A075263 -id:A075263 - cp's OEIS frontend

A123125 Triangle of Eulerian numbers T(n,k), 0 <= k <= n, read by rows.

1, 0, 1, 0, 1, 1, 0, 1, 4, 1, 0, 1, 11, 11, 1, 0, 1, 26, 66, 26, 1, 0, 1, 57, 302, 302, 57, 1, 0, 1, 120, 1191, 2416, 1191, 120, 1, 0, 1, 247, 4293, 15619, 15619, 4293, 247, 1, 0, 1, 502, 14608, 88234, 156190, 88234, 14608, 502, 1, 0, 1, 1013, 47840, 455192, 1310354, 1310354, 455192, 47840, 1013, 1

Offset: 0

Philippe Deléham, Sep 30 2006

The beginning of this sequence does not quite agree with the usual version, which is A173018. - N. J. A. Sloane, Nov 21 2010
Each row of A123125 is the reverse of the corresponding row in A173018. - Michael Somos, Mar 17 2011
A008292 (subtriangle for k>=1 and n>=1) is the main entry for these numbers.
Triangle T(n,k), 0 <= k <= n, read by rows given by [0,1,0,2,0,3,0,4,0,5,0,...] DELTA [1,0,2,0,3,0,4,0,5,0,6,...] where DELTA is the operator defined in A084938.
Row sums are the factorials. - Roger L. Bagula and Gary W. Adamson, Aug 14 2008
If the initial zero column is deleted, the result is A008292. - Roger L. Bagula and Gary W. Adamson, Aug 14 2008
This result gives an alternative method of calculating the Eulerian numbers by an Umbral Calculus expansion from Comtet. - Roger L. Bagula, Nov 21 2009
This function seems to be equivalent to the PolyLog expansion. - Roger L. Bagula, Nov 21 2009
A raising operator formed from the e.g.f. of this entry is the generator of a sequence of polynomials p(n,x;t) defined in A046802 that specialize to those for A119879 as p(n,x;-1), A007318 as p(n,x;0), A073107 as p(n,x;1), and A046802 as p(n,0;t). See Copeland link for more associations. - Tom Copeland, Oct 20 2015
The Eulerian numbers in this setup count the permutation trees of power n and width k (see the Luschny link). For the associated combinatorial statistic over permutations see the Sage program below and the example section. - Peter Luschny, Dec 09 2015 [See Elder et al. link. Peter Luschny, Jul 13 2022]
From Wolfdieter Lang, Apr 03 2017: (Start)
The row polynomials R(n, x) = Sum_{k=0..n} T(n, k)*x^k are the numerator polynomials of the o.g.f. G(n, x) of n-powers {m^n}_{m>=0} (with 0^0 = 1): G(n, x) = R(n, x)/(1-x)^(n+1). See the Aug 14 2008 formula, where f(x,n) = R(n, x). The e.g.f. of R(n, t) is given in Copeland's Oct 14 2015 formula below.
The first nine column sequences are A000007, A000012, A000295, A000460, A000498, A000505, A000514, A001243, A001244. (End)
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 this entry, 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 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
Let b(n) = (1/(n+1))*Sum_{k=0..n-1} (-1)^(n-k+1)*T(n, k+1) / binomial(n, k+1). Then b(n) = Bernoulli(n, 1) = -n*Zeta(1 - n) = Integral_{x=0..1} F_n(x) for n >= 1. Here F_n(x) are the signed Fubini polynomials (A278075). (See also Rzadkowski and Urlinska, example 1.) - Peter Luschny, Feb 15 2021
Patrick J. Burchell (see link) describes the following method: To get the k-th row of the triangle write the nonnegative integers with a fixed exponent k as a sequence, 0^k, 1^k, 2^k, ..., and then apply the first differences to them k + 1 times. - Peter Luschny, Apr 02 2023

			The triangle T(n, k) begins:
  n\k 0 1    2     3      4       5       6      7     8    9 10...
  0:  1
  1:  0 1
  2:  0 1    1
  3:  0 1    4     1
  4:  0 1   11    11      1
  5:  0 1   26    66     26       1
  6:  0 1   57   302    302      57       1
  7:  0 1  120  1191   2416    1191     120      1
  8:  0 1  247  4293  15619   15619    4293    247     1
  9:  0 1  502 14608  88234  156190   88234  14608   502    1
 10:  0 1 1013 47840 455192 1310354 1310354 455192 47840 1013  1
...  Reformatted. - _Wolfdieter Lang_, Feb 14 2015
------------------------------------------------------------------
The width statistic over permutations, n=4.
  [1, 2, 3, 4] => 3; [1, 2, 4, 3] => 2; [1, 3, 2, 4] => 2; [1, 3, 4, 2] => 2;
  [1, 4, 2, 3] => 2; [1, 4, 3, 2] => 1; [2, 1, 3, 4] => 3; [2, 1, 4, 3] => 2;
  [2, 3, 1, 4] => 2; [2, 3, 4, 1] => 3; [2, 4, 1, 3] => 2; [2, 4, 3, 1] => 2;
  [3, 1, 2, 4] => 3; [3, 1, 4, 2] => 3; [3, 2, 1, 4] => 2; [3, 2, 4, 1] => 3;
  [3, 4, 1, 2] => 3; [3, 4, 2, 1] => 2; [4, 1, 2, 3] => 4; [4, 1, 3, 2] => 3;
  [4, 2, 1, 3] => 3; [4, 2, 3, 1] => 3; [4, 3, 1, 2] => 3; [4, 3, 2, 1] => 2;
Gives row(4) = [0, 1, 11, 11, 1]. - _Peter Luschny_, Dec 09 2015
------------------------------------------------------------------
From _Wolfdieter Lang_, Apr 03 2017: (Start)
Recurrence: T(5, 3) = (6-3)*T(4, 2) + 3*T(4, 3) = 3*11 + 3*11 = 66.
O.g.f. column k=2: (x/(1 - 2*x))*E_x*(x/(1-x)) = (x/(1-x))^2/(1-2*x).
E.g.f. column k=2: A(2, x) = x*A(1, x) + x*E(1, x) = x*1 + x*(exp(x)-1) = x*exp(x), hence E(2, x) = (1 + int(x*exp(-x),x ))*exp(2*x) = exp(x)*(exp(x) - (1+x)). See A000295. (End)

