A054589 -id:A054589 - cp's OEIS frontend

A042977 Triangle T(n,k) read by rows: coefficients of a polynomial sequence occurring when calculating the n-th derivative of Lambert function W.

1, -2, -1, 9, 8, 2, -64, -79, -36, -6, 625, 974, 622, 192, 24, -7776, -14543, -11758, -5126, -1200, -120, 117649, 255828, 248250, 137512, 45756, 8640, 720, -2097152, -5187775, -5846760, -3892430, -1651480, -445572, -70560, -5040

Offset: 0

Wouter Meeussen

The first derivative of the Lambert W function is given by dW/dz = exp(-W)/(1+W). Further differentiation yields d^2/dz^2(W) = exp(-2*W)*(-2-W)/(1+W)^3, d^3/dz^3(W) = exp(-3*W)*(9+8*W+2*W^2)/(1+W)^5 and, in general, d^n/dz^n(W) = exp(-n*W)*R(n,W)/(1+W)^(2*n-1), where R(n,W) are the row polynomials of this triangle. - Peter Bala, Jul 22 2012

Conjecture: the polynomials have no real roots greater than or equal to -1. This is equivalent to the statement that the derivatives of the 0th branch of the Lambert W function have no real roots greater than -1/e. - Colin Linzer, Jan 29 2025

			Triangle begins:
 n\k |    1    W   W^2   W^3   W^4
==================================
  1  |    1
  2  |   -2   -1
  3  |    9    8     2
  4  |  -64  -79   -36    -6
  5  |  625  974   622   192    24
...
T(5,2) = -4*(-79) - 9*(-36) + 3*(-6) = 622.

G. C. Greubel, Table of n, a(n) for the first 75 rows, flattened
A. F. Beardon, Winding Numbers, Unwinding Numbers, and the Lambert W Function, Computational Methods and Function Theory, 2021.
George C. Greubel, On Szasz-Mirakyan-Jain Operators preserving exponential functions, arXiv:1805.06968 [math.CA], 2018.
Roy M. Howard, Schröder Based Series for the Lambert W Function, Curtin Univ. (Australia), ResearchGate (2025). See p. 8. See also On Schröder-Type Series Expansions for the Lambert W Function, AppliedMath (2025) Vol. 5, No. 2, Art. No. 66.
G. A. Kalugin and D. J. Jeffrey, Unimodal sequences show that Lambert is Bernstein, C. R. Math. Rep. Acad. Sci. Canada Vol. 33 (2) pp. 50-56, 2011; arXiv:1011.5940 [math.CA], 2010.
Vladimir Kruchinin, Derivation of Bell Polynomials of the Second Kind, arXiv:1104.5065 [math.CO], 2011.
Eric Weisstein's World of Mathematics, Lambert W-Function

Cf. A013703 (twice row sums), A000444, A000525, A064781, A064785, A064782.
First column A000169, main diagonal A000142, first subdiagonal A052582.
Cf. A054589.

# After Vladimir Kruchinin, for 0 <= m <= n:
T := (n, m) -> add(add((-1)^(k+n)*binomial(j,k)*binomial(2*n+1,m-j)*(k+n+1)^(n+j), k=0..j)/j!, j=0..m): seq(seq(T(n, k), k=0..n), n=0..7); # Peter Luschny, Feb 23 2018

Table[ Simplify[ (Evaluate[ D[ ProductLog[ z ], {z, n} ] ] /. ProductLog[ z ]->W)*z^n/W^n (1+W)^(2n-1) ], {n, 12} ] // TableForm
Flatten[ Table[ CoefficientList[ Simplify[ (Evaluate[D[ProductLog[z], {z, n}]] /. ProductLog[z] -> W) z^n / W^n (1 + W)^(2 n - 1)], W], {n, 8}]] (* Michael Somos, Jun 07 2012 *)
T[ n_, k_] := If[ n < 1 || k < 0, 0, Coefficient[ Simplify[(Evaluate[D[ProductLog[z], {z, n}]] /. ProductLog[z] -> W) z^n / W^n (1 + W)^(2 n - 1)], W, k]] (* Michael Somos, Jun 07 2012 *)

