A075856 -id:A075856 - cp's OEIS frontend

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.

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

A054589 Table related to labeled rooted trees, cycles and binary trees.

Original entry on oeis.org

1, 1, 1, 2, 4, 3, 6, 18, 25, 15, 24, 96, 190, 210, 105, 120, 600, 1526, 2380, 2205, 945, 720, 4320, 13356, 26488, 34650, 27720, 10395, 5040, 35280, 128052, 305620, 507430, 575190, 405405, 135135

Offset: 1

F. Chapoton, Apr 14 2000

The left column is (n-1)!, the right column is (2n-3)!!, the total of each row is n^(n-1).
Constant terms of polynomials related to Ramanujan psi polynomials (see Zeng reference).
From Peter Bala, Sep 29 2011: (Start)
Differentiating n times the Lambert function W(x) = Sum_{n>=1} n^(n-1)*x^n/n! with respect to x yields (d/dx)^n W(x) = exp(n*W(x))/(1-W(x))^n*R(n,1/(1-W(x))), where R(n,x) is the n-th row polynomial of this triangle. The first few values are R(1,x) = 1, R(2,x) = 1+x, R(3,x) = 2+4*x+3*x^2. The Ramanujan polynomials R(n,x) are strongly x-log-convex [Chen et al.].
Shor and Dumont-Ramamonjisoa have proved independently that the coefficient of x^k in R(n,x) counts rooted labeled trees on n vertices with k improper edges. Drake, Example 1.7.3, gives another combinatorial interpretation for this triangle as counting a family of labeled trees.
(End)

			Triangle begins:
  {1},
  {1,  1},
  {2,  4,  3},
  {6, 18, 25, 15},
  ...

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.
William Y. C. Chen, Larry X. W. Wang, and Arthur L. B. Yang, Recurrence Relations for Strongly q-Log-Convex Polynomials, arXiv:0806.3641v1 [math.CO], 2008.
Diego Dominici, Nested derivatives: A simple method for Computing series expansions of inverse functions. arXiv:math/0501052v2 [math.CA], 2005.
Brian Drake, An inversion theorem for labeled trees and some limits of areas under lattice paths, A dissertation presented to the Faculty of the Graduate School of Arts and Sciences of Brandeis University.
Dominique Dumont and Armand Ramamonjisoa, Grammaire de Ramanujan et Arbres de Cayley, Electr. J. Combinatorics, Volume 3, Issue 2 (1996) R17 (see page 16).
D. J. Jeffrey, G. A. Kalugin, and N. Murdoch, Lagrange inversion and Lambert W, Preprint 2015.
Matthieu Josuat-Vergès, Derivatives of the tree function, arXiv preprint arXiv:1310.7531 [math.CO], 2013.
Peter W. Shor, A new proof of Cayley's formula for counting labeled trees, J. Combin. Theory Ser. A 71 (1995), no. 1, 154-158.
Jiang Zeng, A Ramanujan sequence that refines the Cayley formula for trees, Ramanujan J., 3(1999) 1, 45-54.

Cf. A000169, A001147, A075856, A048159, A048160, A042977.

p[1] = 1; p[n_] := p[n] = Expand[x^2*D[p[n-1], x] + (n-1)(1+x)p[n-1]]; Flatten[ Table[ CoefficientList[ p[n], x], {n, 1, 8}]] (* Jean-François Alcover, Jul 22 2011 *)
Clear[a];
a[1, 0] = 1;
a[n_, k_] /; k < 0 || k >= n := 0
a[n_, k_] /; 0 <= k <= n - 1 :=
a[n, k] = (n - 1) a[n - 1, k] + (n + k - 2) a[n - 1, k - 1]
Table[a[n, k], {n, 20}, {k, 0, n - 1}] (* David Callan, Oct 14 2012 *)