L. Comtet, Advanced Combinatorics, Reidel, Holland, 1978, page 245. [Roger L. Bagula, Nov 21 2009]
Ronald L. Graham, Donald E. Knuth and Oren Patashnik, Concrete Mathematics, 2nd ed.; Addison-Wesley, 1994, p. 268, Row reversed table 268. - Wolfdieter Lang, Apr 03 2017
Douglas C. Montgomery and Lynwood A. Johnson, Forecasting and Time Series Analysis, MaGraw-Hill, New York, 1976, page 91. - Roger L. Bagula and Gary W. Adamson, Aug 14 2008

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
Paul Barry, Eulerian polynomials as moments, via exponential Riordan arrays, arXiv preprint arXiv:1105.3043 [math.CO], 2011, J. Int. Seq. 14 (2011) # 11.9.5
Paul Barry, Generalized Stirling Numbers, Exponential Riordan Arrays, and Toda Chain Equations, Journal of Integer Sequences, 17 (2014), #14.2.3.
Paul Barry, On a transformation of Riordan moment sequences, arXiv:1802.03443 [math.CO], 2018.
Paul Barry, Generalized Eulerian Triangles and Some Special Production Matrices, arXiv:1803.10297 [math.CO], 2018.
V. Batyrev and M. Blume, The functor of toric varieties associated with Weyl chambers and Losev-Manin moduli spaces, p. 11, arXiv:/0911.3607 [math.AG], 2009. [_Tom Copeland_, Oct 16 2015]
Anna Borowiec and Wojciech Mlotkowski, New Eulerian numbers of type D, arXiv:1509.03758 [math.CO], 2015.
Patrick J. Burchell, A Generalisation of Ramanujan's (back of the envelope) Method for Divergent Series, arXiv:2303.14045 math.NT, 2023.
A. Cohen, Eulerian polynomials of spherical type, Münster J. of Math. 1 (2008). [_Tom Copeland_, Oct 16 2015]
Tom Copeland, The Elliptic Lie Triad: Ricatti and KdV Equations, Infinigens, and Elliptic Genera
Jennifer Elder, Nadia Lafrenière, Erin McNicholas, Jessica Striker and Amanda Welch, Homomesies on permutations -- an analysis of maps and statistics in the FindStat database, arXiv:2206.13409 [math.CO], 2022-2023. (Def. 4.20 and Prop. 4.22.)
FindStat - Combinatorial Statistic Finder, The number of descents of a permutation.
F. Hirzebruch, Eulerian polynomials, Münster J. of Math. 1 (2008), pp. 9-12.
P. Hitczenko and S. Janson, Weighted random staircase tableaux, arXiv preprint arXiv:1212.5498 [math.CO], 2012.
Hsien-Kuei Hwang, Hua-Huai Chern, and Guan-Huei Duh, An asymptotic distribution theory for Eulerian recurrences with applications, arXiv:1807.01412 [math.CO], 2018.
Svante Janson, Euler-Frobenius numbers and rounding, arXiv preprint arXiv:1305.3512 [math.PR], 2013.
Katarzyna Kril and Wojciech Mlotkowski, Permutations of Type B with Fixed Number of Descents and Minus Signs, Volume 26(1) of The Electronic Journal of Combinatorics, 2019.
Wolfdieter Lang, On Sums of Powers of Arithmetic Progressions, and Generalized Stirling, Eulerian and Bernoulli numbers, arXiv:1707.04451 [math.NT], 2017.
Huyile Liang, Yanni Pei, and Yi Wang, Analytic combinatorics of coordination numbers of cubic lattices, arXiv:2302.11856 [math.CO], 2023. See p. 22.
A. Losev and Y. Manin, New moduli spaces of pointed curves and pencils of flat connections, arXiv preprint arXiv:math/0001003 [math.AG], 2000 (p. 8). - _Tom Copeland_, Oct 16 2015
Peter Luschny, Permutation Trees
G. Rzadkowski and M. Urlinska, A Generalization of the Eulerian Numbers, arXiv:1612.06635 [math.CO], 2016.

See A008292 (subtriangle for k>=1 and n>=1), which is the main entry for these numbers. Another version has the zeros at the ends of the rows, as in Concrete Mathematics: see A173018.
Cf. A007318, A130850, A028246, A046802, A009006, A019538, A090582, A119879, A248727.
T(2n,n) gives A180056.

a123125 n k = a123125_tabl !! n !! k
a123125_row n = a123125_tabl !! n
a123125_tabl = [1] : zipWith (:) [0, 0 ..] a008292_tabl
-- Reinhard Zumkeller, Nov 06 2013

gf := 1/(1 - t*exp(x)): ser := series(gf, x, 12):
cx := n -> (-1)^(n + 1)*factor(n!*coeff(ser, x, n)*(t - 1)^(n + 1)):
seq(print(seq(coeff(cx(n), t, k), k = 0..n)), n = 0..9); # Peter Luschny, Feb 11 2021
A123125 := proc(n, k) option remember; if k = n then 1 elif k <= 0 or k > n then 0 else k*procname(n-1, k) + (n-k+1)*procname(n-1, k-1) fi end:
seq(print(seq(A123125(n, k), k=0..n)), n=0..10); # Peter Luschny, Mar 28 2021
# Alternative (Patrick J. Burchell):
t := a -> Statistics:-Difference([0, a]): Trow := k -> (t@@(k+1))([seq(n^k, n = 0..k)]):
seq(print(Trow(n)), n = 0..6); # Peter Luschny, Apr 02 2023

