*n-14)*a(n-3) - 5*(n-4)*(n-3)*(n-2)*(n-1)*a(n-4) + (n-5)*(n-4)*(n-3)*(n-2)*(n-1)*a(n-5).
a(n) ~ 2^(1/5) * 5^(-1/2) * exp(1/80 - 2^(-9/5)*n^(2/5)/3 + 5*2^(-8/5)*n^(4/5) - n) * n^(n - 1/10).

A288269 Expansion of e.g.f.: exp(Sum_{k>=1} (k-1)kx^k).

Original entry on oeis.org

1, 0, 4, 36, 336, 3840, 52800, 836640, 14864640, 291755520, 6264276480, 145962432000, 3665362821120, 98604459233280, 2827182573895680, 86016204578304000, 2766450467708928000, 93741871082943283200, 3336807307530977280000, 124443669133537276723200

Offset: 0

Seiichi Manyama, Oct 20 2017

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..420

Cf. A052887, A288268.

R:=PowerSeriesRing(Rationals(), 30);
Coefficients(R!(Laplace( Exp(2*x^2/(1-x)^3) ))); // G. C. Greubel, Mar 10 2021

With[{m = 30}, CoefficientList[Series[Exp[2*x^2/(1-x)^3], {x, 0, m}], x]*Range[0, m]!] (* G. C. Greubel, Mar 10 2021 *)

{a(n) = n!*polcoeff(exp(sum(k=1, n, (k-1)*k*x^k)+x*O(x^n)), n)}

[factorial(n)*( exp(2*x^2/(1-x)^3) ).series(x,n+1).list()[n] for n in (0..30)] # G. C. Greubel, Mar 10 2021

a(0) = 1 and a(n) = (n-1)! * Sum_{k=1..n} (k-1)*k^2*a(n-k)/(n-k)! for n > 0.
E.g.f.: exp(2*x^2/(1 - x)^3). - Ilya Gutkovskiy, Jul 27 2020
a(n) ~ 3^(1/8) * exp(2/27 - (n/6)^(1/4)/12 - (n/6)^(1/2) + 8*(n/6)^(3/4) - n) * n^(n - 1/8) / 2^(7/8) * (1 - 3203/34560 * (6/n)^(1/4)). - Vaclav Kotesovec, Mar 10 2021
a(n) = 4*(n-1)*a(n-1) - 2*(n-1)*(3*n-8)*a(n-2) + 2*(n-2)*(n-1)*(2*n-5)*a(n-3) - (n-4)*(n-3)*(n-2)*(n-1)*a(n-4). - Vaclav Kotesovec, Dec 01 2021

A288270 E.g.f.: exp(Sum_{k>=1} (k-1)^2*x^k).

Original entry on oeis.org

1, 0, 2, 24, 228, 2400, 30360, 453600, 7702800, 144910080, 2981089440, 66561264000, 1603358729280, 41434803970560, 1142808612865920, 33485770103385600, 1038238875100627200, 33945895488708403200, 1166858228814204326400, 42055660151648798054400

Offset: 0

Seiichi Manyama, Oct 20 2017

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..421

E.g.f.: exp(Sum_{k>=1} (k-1)^m*x^k): A000262 (m=0), A052887 (m=1), this sequence (m=2), A290690 (m=3).

Cf. A255807.

{a(n) = n!*polcoeff(exp(sum(k=1, n, (k-1)^2*x^k)+x*O(x^n)), n)}

a(0) = 1 and a(n) = (n-1)! * Sum_{k=1..n} (k-1)^2*k*a(n-k)/(n-k)! for n > 0.
E.g.f.: exp(x^2*(1 + x)/(1 - x)^3). - Ilya Gutkovskiy, Jul 27 2020
a(n) ~ 2^(-7/8) * 3^(1/8) * n^(n - 1/8) / exp(n - 2^(9/4)*n^(3/4)/3^(3/4) + sqrt(2*n/3) - 2^(3/4)*n^(1/4)/3^(5/4) + 13/54). - Vaclav Kotesovec, Jul 31 2021

A290690 E.g.f.: exp(Sum_{k>=1} (k-1)^3*x^k).

Original entry on oeis.org

1, 0, 2, 48, 660, 8640, 132600, 2520000, 56046480, 1375557120, 36456769440, 1041522451200, 32083867126080, 1061964845061120, 37543201808112000, 1409292653408640000, 55917035430800544000, 2337184142686903910400, 102624865930477758067200

Offset: 0

Seiichi Manyama, Oct 20 2017

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..407

E.g.f.: exp(Sum_{k>=1} (k-1)^m*x^k): A000262 (m=0), A052887 (m=1), A288270 (m=2), this sequence (m=3).