The polynomials p_n = Sum a[n, k]x^k satisfy p_1=1 and p_(n+1) = x*x*dp_n/dx+n*(1+x)*p_n.
From Peter Bala, Sep 29 2011: (Start)
E.g.f.: series reversion with respect to x of (1-t+(t-1+x*t)*exp(-x)) = x + (1+t)*x^2/2! + (2+4*t+3*t^2)*x^3/3! + ....
The sequence of shifted row polynomials {p_n(1+t)}n>=1 begins [1,2+t,9+10*t+3*t^2,...]. These are the row polynomials of A048160.
(End)
Let f(x) = exp(x)/(1-t*x). The e.g.f. A(x,t) = x + (1+t)*x^2/2! + (2+4*t+3*t^2)*x^3/3! + ... satisfies the autonomous differential equation dA/dx = f(A). The n-th row polynomial (n>=1) equals D^(n-1)(f(x)) evaluated at x = 0, where D is the operator f(x)*d/dx (apply [Dominici, Theorem 4.1]). - Peter Bala, Nov 09 2011
The polynomials (1+t)^(n-1)*p_n(1/(1+t)) are (up to sign) the row polynomials of A042977. - Peter Bala, Jul 23 2012
Let q_n = Sum_{k>=0} a(n,k)*t^(n-k), with q_0 = 1. (So q_1=t, q_2 = t+t^2, and q_3 = 3*t + 4*t^2 + 2*t^3.)  Then Sum_{n>=0} q_n*x^n/n! = t - W((t-1-t^2*x)*exp(t-1)), where W is the Lambert function. - Ira M. Gessel, Jan 06 2012

A185164 Coefficients of a set of polynomials associated with the derivatives of x^x.

Original entry on oeis.org

1, 1, 2, 3, 6, 10, 24, 40, 15, 120, 196, 105, 720, 1148, 700, 105, 5040, 7848, 5068, 1260, 40320, 61416, 40740, 12600, 945, 362880, 541728, 363660, 126280, 17325, 3628800, 5319072, 3584856, 1332100, 242550, 10395, 39916800, 57545280, 38764440, 15020720, 3213210, 270270

Offset: 2

Peter Bala, Mar 12 2012

Gould shows that the derivatives of x^x are given by (d/dx)^n(x^x) = (x^x)*Sum_{k = 0..n} (-1)^k*binomial(n,k)*(1 + log(x))^(n-k)*x^(-k)*R(k,x), where R(n,x) is a polynomial in x of degree floor(n/2). The first few values are R(0,x) = 1, R(1,x) = 0, R(2,x) = x, R(3,x) = x and R(4,x) = 2*x + 3*x^2. The coefficients of these polynomials are listed in the table for n >= 2. Gould gives an explicit formula for R(n,x) as a triple sum, and also an expression in terms of the Comtet numbers A008296.

This table read by diagonals gives A075856.

			Triangle begins
n\k.|.....1.....2.....3.....4
= = = = = = = = = = = = = = =
..2.|.....1
..3.|.....1
..4.|.....2.....3
..5.|.....6....10
..6.|....24....40....15
..7.|...120...196...105
..8.|...720..1148...700...105
..9.|..5040..7848..5068..1260
...
Fourth derivative of x^x:
x^(-x)*(d/dx)^4(x^x) = (1+log(x))^4 + C(4,2)/x^2*(1+log(x))^2*x - C(4,3)/x^3*(1+log(x)) + C(4,4)/x^4*(2*x + 3*x^2).
Example of recurrence relation for table entries:
T(7,2) = 4*T(6,2) + 6*T(5,1) = 4*40 + 6*6 = 196.

Robert Israel, Table of n, a(n) for n = 2..10001 (rows 2 to 200, flattened)
Peter Bala, Diagonals of triangles with generating function exp(t*F(x)).
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.

Cf. A008296, A075856, A203852 (row sums).