f[x_, n_] := f[x, n] = (1 - x)^(n + 1)*Sum[k^n*x^k, {k, 0, Infinity}];
Table[CoefficientList[f[x, n], x], {n,0,9}] // Flatten (* Roger L. Bagula, Aug 14 2008 *)
t[n_ /; n >= 0, 0] = 1; t[n_, k_] /; k<0 || k>n = 0; t[n_, k_] := t[n, k] = (n-k) t[n-1, k-1] + (k+1) t[n-1, k]; T[n_, k_] := t[n, n-k];
Table[T[n, k], {n,0,10}, {k, 0, n}] // Flatten (* Jean-François Alcover, May 26 2019 *)
A123125[n_, k_] := Sum[(-1)^j*(n - j - k + 1)^n * Binomial[n + 1, j], {j, 0, n - k}];
Table[A123125[n, k], {n, 0, 9}, {k, 0, n}] // TableForm  (* Peter Luschny, Aug 12 2022 *)

from math import isqrt, comb
def A123125(n):
    a = (m:=isqrt(k:=n+1<<1))+(k>m*(m+1))
    b = comb(a+1,2)-n
    return sum(-(b-j)**(a-1)*comb(a,j) if j&1 else (b-j)**(a-1)*comb(a,j) for j in range(b)) # Chai Wah Wu, Nov 13 2024

def statistic_eulerian(pi):
    if not pi: return 0
    h, i, branch, next = 0, len(pi), [0], pi[0]
    while True:
        while next < branch[len(branch)-1]:
            del(branch[len(branch)-1])
        current = 0
        h += 1
        while next > current:
            i -= 1
            if i == 0: return h
            branch.append(next)
            current, next = next, pi[i]
def A123125_row(n):
    L = [0]*(n+1)
    for p in Permutations(n):
        L[statistic_eulerian(p)] += 1
    return L
[A123125_row(n) for n in range(7)] # Peter Luschny, Dec 09 2015

Sum_{k=0..n} T(n,k) = n! = A000142(n).
Sum_{k=0..n} 2^k*T(n,k) = A000629(n).
Sum_{k=0..n} 3^k*T(n,k) = abs(A009362(n+1)).
Sum_{k=0..n} 2^(n-k)*T(n,k) = A000670(n).
Sum_{k=0..n} T(n,k)*3^(n-k) = A122704(n). - Philippe Deléham, Nov 07 2007
G.f.: f(x,n) = (1 - x)^(n + 1)*Sum_{k>=0} k^n*x^k. - Roger L. Bagula and Gary W. Adamson, Aug 14 2008. f is not the g.f. of the triangle, it is the polynomial of row n. See an Apr 03 2017 comment above - Wolfdieter Lang, Apr 03 2017
Sum_{k=0..n} T(n,k)*x^k = A000007(n), A000142(n), A000629(n), A123227(n), A201355(n), A201368(n) for x = 0, 1, 2, 3, 4, 5 respectively. - Philippe Deléham, Dec 01 2011
E.g.f. (1-t)/(1-t*exp((1-t)x)). A123125 * A007318 = A130850 = unsigned A075263, related to reversed A028246. A007318 * A123125 = A046802. Evaluating the row polynomials at -1, giving the alternating-sign row sum, generates A009006. - Tom Copeland, Oct 14 2015
From Wolfdieter Lang, Apr 03 2017: (Start)
T(n, k) = A173018(n, n-k), 0 <= k <= n. Row reversed Euler's triangle. See Graham et al., p. 268.
Recurrence (from A173018): T(n, 0) = 1 if n=0 else 0; T(n, k) = 0 if n < k and T(n, k) = (n+1-k)*T(n-1, k-1) + k*T(n-1, k) else.
T(n, k) = Sum_{j=0..k} (-1)^(k-j)*binomial(n-j, k-j)*S2(n, j)*j!, 0 <= k <= n, else 0. For S2(n, k)*k! see A131689.
The recurrence for the o.g.f. of the sequence of column k is
  G(k, x) = (x/(1 - k*x))*(E_x - (k-2))*G(k-1, x), with the Euler operator E_x  = x*d_x, for k >= 1, with G(0, x) = 1. (Proof from the recurrence of T(n, k)).
The e.g.f of the sequence of column k is found from E(k, x) = (1 + int(A(k, x),x)*exp(-k*x))*exp(k*x), k >= 1, with the recurrence
  A(k, x) = x*A(k-1, x) +(1 + (1-k)*(1-x))*E(k-1, x) for k >= 1, with  A(0,x)= 0. (Proof from the recurrence of T(n, k)). (End)
T(n, k) = Sum_{j=0..n-k} (-1)^j*(n-j-k+1)^n*binomial(n + 1, j). - Peter Luschny, Aug 12 2022
G.f.: Sum_{m >= 0} x^m/(1/(1-x)-m*t). - Mamuka Jibladze, Mar 12 2025

A028246 Triangular array a(n,k) = (1/k)Sum_{i=0..k} (-1)^(k-i)binomial(k,i)*i^n; n >= 1, 1 <= k <= n, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 3, 2, 1, 7, 12, 6, 1, 15, 50, 60, 24, 1, 31, 180, 390, 360, 120, 1, 63, 602, 2100, 3360, 2520, 720, 1, 127, 1932, 10206, 25200, 31920, 20160, 5040, 1, 255, 6050, 46620, 166824, 317520, 332640, 181440, 40320, 1, 511, 18660, 204630, 1020600, 2739240, 4233600, 3780000, 1814400, 362880