B(n):=(if n=1 then 1/(1+x)*exp(-x) else -n!*sum((sum((-1)^(m-j)*binomial(m,j)*sum((j^(n-i)*binomial(j,i)*x^(m-i))/(n-i)!,i,0,n),j,1,m))*B(m)/m!,m,1,n-1)/(1+x)^n);
a(n):=B(n)*(1+x)^(2*n-1);
/* Vladimir Kruchinin, Apr 07 2011 */

a(n):=if n=1 then 1 else (n-1)!*(sum((binomial(n+k-1, n-1)*sum(binomial(k, j)*(x+1)^(n-j-1)*sum(binomial(j, l)*(-1)^(l)*sum((l^(n+j-i-1)*binomial(l, i)*x^(j-i))/(n+j-i-1)!, i, 0, l), l, 1, j), j, 1, k)), k, 1, n-1));
T(n, k):=coeff(ratsimp(a(n)), x, k);
for n: 1 thru 12 do print(makelist(T(n, k), k, 0, n-1));
/* Vladimir Kruchinin, Oct 09 2012 */
T(n,m):=sum(binomial(2*n+1,m-j)*sum(((n+k+1)^(n+j)*(-1)^(n+k))/((j-k)!*k!),k,0,j),j,0,m); /* Vladimir Kruchinin, Feb 20 2018 */

E.g.f.: (LambertW(exp(x)*(x+y*(1+x)^2))-x)/(1+x). - Vladeta Jovovic, Nov 19 2003
a(n) = B(n)*(1+x)^(2*n-1), where B(1) = 1/(1+x), and for n>=2, B(n) = -(n!/(1+x)^n)*Sum_{m=1..n-1} (B(m)/m!)*Sum_{j=1..m} (-1)^(m-j)*binomial(m,j)*Sum_{i=0..n} j^(n-i)*binomial(j,i)*x^(m-i)/(n-i)!. - Vladimir Kruchinin, Apr 07 2011
Recurrence equation: T(n+1,k) = -n*T(n,k-1) - (3*n-k-1)*T(n,k) + (k+1)*T(n,k+1). - Peter Bala, Jul 22 2012
T(n,m) = Sum_{j=0..m} C(2*n+1,m-j)*(Sum_{k=0..j} (n+k+1)^(n+j)*(-1)^(n+k)/((j-k)!*k!)). - Vladimir Kruchinin, Feb 20 2018

A112493 Triangle read by rows, T(n, k) = Sum_{j=0..n} C(n-j, n-k)*E2(n, j), where E2 are the second-order Eulerian numbers A201637, for n >= 0 and 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 4, 3, 1, 11, 25, 15, 1, 26, 130, 210, 105, 1, 57, 546, 1750, 2205, 945, 1, 120, 2037, 11368, 26775, 27720, 10395, 1, 247, 7071, 63805, 247555, 460845, 405405, 135135, 1, 502, 23436, 325930, 1939630, 5735730, 8828820, 6756750, 2027025, 1

Offset: 0

Wolfdieter Lang, Oct 14 2005

Previous name was: Coefficient triangle of polynomials used for e.g.f.s of Stirling2 diagonals.
For the o.g.f. of diagonal k of the Stirling2 triangle one has a similar result. See A008517 (second-order Eulerian triangle).
A(m,x), the o.g.f. for column m, satisfies the recurrence A(m,x) = x*(x*(d/dx)A(m-1,x) + m*A(m-1,x))/(1-(m+1)*x), for m >= 1 and A(0,x) = 1/(1-x).
The e.g.f. for the sequence in column k+1, k >= 0, of A008278, i.e., for the diagonal k >= 0 of the Stirling2 triangle A048993, is exp(x)*Sum_{m=0..k} a(k,m)*(x^(m+k))/(m+k)!.
It appears that the triangles in this sequence and A124324 have identical columns, except for shifts. - Jörgen Backelin, Jun 20 2022
A refined version of this triangle is given in A356145, which contains a link providing the precise relationship between A124324 and this entry, confirming Jörgen Backelin's observation above. - Tom Copeland, Sep 24 2022