Cf. A255819.

nmax = 20; CoefficientList[Series[Exp[x^2*(1 + 4*x + x^2)/(1-x)^4], {x, 0, nmax}], x] * Range[0, nmax]! (* Vaclav Kotesovec, Jul 31 2021 *)

{a(n) = n!*polcoeff(exp(sum(k=1, n, (k-1)^3*x^k)+x*O(x^n)), n)}

a(0) = 1 and a(n) = (n-1)! * Sum_{k=1..n} (k-1)^3*k*a(n-k)/(n-k)! for n > 0.
From Vaclav Kotesovec, Jul 31 2021: (Start)
E.g.f.: exp(x^2*(1 + 4*x + x^2)/(1-x)^4).
a(n) ~ exp(65/384 - 101 * 2^(2/5) * 3^(4/5) * n^(1/5) / 1200 + 11 * 2^(4/5) * 3^(3/5) * n^(2/5) / 80 - 2^(-4/5) * 3^(2/5) * n^(3/5) + 5 * 2^(-7/5) * 3^(1/5) * n^(4/5) - n) * 2^(3/10) * 3^(1/10) * n^(n - 1/10) / sqrt(5).
(End)

A361576 Expansion of e.g.f. exp((x / (1-x))^4).

Original entry on oeis.org

1, 0, 0, 0, 24, 480, 7200, 100800, 1431360, 21772800, 370137600, 7185024000, 158150361600, 3848298854400, 100865282918400, 2799294930432000, 81599752346112000, 2492894621048832000, 79852538982408192000, 2684220785621286912000

Offset: 0

Seiichi Manyama, Mar 16 2023

nonn

Cf. A000262, A052887, A361572.

Table[n! * Sum[Binomial[n-1,n-4*k]/k!, {k,0,n/4}], {n,0,20}] (* Vaclav Kotesovec, Mar 17 2023 *)
With[{nn=20},CoefficientList[Series[Exp[(x/(1-x))^4],{x,0,nn}],x] Range[0,nn]!] (* Harvey P. Dale, Aug 01 2025 *)

my(N=20, x='x+O('x^N)); Vec(serlaplace(exp((x/(1-x))^4)))

a_vector(n) = my(v=vector(n+1)); v[1]=1; for(i=1, n, v[i+1]=(i-1)!*sum(j=4, i, (-1)^(j-4)*j*binomial(-4, j-4)*v[i-j+1]/(i-j)!)); v;

E.g.f.: exp( (x / (1-x))^4 ).
a(n) = n! * Sum_{k=0..floor(n/4)} binomial(n-1,n-4*k)/k!.
a(0) = 1; a(n) = (n-1)! * Sum_{k=4..n} (-1)^(k-4) * k * binomial(-4,k-4) * a(n-k)/(n-k)!.
a(n) = 5*(n-1)*a(n-1) - 10*(n-2)*(n-1)*a(n-2) + 10*(n-3)*(n-2)*(n-1)*a(n-3) - (n-3)*(n-2)*(n-1)*(5*n - 24)*a(n-4) + (n-5)*(n-4)*(n-3)*(n-2)*(n-1)*a(n-5). - Vaclav Kotesovec, Mar 17 2023

A386514 Expansion of e.g.f. exp(x^2/(1-x)^3).

Original entry on oeis.org

1, 0, 2, 18, 156, 1560, 18480, 254520, 3973200, 68947200, 1312748640, 27175024800, 607314818880, 14566195163520, 373027570755840, 10154293067318400, 292659790712889600, 8899747730037964800, 284685195814757337600, 9553060139009702515200, 335468448755976164428800

Offset: 0

Enrique Navarrete, Aug 23 2025

For n > 0, a(n) is the number of ways to linearly order n distinguishable objects into one or several lines and then choose 2 objects from each line. If the lines are also linearly ordered see A364524.

A001804(n) is the number of ways if only 1 line is used.

			a(6)=18480 since there are 10800 ways using one line, 4320 ways with 2 lines using 2 and 4 objects, 3240 ways with 2 lines of 3 objects each, and 120 ways with 3 lines of 2 objects each.