T[2,1]:= 1:
for n from 3 to 15 do
  for k from 1 to floor(n/2) do
    T[n,k]:= (n-1-k)*`if`(k<= floor((n-1)/2),T[n-1,k],0) + `if`(k>=2 and k-1 <= floor((n-2)/2),(n-1)*T[n-2,k-1],0)
od od:
seq(seq(T[n,k],k=1..floor(n/2)),n=2..15); # Robert Israel, Jan 13 2016

m = 14; F = Exp[t (x + (1-x) Log[1-x])];
cc = CoefficientList[# + O[t]^m, t]& /@ CoefficientList[F + O[x]^m, x]* Range[0, m - 1]!;
Rest /@ Drop[cc, 2] (* Jean-François Alcover, Jun 26 2019 *)

# uses[bell_transform from A264428]
# Computes the full triangle for n>=0 and 0<=k<=n.
def A185164_row(n):
    g = lambda k: factorial(k-1) if k>0 else 0
    s = [g(k) for k in (0..n)]
    return bell_transform(n, s)
[A185164_row(n) for n in (0..10)] # Peter Luschny, Jan 13 2016

Recurrence relation: T(n+1,k) = (n - k)*T(n,k) + n*T(n-1,k-1).
The diagonal entries D(n,k) := T(n+k,k) satisfy the recurrence D(n+1,k) = n*D(n,k) + (n + k)*D(n,k-1) so this table read by diagonals is A075856.
E.g.f.: F(x,t) = exp(t*(x + (1 - x)*log(1 - x))) = Sum_{n = 0..oo} R(n,t)*x^n/n! = 1 + t*x^2/2! + t*x^3/3! + (2*t + 3*t^2)*x^4/4! + .... The e.g.f. F(x,t) satisfies the partial differential equation (1 - x)*dF/dx + t*dF/dt = x*t*F.
This gives the recurrence relation for the row generating polynomials: R(n+1,x) = n*R(n,x) - x*d/dx(R(n,x)) + n*x*R(n-1,x) for n >= 1, with initial conditions R(0,x) = 1, R(1,x) = 0.
The e.g.f. for the triangle read by diagonals is given by the series reversion (with respect to x) (x - t*(x + (1 - x)*log(1 - 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! + ....
Diagonal sums: Sum_{k = 1..n} T(n+k,k) = n^n , n >= 1.
Row sums A203852.
Also the Bell transform of the sequence g(k) = (k-1)! if k>0 else 0. For the definition of the Bell transform see A264428. - Peter Luschny, Jan 13 2016

More terms from Jean-François Alcover, Jun 26 2019

A209937 E.g.f. A(x) satisfies: A( x - Sum_{n>=2} x^n/(n*(n-1)/2) ) = x.

Original entry on oeis.org

1, 2, 14, 164, 2692, 56832, 1466656, 44735392, 1574507104, 62807331520, 2800211567936, 137988945383808, 7447519816062848, 436913348544200192, 27682538499602786816, 1883880221782019929088, 137046014280583363879936, 10612885049611654523670528

Offset: 1

Paul D. Hanna, Mar 15 2012

nonn

Compare e.g.f. to the identity: let W(x) = Sum_{n>=1} (n-1)^(n-1)*x^n/n!, then W( x - Sum_{n>=2} x^n/(n*(n-1)) ) = x.

			E.g.f.: A(x) = x + 2*x^2/2! + 14*x^3/3! + 164*x^4/4! + 2692*x^5/5! + 56832*x^6/6! + 1466656*x^7/7! + 44735392*x^8/8! +...
Let R(x) be the series reversion of e.g.f. A(x), then R(x) begins:
R(x) = x - x^2/1 - x^3/3 - x^4/6 - x^5/10 - x^6/15 - x^7/21 - x^8/28 -...
...
Compare to the series reversion of the function W(x) defined by:
W(x) = x + x^2/2! + 2^2*x^3/3! + 3^3*x^4/4! + 4^4*x^5/5! + 5^5*x^6/6! +...
where W(x - x^2/2 - x^3/6 - x^4/12 - x^5/20 - x^6/30 - x^7/42 -...) = x.

Cf. A209923, A075856.