			Triangle starts:
  [1]
  [1, 1]
  [1, 4,  3]
  [1, 11, 25,  15]
  [1, 26, 130, 210,  105]
  [1, 57, 546, 1750, 2205, 945]
  ...
The e.g.f. of [0,0,1,7,25,65,...], the k=3 column of A008278, but with offset n=0, is exp(x)*(1*(x^2)/2! + 4*(x^3)/3! + 3*(x^4)/4!).
Third row [1,4,3]: There are three plane increasing trees on 3 vertices. The number of colors are shown to the right of a vertex.
...................................................
....1o.(1+t)...........1o.t*(1+t).....1o.t*(1+t)...
....|................. /.\............/.\..........
....|................ /...\........../...\.........
....2o.(1+t)........2o.....3o......3o....2o........
....|..............................................
....|..............................................
....3o.............................................
...................................................
The total number of trees is (1+t)^2 + t*(1+t) + t*(1+t) = 1+4*t+3*t^2 = R(2,t).

F. Bergeron, Ph. Flajolet and B. Salvy, Varieties of Increasing Trees, Lecture Notes in Computer Science vol. 581, ed. J.-C. Raoult, Springer 1992, pp. 24-48.
D. Dominici, Nested derivatives: A simple method for computing series expansions of inverse functions. arXiv:math/0501052v2 [math.CA], 2005.
Wolfdieter Lang, First ten rows.
MathOverflow, Recursion for row polynomials of A112493, (2025).
Andrew Elvey Price and Alan D. Sokal, Phylogenetic trees, augmented perfect matchings, and a Thron-type continued fraction (T-fraction) for the Ward polynomials, arXiv:2001.01468 [math.CO], 2020.

Row sums give A006351(k+1), k>=0.
The column sequences start with A000012 (powers of 1), A000295 (Eulerian numbers), A112495, A112496, A112497.
Cf. A008278, A008517, A048993, A054589, A075856, A134991, A201637, A124324.
Antidiagonal sums give A000110.
Cf. A356145.

T := (n, k) -> add(combinat:-eulerian2(n, j)*binomial(n-j, n-k), j=0..n):
seq(seq(T(n, k), k=0..n), n=0..9); # Peter Luschny, Apr 11 2016

max = 11; f[x_, t_] := -1 - (1 + t)/t*ProductLog[-t/(1 + t)*Exp[(x - t)/(1 + t)]]; coes = CoefficientList[ Series[f[x, t], {x, 0, max}, {t, 0, max}], {x, t}]* Range[0, max]!; Table[coes[[n, k]], {n, 0, max}, {k, 1, n - 1}] // Flatten (* Jean-François Alcover, Nov 22 2012, from e.g.f. *)