Offset: 1

N. J. A. Sloane, Doug McKenzie (mckfam4(AT)aol.com)

Let M = n X n matrix with (i,j)-th entry a(n+1-j, n+1-i), e.g., if n = 3, M = [1 1 1; 3 1 0; 2 0 0]. Given a sequence s = [s(0)..s(n-1)], let b = [b(0)..b(n-1)] be its inverse binomial transform and let c = [c(0)..c(n-1)] = M^(-1)*transpose(b). Then s(k) = Sum_{i=0..n-1} b(i)*binomial(k,i) = Sum_{i=0..n-1} c(i)*k^i, k=0..n-1. - Gary W. Adamson, Nov 11 2001
From Gary W. Adamson, Aug 09 2008: (Start)
Julius Worpitzky's 1883 algorithm generates Bernoulli numbers.
By way of example [Wikipedia]:
  B0 =  1;
  B1 = 1/1 -  1/2;
  B2 = 1/1 -  3/2 +   2/3;
  B3 = 1/1 -  7/2 +  12/3 -    6/4;
  B4 = 1/1 - 15/2 +  50/3 -   60/4 +   24/5;
  B5 = 1/1 - 31/2 + 180/3 -  390/4 +  360/5 -  120/6;
  B6 = 1/1 - 63/2 + 602/3 - 2100/4 + 3360/5 - 2520/6 + 720/7;
  ...
Note that in this algorithm, odd n's for the Bernoulli numbers sum to 0, not 1, and the sum for B1 = 1/2 = (1/1 - 1/2). B3 = 0 = (1 - 7/2 + 13/3 - 6/4) = 0. The summation for B4 = -1/30. (End)
Pursuant to Worpitzky's algorithm and given M = A028246 as an infinite lower triangular matrix, M * [1/1, -1/2, 1/3, ...] (i.e., the Harmonic series with alternate signs) = the Bernoulli numbers starting [1/1, 1/2, 1/6, ...]. - Gary W. Adamson, Mar 22 2012
From Tom Copeland, Oct 23 2008: (Start)
G(x,t) = 1/(1 + (1-exp(x*t))/t) = 1 + 1 x + (2 + t)*x^2/2! + (6 + 6t + t^2)*x^3/3! + ... gives row polynomials for A090582, the f-polynomials for the permutohedra (see A019538).
G(x,t-1) = 1 + 1*x + (1 + t)*x^2 / 2! + (1 + 4t + t^2)*x^3 / 3! + ... gives row polynomials for A008292, the h-polynomials for permutohedra.
G[(t+1)x,-1/(t+1)] = 1 + (1+ t) x + (1 + 3t + 2 t^2) x^2 / 2! + ... gives row polynomials for the present triangle. (End)
The Worpitzky triangle seems to be an apt name for this triangle. - Johannes W. Meijer, Jun 18 2009
If Pascal's triangle is written as a lower triangular matrix and multiplied by A028246 written as an upper triangular matrix, the product is a matrix where the (i,j)-th term is (i+1)^j. For example,
  1,0,0,0     1,1,1, 1      1,1, 1, 1
  1,1,0,0  *  0,1,3, 7   =  1,2, 4, 8
  1,2,1,0     0,0,2,12      1,3, 9,27
  1,3,3,1     0,0,0, 6      1,4,16,64
  So, numbering all three matrices' rows and columns starting at 0, the (i,j) term of the product is (i+1)^j. - Jack A. Cohen (ProfCohen(AT)comcast.net), Aug 03 2010
The Fi1 and Fi2 triangle sums are both given by sequence A000670. For the definition of these triangle sums see A180662. The mirror image of the Worpitzky triangle is A130850. - Johannes W. Meijer, Apr 20 2011
Let S_n(m) = 1^m + 2^m + ... + n^m. Then, for n >= 0, we have the following representation of S_n(m) as a linear combination of the binomial coefficients:
  S_n(m) = Sum_{i=1..n+1} a(i+n*(n+1)/2)*C(m,i). E.g., S_2(m) = a(4)*C(m,1) + a(5)*C(m,2) + a(6)*C(m,3) = C(m,1) + 3*C(m,2) + 2*C(m,3). - Vladimir Shevelev, Dec 21 2011
Given the set X = [1..n] and 1 <= k <= n, then a(n,k) is the number of sets T of size k of subset S of X such that S is either empty or else contains 1 and another element of X and such that any two elemements of T are either comparable or disjoint. - Michael Somos, Apr 20 2013
Working with the row and column indexing starting at -1, a(n,k) gives the number of k-dimensional faces in the first barycentric subdivision of the standard n-dimensional simplex (apply Brenti and Welker, Lemma 2.1). For example, the barycentric subdivision of the 2-simplex (a triangle) has 1 empty face, 7 vertices, 12 edges and 6 triangular faces giving row 4 of this triangle as (1,7,12,6). Cf. A053440. - Peter Bala, Jul 14 2014
See A074909 and above g.f.s for associations among this array and the Bernoulli polynomials and their umbral compositional inverses. - Tom Copeland, Nov 14 2014
An e.g.f. G(x,t) = exp[P(.,t)x] = 1/t - 1/[t+(1-t)(1-e^(-xt^2))] = (1-t) * x + (-2t + 3t^2 - t^3) * x^2/2! + (6t^2 - 12t^3 + 7t^4 - t^5) * x^3/3! + ... for the shifted, reverse, signed polynomials with the first element nulled, is generated by the infinitesimal generator g(u,t)d/du = [(1-u*t)(1-(1+u)t)]d/du, i.e., exp[x * g(u,t)d/du] u eval. at u=0 generates the polynomials. See A019538 and the G. Rzadkowski link below for connections to the Bernoulli and Eulerian numbers, a Ricatti differential equation, and a soliton solution to the KdV equation. The inverse in x of this e.g.f. is Ginv(x,t) = (-1/t^2)*log{[1-t(1+x)]/[(1-t)(1-tx)]} = [1/(1-t)]x + [(2t-t^2)/(1-t)^2]x^2/2 + [(3t^2-3t^3+t^4)/(1-t)^3]x^3/3  + [(4t^3-6t^4+4t^5-t^6)/(1-t)^4]x^4/4 + ... . The numerators are signed, shifted A135278 (reversed A074909), and the rational functions are the columns of A074909. Also, dG(x,t)/dx = g(G(x,t),t) (cf. A145271). (Analytic G(x,t) added, and Ginv corrected and expanded on Dec 28 2015.) - Tom Copeland, Nov 21 2014
The operator R = x + (1 + t) + t e^{-D} / [1 + t(1-e^(-D))] = x + (1+t) + t - (t+t^2) D + (t+3t^2+2t^3) D^2/2! - ... contains an e.g.f. of the reverse row polynomials of the present triangle, i.e., A123125 * A007318 (with row and column offset 1 and 1). Umbrally, R^n 1 = q_n(x;t) = (q.(0;t)+x)^n, with q_m(0;t) = (t+1)^(m+1) - t^(m+1), the row polynomials of A074909, and D = d/dx. In other words, R generates the Appell polynomials associated with the base sequence A074909. For example, R 1 = q_1(x;t) = (q.(0;t)+x) = q_1(0;t) + q__0(0;t)x = (1+2t) + x, and R^2 1 = q_2(x;t) = (q.(0;t)+x)^2 = q_2(0:t) + 2q_1(0;t)x + q_0(0;t)x^2 = 1+3t+3t^2 + 2(1+2t)x + x^2. Evaluating the polynomials at x=0 regenerates the base sequence. With a simple sign change in R, R generates the Appell polynomials associated with A248727. - Tom Copeland, Jan 23 2015
For a natural refinement of this array, see A263634. - Tom Copeland, Nov 06 2015
From Wolfdieter Lang, Mar 13 2017: (Start)
The e.g.f. E(n, x) for {S(n, m)}{m>=0} with S(n, m) = Sum{k=1..m} k^n, n >= 0, (with undefined sum put to 0) is exp(x)*R(n+1, x) with the exponential row polynomials R(n, x) = Sum_{k=1..n} a(n, k)*x^k/k!. E.g., e.g.f. for n = 2, A000330: exp(x)*(1*x/1!+3*x^2/2!+2*x^3/3!).
The o.g.f. G(n, x) for {S(n, m)}{m >=0} is then found by Laplace transform to be G(n, 1/p) = p*Sum{k=1..n} a(n+1, k)/(p-1)^(2+k).
  Hence G(n, x) = x/(1 - x)^(n+2)*Sum_{k=1..n} A008292(n,k)*x^(k-1).
  E.g., n=2: G(2, 1/p) = p*(1/(p-1)^2 + 3/(p-1)^3 + 2/(p-1)^4) = p^2*(1+p)/(p-1)^4; hence G(2, x) = x*(1+x)/(1-x)^4.
This works also backwards: from the o.g.f. to the e.g.f. of {S(n, m)}_{m>=0}. (End)
a(n,k) is the number of k-tuples of pairwise disjoint and nonempty subsets of a set of size n. - Dorian Guyot, May 21 2019
From Rajesh Kumar Mohapatra, Mar 16 2020: (Start)
a(n-1,k) is the number of chains of length k in a partially ordered set formed from subsets of an n-element set ordered by inclusion such that the first term of the chains is either the empty set or an n-element set.
Also, a(n-1,k) is the number of distinct k-level rooted fuzzy subsets of an n-set ordered by set inclusion. (End)
The relations on p. 34 of Hasan (also p. 17 of Franco and Hasan) agree with the relation between A019538 and this entry given in the formula section. - Tom Copeland, May 14 2020
T(n,k) is the size of the Green's L-classes in the D-classes of rank (k-1) in the semigroup of partial transformations on an (n-1)-set. - Geoffrey Critzer, Jan 09 2023
T(n,k) is the number of strongly connected binary relations on [n] that have period k (A367948) and index 1.  See Theorem 5.4.25(6) in Ki Hang Kim reference. - Geoffrey Critzer, Dec 07 2023