Cf. A001804, A364524.
Cf. A293012, A000262, A082579, A091695, A361283.
Cf. A052845, A052887, A387244.

nmax = 20; CoefficientList[Series[E^(x^2/(1-x)^3), {x, 0, nmax}], x] * Range[0, nmax]! (* or *)
nmax = 20; Join[{1}, Table[n!*Sum[Binomial[n + k - 1, 3*k - 1]/k!, {k, 1, n}], {n, 1, nmax}]] (* Vaclav Kotesovec, Aug 24 2025 *)

From Vaclav Kotesovec, Aug 24 2025: (Start)
For n > 0, a(n) = n! * Sum_{k=1..n} binomial(n+k-1, 3*k-1) / k!.
a(n) = 4*(n-1)*a(n-1) - 2*(n-1)*(3*n-7)*a(n-2) + (n-2)*(n-1)*(4*n-11)*a(n-3) - (n-4)*(n-3)*(n-2)*(n-1)*a(n-4).
a(n) ~ 3^(1/8) * exp(1/27 - 3^(-5/4)*n^(1/4)/8 - 3^(-1/2)*n^(1/2)/2 + 4*3^(-3/4)*n^(3/4) - n) * n^(n-1/8) / 2. (End)

A123081 Infinite square array read by antidiagonals: T(n,k) = Bell(n+k) = A000110(n+k).

Original entry on oeis.org

1, 1, 1, 2, 2, 2, 5, 5, 5, 5, 15, 15, 15, 15, 15, 52, 52, 52, 52, 52, 52, 203, 203, 203, 203, 203, 203, 203, 877, 877, 877, 877, 877, 877, 877, 877, 4140, 4140, 4140, 4140, 4140, 4140, 4140, 4140, 4140, 21147, 21147, 21147, 21147, 21147, 21147, 21147, 21147, 21147, 21147, 115975, 115975, 115975, 115975

Offset: 0

Gary W. Adamson, Jan 19 2007

Alternatively, triangle read by rows in which row n (n >= 0) contains A000110(n) repeated n+1 times.

Row sums = A052887: 1, 2, 6, 20, 75, 312, ... A127568 = Q * M n-th row is composed of n+1 terms of A000110(n).

			Square array begins:
    1,    1,     2,      5,     15,      52,      203,       877, ...;
    1,    2,     5,     15,     52,     203,      877,      4140, ...;
    2,    5,    15,     52,    203,     877,     4140,     21147, ...;
    5,   15,    52,    203,    877,    4140,    21147,    115975, ...;
   15,   52,   203,    877,   4140,   21147,   115975,    678570, ...;
   52,  203,   877,   4140,  21147,  115975,   678570,   4213597, ...;
  203,  877,  4140,  21147, 115975,  678570,  4213597,  27644437, ...;
  877, 4140, 21147, 115975, 678570, 4213597, 27644437, 190899322, ...;
First few rows of the triangle:
    1;
    1,   1;
    2,   2,   2;
    5,   5,   5,   5;
   15,  15,  15,  15,  15;
   52,  52,  52,  52,  52,  52;
  203, 203, 203, 203, 203, 203, 203;

G. C. Greubel, Antidiagonal rows n = 0..50, flattened
W. F. Lunnon et al., Arithmetic properties of Bell numbers to a composite modulus I, Acta Arith., 35 (1979), 1-16. [From _N. J. A. Sloane_, Feb 07 2009]

Cf. A000110, A052887, A127568.

[Bell(n): k in [0..n], n in [0..12]]; // G. C. Greubel, Jul 21 2021

Table[BellB[n], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Jul 21 2021 *)

B(n)=sum(k=0,n,stirling(n,k,2));
for(n=0,20,for(k=0,n,print1(B(n),", "))); \\ Joerg Arndt, Apr 21 2014

flatten([[bell_number(n) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Jul 21 2021

M * Q, as infinite lower triangular matrices; M = the Bell sequence, A000110 in the main diagonal and the rest zeros. Q = (1; 1, 1; 1, 1, 1; ...)

Edited by N. J. A. Sloane, Feb 07 2009

Added more terms, Joerg Arndt, Apr 21 2014

cp's OEIS Frontend

A288268 Expansion of e.g.f.: exp(Sum_{k>=1} (k-1)*x^k/k).

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A335344 Expansion of e.g.f. exp(x^2/(2*(1 - x)^2)).

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A361572 Expansion of e.g.f. exp( (x / (1-x))^3 ).

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A387244 Expansion of e.g.f. exp(x^2/(1-x)^4).

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A288269 Expansion of e.g.f.: exp(Sum_{k>=1} (k-1)*k*x^k).

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A288270 E.g.f.: exp(Sum_{k>=1} (k-1)^2*x^k).

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A290690 E.g.f.: exp(Sum_{k>=1} (k-1)^3*x^k).

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Formula

A361576 Expansion of e.g.f. exp((x / (1-x))^4).

A288269 Expansion of e.g.f.: exp(Sum_{k>=1} (k-1)kx^k).