a(k, m) = 0 if k < m, a(k, -1):=0, a(0, 0)=1, a(k, m)=(m+1)*a(k-1, m) + (k+m-1)*a(k-1, m-1) else.
From Peter Bala, Sep 30 2011: (Start)
E.g.f.: A(x,t) = -1-((1+t)/t)*LambertW(-(t/(1+t))*exp((x-t)/(1+t))) = x + (1+t)*x^2/2! + (1+4*t+3*t^2)*x^3/3! + .... A(x,t) is the inverse function of (1+t)*log(1+x)-t*x.
A(x,t) satisfies the partial differential equation (1-x*t)*dA/dx = 1 + A + t*(1+t)*dA/dt. It follows that the row generating polynomials R(n,t) satisfy the recurrence R(n+1,t) =(n*t+1)*R(n,t) + t*(1+t)*dR(n,t)/dt. Cf. A054589 and A075856. The polynomials t/(1+t)*R(n,t) are the row polynomials of A134991.
The generating function A(x,t) satisfies the autonomous differential equation dA/dx = (1+A)/(1-t*A). Applying [Bergeron et al., Theorem 1] gives a combinatorial interpretation for the row generating polynomials R(n,t): R(n,t) counts plane increasing trees on n+1 vertices where the non-leaf vertices of outdegree k come in t^(k-1)*(1+t) colors. An example is given below. Cf. A006351, which corresponds to the case t = 1. Applying [Dominici, Theorem 4.1] gives the following method for calculating the row polynomials R(n,t): Let f(x) = (1+x)/(1-x*t). Then R(n,t) = (f(x)*d/dx)^n(f(x)) evaluated at x = 0. (End)
Sum_{j=0..n} T(n-j,j) = A000110(n). - Alois P. Heinz, Jun 20 2022
From Mikhail Kurkov, Apr 01 2025: (Start)
E.g.f.: B(y) = -w/(x*(1+w)) where w = LambertW(-x/(1+x)*exp((y-x)/(1+x))) satisfies the first-order ordinary differential equation (1+x)*B'(y) = B(y)*(1+x*B(y))^2, hence row polynomials are P(n,x) = P(n-1,x) + x*Sum_{j=0..n-1} binomial(n, j)*P(j,x)*P(n-j-1,x) for n > 0 with P(0,x) = 1 (see MathOverflow link).
Conjecture: row polynomials are P(n,x) = Sum_{i=0..n} Sum_{j=0..i} Sum_{k=0..j} (n+i)!*Stirling1(n+j-k,j-k)*x^k*(x+1)^(j-k)*(-1)^(j+k)/((n+j-k)!*(i-j)!*k!). (End)
Conjecture: g.f. satisfies 1/(1 - x - x*y/(1 - 2*x - 2*x*y/(1 - 3*x - 3*x*y/(1 - 4*x - 4*x*y/(1 - 5*x - 5*x*y/(1 - ...)))))) (see A383019 for conjectures about combinatorial interpretation and algorithm for efficient computing). - Mikhail Kurkov, Apr 21 2025

New name from Peter Luschny, Apr 11 2016

A048160 Triangle giving T(n,k) = number of (n,k) labeled rooted Greg trees (n >= 1, 0<=k<=n-1).

Original entry on oeis.org

1, 2, 1, 9, 10, 3, 64, 113, 70, 15, 625, 1526, 1450, 630, 105, 7776, 24337, 31346, 20650, 6930, 945, 117649, 450066, 733845, 650188, 329175, 90090, 10395, 2097152, 9492289, 18760302, 20925065, 14194180, 5845455, 1351350, 135135, 43046721

Offset: 1

N. J. A. Sloane

An (n,k) rooted Greg tree can be described as a rooted tree with n black nodes and k white nodes where only the black nodes are labeled and the white nodes have at least 2 children. - Christian G. Bower, Nov 15 1999

			Triangle begins:
  1;
  2, 1;
  9, 10, 3;
  64, 113, 70, 15;
  ...

C. Flight, How many stemmata?, Manuscripta, 34 (1990), 122-128.
C. Flight, How many stemmata?, Manuscripta, 34 (1990), 122-128. (Annotated scanned copy)
C. Flight, Letter to N. J. A. Sloane, Nov 1990
D. J. Jeffrey, G. A. Kalugin, N. Murdoch, Lagrange inversion and Lambert W, Preprint 2015.
M. Josuat-Vergès, Derivatives of the tree function, arXiv preprint arXiv:1310.7531 [math.CO], 2013.
Index entries for sequences related to rooted trees
Index entries for sequences related to trees

Row sums give A005264. Cf. A005263, A048159, A052300-A052303. A054589.

t[n_ /; n >= 1, k_ /; k >= 0] /; 0 <= k <= n-1 := t[n, k] = (n+k-2) t[n-1, k-1] + (2n + 2k - 2)*t[n-1, k] + (k+1) t[n-1, k+1]; t[1, 0] = 1; t[, ] = 0; Flatten[Table[t[n, k], {n, 1, 9}, {k, 0, n-1}]] (* Jean-François Alcover, Jul 20 2011, after formula *)