			The triangle a(n, k) starts:
n\k 1   2    3     4      5      6      7      8     9
1:  1
2:  1   1
3:  1   3    2
4:  1   7   12     6
5:  1  15   50    60     24
6:  1  31  180   390    360    120
7:  1  63  602  2100   3360   2520    720
8:  1 127 1932 10206  25200  31920  20160   5040
9:  1 255 6050 46620 166824 317520 332640 181440 40320
... [Reformatted by _Wolfdieter Lang_, Mar 26 2015]
-----------------------------------------------------
Row 5 of triangle is {1,15,50,60,24}, which is {1,15,25,10,1} times {0!,1!,2!,3!,4!}.
From _Vladimir Shevelev_, Dec 22 2011: (Start)
Also, for power sums, we have
S_0(n) = C(n,1);
S_1(n) = C(n,1) +    C(n,2);
S_2(n) = C(n,1) +  3*C(n,2) +  2*C(n,3);
S_3(n) = C(n,1) +  7*C(n,2) + 12*C(n,3) +  6*C(n,4);
S_4(n) = C(n,1) + 15*C(n,2) + 50*C(n,3) + 60*C(n,4) + 24*C(n,5); etc.
(End)
For X = [1,2,3], the sets T are {{}}, {{},{1,2}}, {{},{1,3}}, {{},{1,2,3}}, {{},{1,2},{1,2,3}}, {{},{1,3},{1,2,3}} and so a(3,1)=1, a(3,2)=3, a(3,3)=2. - _Michael Somos_, Apr 20 2013

Ki Hang Kim, Boolean Matrix Theory and Applications, Marcel Dekker, New York and Basel (1982).