A209937_list := proc(len) local A, n; A[1] := 1; for n from 2 to len do
A[n] := (n-2)*A[n-1] + add(binomial(n,j)*A[j]*A[n-j], j=1..n-1) od:
convert(A,list) end: A209937_list(18); # Peter Luschny, May 24 2017

Rest[CoefficientList[InverseSeries[Series[-x-2*(1-x)*Log[1-x], {x, 0, 20}], x],x]*Range[0, 20]!] (* Vaclav Kotesovec, Jan 23 2014 *)

{a(n)=n!*polcoeff(serreverse(x-sum(m=2, n, x^m/(m*(m-1)/2)) +x*O(x^n)), n)}

{a(n)=n!*polcoeff(serreverse(-x-2*(1-x)*log(1-x +x*O(x^n))), n)}
for(n=1, 25, print1(a(n), ", "))

/* From a formula of Michael Somos for triangle A075856: */
{A075856(n, k)=if(k<1|nA075856(n-1,k)+(n+k-1)*A075856(n-1,k-1)))}
{a(n)=if(n<1,0,if(n==1,1,sum(k=1,n-1,A075856(n-1, k)*2^k)))}
for(n=1,20,print1(a(n),", "))

E.g.f. A(x) satisfies: A( -x - 2*(1-x)*log(1-x) ) = x.
a(n) = Sum_{k=1..n-1} A075856(n-1,k)*2^k for n>1 with a(1)=1.
a(n) ~ n^(n-1) / (sqrt(2) * (2-exp(1/2))^(n-1/2) * exp(n/2+1/2)). - Vaclav Kotesovec, Jan 23 2014
a(n) = (n-2)*a(n-1) + Sum_{j=1..n-1} binomial(n,j)*a(j)*a(n-j) for n>1, a(1)=1. - Peter Luschny, May 24 2017

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

A239098 Triangle read by rows: T(0,0)=1; T(m,0)=0; otherwise T(m,n) = (m-1)T(m-1,n)+(m-1+n)T(m-1,n-1).

Original entry on oeis.org

1, 0, 1, 0, 1, 3, 0, 2, 10, 15, 0, 6, 40, 105, 105, 0, 24, 196, 700, 1260, 945, 0, 120, 1148, 5068, 12600, 17325, 10395, 0, 720, 7848, 40740, 126280, 242550, 270270, 135135, 0, 5040, 61416, 363660, 1332100, 3213210, 5045040, 4729725, 2027025, 0, 40320, 541728, 3584856, 15020720, 43022980, 85345260, 113513400, 91891800, 34459425

Offset: 0

N. J. A. Sloane, Mar 23 2014

If the first column is omitted we get A075856, which has much more information about this triangle.

			Triangle begins:
1,
0, 1,
0, 1, 3,
0, 2, 10, 15,
0, 6, 40, 105, 105,
0, 24, 196, 700, 1260, 945,
0, 120, 1148, 5068, 12600, 17325, 10395,
0, 720, 7848, 40740, 126280, 242550, 270270, 135135,
...

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).

Elena L. Wang and Guoce Xin, On Ward Numbers and Increasing Schröder Trees, arXiv:2507.15654 [math.CO], 2025. See pp. 12-13.

Cf. A075856.

T:=proc(m,n) option remember;
if (m=0) and (n=0) then 1;
elif (m=0) or (n=0) then 0;
else (m-1)*T(m-1,n)+(m-1+n)*T(m-1,n-1); fi; end;
M:=20;
for m from 0 to M do
lprint([seq(T(m,n),n=0..m)]); od:

cp's OEIS Frontend

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A239098 Triangle read by rows: T(0,0)=1; T(m,0)=0; otherwise T(m,n) = (m-1)*T(m-1,n)+(m-1+n)*T(m-1,n-1).

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A239098 Triangle read by rows: T(0,0)=1; T(m,0)=0; otherwise T(m,n) = (m-1)T(m-1,n)+(m-1+n)T(m-1,n-1).