T(n, 0)=n^(n-1), T(n, k)=(n+k-2)*T(n-1, k-1)+(2*n+2*k-2)*T(n-1, k)+(k+1)*T(n-1, k+1).
From Peter Bala, Sep 29 2011: (Start)
E.g.f.: compositional inverse with respect to x of t*(exp(-x)-1) + (1+t)*x*exp(-x) = compositional inverse with respect to x of (x - (2+t)*x^2/2! + (3+2*t)*x^3/3! - (4+3*t)*x^4/4! + ...) = x + (2+t)*x^2/2! + (9+10*t+3*t^2)*x^3/3! + ....
The row generating polynomials R(n,t) satisfy the recurrence R(n+1,t) = (1+t)^2*R'(n,t)+n*(2+t)*R(n,t) with R(1,t) = 1.
The shifted row polynomials R(n,t-1) are the row generating polynomials of A054589. (End)
From Peter Bala, Sep 12 2012: (Start)
It appears that the entries in column k = 1 are given by T(n,1) = (n+1)^n - 2*n^n (checked up to n = 15) - see A176824.
Assuming this, we could then use the recurrence equation to obtain explicit formulas for columns k = 2,3,....
For example, T(n,2) = 1/2*{(n+2)^(n+1) - 4*(n+1)^(n+1) + (4*n+3)*n^n}. (End)

More terms from Larry Reeves (larryr(AT)acm.org), Apr 07 2000

A075856 Triangle formed from coefficients of the polynomials p(1)=x, p(n+1) = (n + x(n+1))p(n) + xx(d/dx)p(n).

Original entry on oeis.org

1, 1, 3, 2, 10, 15, 6, 40, 105, 105, 24, 196, 700, 1260, 945, 120, 1148, 5068, 12600, 17325, 10395, 720, 7848, 40740, 126280, 242550, 270270, 135135, 5040, 61416, 363660, 1332100, 3213210, 5045040, 4729725, 2027025

Offset: 1

F. Chapoton, Oct 15 2002

Constant terms of polynomials related to Ramanujan psi polynomials (see Zeng reference).

			Triangle begins
    1;
    1,    3;
    2,   10,   15;
    6,   40,  105,   105;
   24,  196,  700,  1260,   945;
  120, 1148, 5068, 12600, 17325, 10395;
  ...
p(1) = x, p(2) = 3*x^2 + x, p(3) = 15*x^3 + 10*x^2 + 2*x, etc. - _Michael Somos_, Mar 17 2011

P. Bala, Diagonals of triangles with generating function exp(t*F(x)).
J. Fernando Barbero G., Jesús Salas, and Eduardo J. S. Villaseñor, Bivariate Generating Functions for a Class of Linear Recurrences. I. General Structure, arXiv:1307.2010 [math.CO], 2013.
Brian Drake, Ira M. Gessel, and Guoce Xin, Three Proofs and a Generalization of the Goulden-Litsyn-Shevelev Conjecture on a Sequence Arising in Algebraic Geometry, J. of Integer Sequences, Vol. 10 (2007), #07.3.7.
Dominique Dumont and Armand Ramamonjisoa, Grammaire de Ramanujan et Arbres de Cayley, Electr. J. Combinatorics, Volume 3, Issue 2 (1996) R17 (see page 16).
H. W. Gould, A Set of Polynomials Associated with the Higher Derivatives of y = x^x, Rocky Mountain J. Math. Volume 26, Number 2 (1996), 615-625.
M. Josuat-Vergès, Derivatives of the tree function, arXiv preprint arXiv:1310.7531 [math.CO], 2013.
S. Ramanujan, Notebook entry
P. W. Shor, Problem 78-6: A combinatorial identity, in Problems and Solutions column, SIAM Review; problem in 20, p. 394 (1978); solution in 21, pp. 258-260 (1979). [N. Sato, Feb 19 2010]
P. W. Shor, A = B (but not quite); 3-d array with multiple recurrences, MathOverflow, Nov 2010-Nov 2011.
J. Zeng, A Ramanujan sequence that refines the Cayley formula for trees, manuscript, 1996.
J. Zeng, A Ramanujan sequence that refines the Cayley formula for trees, Ramanujan J., 3(1999) 1, 45-54.