Seiichi Manyama, Table of n, a(n) for n = 1..10000
V. S. Abramovich, Power sums of natural numbers, Kvant, no. 5 (1973), 22-25. (in Russian)
Peter Bala, Deformations of the Hadamard product of power series
Paul Barry, Three Études on a sequence transformation pipeline, arXiv:1803.06408 [math.CO], 2018.
Melike Baykal-Gürsoy, Marcelo Figueroa-Candia, and Zhe Duan, Completion times of jobs on two-state service processes and their asymptotic behavior, Probability in the Engineering and Information Sciences (2024), 1-29. See p. 23.
H. Belbachir, M. Rahmani, and B. Sury, Sums Involving Moments of Reciprocals of Binomial Coefficients, J. Int. Seq. 14 (2011) #11.6.6.
Hacene Belbachir and Mourad Rahmani, Alternating Sums of the Reciprocals of Binomial Coefficients, Journal of Integer Sequences, Vol. 15 (2012), #12.2.8.
F. Brenti and V. Welker, f-vectors of barycentric subdivisions, arXiv:math/0606356v1 [math.CO], Math. Z., 259(4), 849-865, 2008.
Patibandla Chanakya and Putla Harsha, Generalized Nested Summation of Powers of Natural Numbers, arXiv:1808.08699 [math.NT], 2018. See Table 1.
Tom Copeland, Generators, Inversion, and Matrix, Binomial, and Integral Transforms
Colin Defant, Troupes, Cumulants, and Stack-Sorting, arXiv:2004.11367 [math.CO], 2020.
E. Delucchi, A. Pixton and L. Sabalka. Face vectors of subdivided simplicial complexes arXiv:1002.3201v3 [math.CO], Discrete Mathematics, Volume 312, Issue 2, January 2012, Pages 248-257.
G. H. E Duchamp, N. Hoang-Nghia, and A. Tanasa, A word Hopf algebra based on the selection/quotient principle, arXiv:1207.6522 [math.CO], 2012-2013; Séminaire Lotharingien de Combinatoire 68 (2013), Article B68c.
M. Dukes and C. D. White, Web Matrices: Structural Properties and Generating Combinatorial Identities, arXiv:1603.01589 [math.CO], 2016.
Nick Early, Honeycomb tessellations and canonical bases for permutohedral blades, arXiv:1810.03246 [math.CO], 2018.
S. Franco and A. Hasan, Graded Quivers, Generalized Dimer Models and Toric Geometry , arXiv preprint arXiv:1904.07954 [hep-th], 2019
A. Hasan, Physics and Mathematics of Graded Quivers, dissertation, Graduate Center, City University of New York, 2019.
H. Hasse, Ein Summierungsverfahren für die Riemannsche Zeta-Reihe, Math. Z. 32, 458-464 (1930).
Guy Louchard, Werner Schachinger, and Mark Daniel Ward, The number of distinct adjacent pairs in geometrically distributed words: a probabilistic and combinatorial analysis, arXiv:2203.14773 [math.PR], 2022. See p. 5.
Shi-Mei Ma, A family of two-variable derivative polynomials for tangent and secant, El J. Combinat. 20(1) (2013), P11.
Richard J. Mathar, Integrals Associated with the Digamma Integral Representation, arXiv:2308.14154 [math.GM], 2023. See p. 5.
Rajesh Kumar Mohapatra and Tzung-Pei Hong, On the Number of Finite Fuzzy Subsets with Analysis of Integer Sequences, Mathematics (2022) Vol. 10, No. 7, 1161.
A. Riskin and D. Beckwith, Problem 10231, Amer. Math. Monthly, 102 (1995), 175-176.
G. Rzadkowski, Bernoulli numbers and solitons revisited, Journal of Nonlinear Mathematical Physics, Volume 17, Issue 1, 2010.
John K. Sikora, On Calculating the Coefficients of a Polynomial Generated Sequence Using the Worpitzky Number Triangles, arXiv:1806.00887 [math.NT], 2018.
G. J. Simmons, A combinatorial problem associated with a family of combination locks, Math. Mag., 37 (1964), 127-132 (but there are errors). The triangle is on page 129.
N. J. A. Sloane, Transforms
Sam Vandervelde, The Worpitzky Numbers Revisited, Amer. Math. Monthly, 125:3 (2018), 198-206.
Wikipedia, Bernoulli number.
Wikipedia, Barycentric subdivision
David C. Wood, The computation of polylogarithms (2014).

Dropping the column of 1's gives A053440.
Without the k in the denominator (in the definition), we get A019538. See also the Stirling number triangle A008277.
Cf. A087127, A087107, A087108, A087109, A087110, A087111, A084938 A075263.
Row sums give A000629(n-1) for n >= 1.
Cf. A027642, A002445. - Gary W. Adamson, Aug 09 2008
Appears in A161739 (RSEG2 triangle), A161742 and A161743. - Johannes W. Meijer, Jun 18 2009
Binomial transform is A038719. Cf. A131689.
Cf. A007318, A008292, A046802, A074909, A090582, A123125, A130850, A135278, A141618, A145271, A163626, A248727, A263634.
Cf. A119879.
From Rajesh Kumar Mohapatra, Mar 29 2020: (Start)
A000007(n-1) (column k=1), A000225(n-1) (column k=2), A028243(n-1) (column k=3), A028244(n-1) (column k=4), A028245(n-1) (column k=5), for n > 0.
Diagonal gives A000142(n-1), for n >=1.
Next-to-last diagonal gives A001710,
Third, fourth, fifth, sixth, seventh external diagonal respectively give A005460, A005461, A005462, A005463, A005464. (End)

Flat(List([1..10], n-> List([1..n], k-> Stirling2(n,k)* Factorial(k-1) ))); # G. C. Greubel, May 30 2019

[[StirlingSecond(n,k)*Factorial(k-1): k in [1..n]]: n in [1..10]]; // G. C. Greubel, May 30 2019

a := (n,k) -> add((-1)^(k-i)*binomial(k,i)*i^n, i=0..k)/k;
seq(print(seq(a(n,k),k=1..n)),n=1..10);
T := (n,k) -> add(eulerian1(n,j)*binomial(n-j,n-k), j=0..n):
seq(print(seq(T(n,k),k=0..n)),n=0..9); # Peter Luschny, Jul 12 2013

a[n_, k_] = Sum[(-1)^(k-i) Binomial[k,i]*i^n, {i,0,k}]/k; Flatten[Table[a[n, k], {n, 10}, {k, n}]] (* Jean-François Alcover, May 02 2011 *)