Cf. A000142, A000312, A001147, A054589. A185164, A209937.

See A239098 for another version.

p[1] = x; p[n_] := p[n] = (n - 1 + x*n)*p[n - 1] + x*x*D[p[n - 1], x]; Flatten[Rest[CoefficientList[#1, x]] & /@ Table[p[n], {n, 8}]] (* Jean-François Alcover, May 31 2011 *)

{T(n, k) = if( k<1 || nMichael Somos, Mar 17 2011 */

T(n, k) = (n-1) * T(n-1, k) + (n+k-1) * T(n-1, k-1). - Michael Somos, Mar 17 2011
G.f.: A(x, t) = Sum_{n>0} p[n] t^n / n! satisfies (dA / dt) * (x + t - 1) = x * (1 + A)^2 * (x * (1 + A) - 1). - Michael Somos, Mar 17 2011
T(n, 1) = (n-1)! = A000142(n-1). T(n, n) = A001147(n). Sum_{k>0} T(n, k) = n^n = A000312(n). Sum_{k>0} T(n, k) x^k = p[n].
From Peter Bala, Mar 14 2012: (Start)
This triangle is A185164 read by diagonals.
Let F(x) = x + (1-x)*log(1-x). The e.g.f. is given by the compositional inverse
(x - t*F(x))^(-1) = x + t*x^2/2! + (t + 3*t^2)x^3/3! + (2*t + 10*t^2 + 15*t^3)*x^4/4! + ....
Let f(x) = 1/log(1+x) and define inductively D^(n+1)(f(x)) = f(x)*(d/dx)(D^n(f(x))) with D^(0)f(x) = f(x). Then D^(n)f = (-1)^n*Sum_{k = 1..n} T(n,k)*f^(n-k)/((1+x)^n*f^(2n+1)).
(End)

A181415 Irregular triangle a(n,k) = A049009(n,k)/n, read by rows 1<=k<=A000041(n).

Original entry on oeis.org

1, 1, 1, 1, 6, 2, 1, 12, 9, 36, 6, 1, 20, 40, 120, 180, 240, 24, 1, 30, 75, 50, 300, 1200, 300, 1200, 2700, 1800, 120, 1, 42, 126, 210, 630, 3150, 2100, 3150, 4200, 25200, 12600, 12600, 37800, 15120, 720, 1, 56, 196, 392, 245, 1176, 7056, 11760, 8820, 11760, 11760, 88200

Offset: 1

Alford Arnold, Oct 20 2010

			Row three is calculated as follows:
( 3 18 6) divided by (3 3 3) yielding (1 6 2)
1;
1,1;
1,6,2;
1,12,9,36,6;
1,20,40,120,180,240,24;
1,30,75,50,300,1200,300,1200,2700,1800,120;
1,42,126,210,630,3150,2100,3150,4200,25200,12600,12600,37800,15120,720;

Cf. A000169 (row sums), A000081 (unlabeled rooted trees) A179438 (a similar refinement), A054589, A135278, A019538, A101817, A101818

Sum_{k=1.. A000041(n)} a(n,k) = A000169(n). (Row sums)

a(n,k) = A098546(n,k) *A049019(n,k) /n. - Compare with the formula in A101818.

Edited by R. J. Mathar, May 17 2016

A217922 Triangle read by rows: labeled trees counted by improper edges.

Original entry on oeis.org

1, 1, 2, 1, 6, 7, 3, 24, 46, 40, 15, 120, 326, 430, 315, 105, 720, 2556, 4536, 4900, 3150, 945, 5040, 22212, 49644, 70588, 66150, 38115, 10395, 40320, 212976, 574848, 1011500, 1235080, 1032570, 540540, 135135

Offset: 1

David Callan, Oct 14 2012

T(n,k) is the number of labeled trees on [n], rooted at 1, with k improper edges, for n >= 1, k >= 0. See Zeng link for definition of improper edge.