{T(n, k) = if( k<0 || k>n, 0, n! * polcoeff( (x / log(1 + x + x^2 * O(x^n) ))^(n+1), n-k))}; /* Michael Somos, Oct 02 2002 */

{T(n,k) = stirling(n,k,2)*(k-1)!}; \\ G. C. Greubel, May 31 2019

# Assuming offset (n, k) = (0, 0).
def T(n, k):
    if k >  n: return 0
    if k == 0: return 1
    return k*T(n - 1, k - 1) + (k + 1)*T(n - 1, k)
for n in range(9):
    print([T(n, k) for k in range(n + 1)])  # Peter Luschny, Apr 26 2022

def A163626_row(n) :
    x = polygen(ZZ,'x')
    A = []
    for m in range(0, n, 1) :
        A.append((-x)^m)
        for j in range(m, 0, -1):
            A[j - 1] = j * (A[j - 1] - A[j])
    return list(A[0])
for i in (1..7) : print(A163626_row(i))  # Peter Luschny, Jan 25 2012

[[stirling_number2(n,k)*factorial(k-1) for k in (1..n)] for n in (1..10)] # G. C. Greubel, May 30 2019

E.g.f.: -log(1-y*(exp(x)-1)). - Vladeta Jovovic, Sep 28 2003
a(n, k) = S2(n, k)*(k-1)! where S2(n, k) is a Stirling number of the second kind (cf. A008277). Also a(n,k) = T(n,k)/k, where T(n, k) = A019538.
Essentially same triangle as triangle [1, 0, 2, 0, 3, 0, 4, 0, 5, 0, 6, 0, 7, ...] DELTA [1, 1, 2, 2, 3, 3, 4, 4, 5, 5, ...] where DELTA is Deléham's operator defined in A084938, but the notation is different.
Sum of terms in n-th row = A000629(n) - Gary W. Adamson, May 30 2005
The row generating polynomials P(n, t) are given by P(1, t)=t, P(n+1, t) = t(t+1)(d/dt)P(n, t) for n >= 1 (see the Riskin and Beckwith reference). - Emeric Deutsch, Aug 09 2005
From Gottfried Helms, Jul 12 2006: (Start)
Delta-matrix as can be read from H. Hasse's proof of a connection between the zeta-function and Bernoulli numbers (see link below).
Let P = lower triangular matrix with entries P[row,col] = binomial(row,col).
Let J = unit matrix with alternating signs J[r,r]=(-1)^r.
Let N(m) = column matrix with N(m)(r) = (r+1)^m, N(1)--> natural numbers.
Let V = Vandermonde matrix with V[r,c] = (r+1)^c.
V is then also N(0)||N(1)||N(2)||N(3)... (indices r,c always beginning at 0).
Then Delta = P*J * V and B' = N(-1)' * Delta, where B is the column matrix of Bernoulli numbers and ' means transpose, or for the single k-th Bernoulli number B_k with the appropriate column of Delta,
B_k = N(-1)' * Delta[ *,k ] = N(-1)' * P*J * N(k).
Using a single column instead of V and assuming infinite dimension, H. Hasse showed that in x = N(-1) * P*J * N(s), where s can be any complex number and s*zeta(1-s) = x.
His theorem reads: s*zeta(1-s) = Sum_{n>=0..inf} (n+1)^-1*delta(n,s), where delta(n,s) = Sum_{j=0..n} (-1)^j * binomial(n,j) * (j+1)^s.
(End)
a(n,k) = k*a(n-1,k) + (k-1)*a(n-1,k-1) with a(n,1) = 1 and a(n,n) = (n-1)!. - Johannes W. Meijer, Jun 18 2009
Rephrasing the Meijer recurrence above: Let M be the (n+1)X(n+1) bidiagonal matrix with M(r,r) = M(r,r+1) = r, r >= 1, in the two diagonals and the rest zeros. The row a(n+1,.) of the triangle is row 1 of M^n. - Gary W. Adamson, Jun 24 2011
From Tom Copeland, Oct 11 2011: (Start)
With e.g.f.. A(x,t) = G[(t+1)x,-1/(t+1)]-1 (from 2008 comment) = -1 + 1/[1-(1+t)(1-e^(-x))] = (1+t)x + (1+3t+2t^2)x^2/2! + ..., the comp. inverse in x is
B(x,t)= -log(t/(1+t)+1/((1+t)(1+x))) = (1/(1+t))x - ((1+2t)/(1+t)^2)x^2/2 + ((1+3t+3t^2)/(1+t)^3)x^3/3 + .... The numerators are the row polynomials of A074909, and the rational functions are (omitting the initial constants) signed columns of the re-indexed Pascal triangle A007318.
Let h(x,t)= 1/(dB/dx) = (1+x)(1+t(1+x)), then the row polynomial P(n,t) = (1/n!)(h(x,t)*d/dx)^n x, evaluated at x=0, A=exp(x*h(y,t)*d/dy) y, eval. at y=0, and dA/dx = h(A(x,t),t), with P(1,t)=1+t. (Series added Dec 29 2015.)(End)
Let  denote the Eulerian numbers A173018(n,k), then T(n,k) = Sum_{j=0..n} *binomial(n-j,n-k). - Peter Luschny, Jul 12 2013
Matrix product A007318 * A131689. The n-th row polynomial R(n,x) = Sum_{k >= 1} k^(n-1)*(x/(1 + x))^k, valid for x in the open interval (-1/2, inf). Cf A038719. R(n,-1/2) = (-1)^(n-1)*(2^n - 1)*Bernoulli(n)/n. - Peter Bala, Jul 14 2014
a(n,k) = A141618(n,k) / C(n,k-1). - Tom Copeland, Oct 25 2014
For the row polynomials, A028246(n,x) = A019538(n-1,x) * (1+x). - Tom Copeland, Dec 28 2015
A248727 = A007318*(reversed A028246) = A007318*A130850 = A007318*A123125*A007318 = A046802*A007318. - Tom Copeland, Nov 14 2016
n-th row polynomial R(n,x) = (1+x) o (1+x) o ... o (1+x) (n factors), where o denotes the black diamond multiplication operator of Dukes and White. See example E11 in the Bala link. - Peter Bala, Jan 12 2018
From Dorian Guyot, May 21 2019: (Start)
Sum_{i=0..k} binomial(k,i) * a(n,i) = (k+1)^n.
Sum_{k=0..n} a(n,k) = 2*A000670(n).
(End)
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 this entry, A028246, are given by x^n * A_n(1 + 1/x;0). Other specializations of A_n(x;y) give A046802, A090582, A119879, A130850, and A248727. - Tom Copeland, Jan 24 2020
The row generating polynomials R(n,x) = Sum_{i=1..n} a(n,i) * x^i satisfy the recurrence equation R(n+1,x) = R(n,x) + Sum_{k=0..n-1} binomial(n-1,k) * R(k+1,x) * R(n-k,x) for n >= 1 with initial value R(1,x) = x. - Werner Schulte, Jun 17 2021