			Triangle begins:
  \ k  0....1....2....3....4......
  n
  1 |..1
  2 |..1
  3 |..2....1
  4 |..6....7....3
  5 |.24...46...40....15
  6 |120..326..430...315...105
T(4,2) = 3 because we have 1->3->4->2, 1->4->2->3, 1->4->3->2, in each of which the last 2 edges are improper.

G. C. Greubel, Rows n = 1..50 of the irregular triangle, flattened
J. Fernando Barbero G., Jesús Salas, and Eduardo J. S. Villaseñor, Bivariate Generating Functions for a Class of Linear Recurrences. I. General Structure, arXiv:1307.2010 [math.CO], 2013-2014.
William Y. C. Chen, Amy M. Fu, and Elena L. Wang, A Grammatical Calculus for the Ramanujan Polynomials, arXiv:2506.01649 [math.CO], 2025. See p. 3.
Dominique Dumont and Armand Ramamonjisoa, Grammaire de Ramanujan et Arbres de Cayley, Electr. J. Combinatorics, Volume 3, Issue 2 (1996) R17 (see page 17).
Matthieu Josuat-Vergès, Derivatives of the tree function, arXiv preprint arXiv:1310.7531 [math.CO], 2013.
Lucas Randazzo, Arboretum for a generalization of Ramanujan polynomials, arXiv:1905.02083 [math.CO], 2019.
Jiang Zeng, A Ramanujan sequence that refines the Cayley formula for trees, Ramanujan Journal 3 (1999) 1, 45-54, [DOI]

Row sums are A000272.

Cf. A054589, A075856.

function T(n,k) // T = A217922
  if k lt 0 or k gt n-2 then return 0;
  elif k eq 0 then return Factorial(n-1);
  else return (n-1)*T(n-1,k) + (n+k-3)*T(n-1,k-1);
  end if;
end function;
[1] cat [T(n,k): k in [0..n-2], n in [2..12]]; // G. C. Greubel, Jan 10 2025

T[n_, k_]:= T[n,k]= If[k<0 || k>n-2, 0, If[k==0, (n-1)!, (n-1)*T[n-1,k] + (n+k-3)*T[n-1, k-1]]];
Join[{1}, Table[T[n,k], {n,12}, {k,0,n-2}]//Flatten] (* modified by G. C. Greubel, May 07 2019 *)

def T(n, k):
    if k==0: return factorial(n-1)
    elif (k<0 or k > n-2): return 0
    else: return (n-1)*T(n-1, k) + (n+k-3)* T(n-1, k-1)
flatten([1] + [[T(n, k) for k in (0..n-2)] for n in (2..12)]) # G. C. Greubel, May 07 2019

T(n, k) = (n-1)*T(n-1, k) + (n+k-3)*T(n-1, k-1), for 1 <= k <= n-2, with T(n, 0) = (n-1)!. - G. C. Greubel, Jan 10 2025

cp's OEIS Frontend

A042977 Triangle T(n,k) read by rows: coefficients of a polynomial sequence occurring when calculating the n-th derivative of Lambert function W.

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Maxima

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A112493 Triangle read by rows, T(n, k) = Sum_{j=0..n} C(n-j, n-k)*E2(n, j), where E2 are the second-order Eulerian numbers A201637, for n >= 0 and 0 <= k <= n.

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A048160 Triangle giving T(n,k) = number of (n,k) labeled rooted Greg trees (n >= 1, 0<=k<=n-1).

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A075856 Triangle formed from coefficients of the polynomials p(1)=x, p(n+1) = (n + x*(n+1))*p(n) + x*x*(d/dx)p(n).

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A181415 Irregular triangle a(n,k) = A049009(n,k)/n, read by rows 1<=k<=A000041(n).

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A217922 Triangle read by rows: labeled trees counted by improper edges.

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A075856 Triangle formed from coefficients of the polynomials p(1)=x, p(n+1) = (n + x(n+1))p(n) + xx(d/dx)p(n).