Definition corrected by Li Guo, Dec 16 2006
Typo in link corrected by Johannes W. Meijer, Oct 17 2009
Error in title corrected by Johannes W. Meijer, Sep 24 2010
Edited by M. F. Hasler, Oct 29 2014

A009006 Expansion of e.g.f.: 1 + tan(x).

Original entry on oeis.org

1, 1, 0, 2, 0, 16, 0, 272, 0, 7936, 0, 353792, 0, 22368256, 0, 1903757312, 0, 209865342976, 0, 29088885112832, 0, 4951498053124096, 0, 1015423886506852352, 0, 246921480190207983616, 0, 70251601603943959887872, 0, 23119184187809597841473536, 0

Offset: 0

R. H. Hardin

R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 1, 1997; See Exercise 1.41(d).

G. C. Greubel, Table of n, a(n) for n = 0..451
Kwang-Wu Chen, An Interesting Lemma for Regular C-fractions, J. Integer Seqs., Vol. 6, 2003.

A000182(n) = a(2n-1).

m:=30; R:=PowerSeriesRing(Rationals(), m); b:=Coefficients(R!(1 + Tan(x))); [Factorial(n-1)*b[n]: n in [1..m]]; // G. C. Greubel, Jul 21 2018

u:=proc(n) if n=0 then 1 else -add(u(k)*binomial(n,k)/2*2^(n-k),k=0..n-1) fi end;seq(u(n),n=0..15); # Robert FERREOL, Dec 30 2006

a[m_] := Abs[Sum[(-2)^(m-k) k! StirlingS2[m,k], {k,0,m}]]; Table[a[i], {i,0,20}] (* Peter Luschny, Apr 29 2009 *)
A009006[n_] :=  Cos[Pi (n-1) / 2] (4^(n+1) - 2^(n+1)) * BernoulliB[n+1] / (n+1); a[0] := 1; Table[A009006[n], {n, 0, 30}] (* Peter Luschny, Jun 14 2021 *)

a(n)=if(n<1,n==0,n!*polcoeff(tan(x+x*O(x^n)),n))

def A009006(n) :
    if n == 0 : return 1
    return add(add((-1)^(n//2+j+1)*binomial(n+1,k-j)*j^n for j in (0..k)) for k in (1..n))
[A009006(n) for n in (0..26)] # Peter Luschny, Jul 23 2012

Let b(n) be a(n) shifted one place to the left with b(2+4k) = -a(3+4k), k=0, 1, .. Then b(n) is the expansion of sech(x)^2. - Mario Catalani (mario.catalani(AT)unito.it), Feb 08 2003
g(x) = x + x^2 - 2*x^4 + 16*x^6 - 272*x^8 + ... satisfies g(x/(1+2x)) = -g(-x).
E.g.f.: 1 + tan(x).
E.g.f. exp(x)*sech(x) is 1,1,0,-2,0,16,0,-272,... (A155585). - Paul Barry, Mar 15 2006
From Robert FERREOL, Dec 30 2006: (Start)
a(n) = 2^n*abs(Euler(n,0)) where Euler(n,x) is the n-th Eulerian polynomial.
a(n) = abs(u(n)) where u(n) = -Sum_{k=0..n-1} u(k)*binomial(n, k)*2^(n-k-1) with u(0) = 1.  (End)
Sum_{k=0..n} A075263(n, k)*2^k = 1,-1,0,2,0,-16,0,272,0,-7936,0,... for n = 0, 1, 2, 3, 4, ..., respectively. - Philippe Deléham, Aug 20 2007
E.g.f. -log(cos(x)), for n > 0. - Vladimir Kruchinin, Aug 09 2010
a(n) = Sum_{k=1..n} Sum_{j=0..k} (-1)^(floor(n/2)+j+1)*binomial(n+1,k-j)*j^n for n > 0. - Peter Luschny, Jul 23 2012
From Sergei N. Gladkovskii, Oct 25 2012 - Dec 20 2013: (Start)
Continued fractions:
G.f.: 1 + x/T(0) where T(k) = 1 - (k+1)*(k+2)*x^2/T(k+1).
E.g.f.: 1 + tan(x) = 1+x/(U(0)-x) where U(k)= 4*k+1 + x/(1+x/(4*k+3 - x/(1- x/U(k+1)))).
E.g.f.: 1+tan(x) = 1 - 3*x/(U(0) + 3*x^2) where U(k) = 64*k^3 + 48*k^2 - 4*k*(2*x^2+1) - 2*x^2 - 3 - x^4*(4*k-1)*(4*k+7)/U(k+1).
E.g.f.: 1+x*G(0) where G(k) = 1 - x^2/(x^2 - (2*k+1)*(2*k+3)/G(k+1)).
G.f.: 1 + x/G(0) where G(k) = 1 - 2*x^2*(4*k^2+4*k+1)-4*x^4*(k+1)^2*(4*k^2+8*k+3) /G(k+1).
G.f.: 1 + x*Q(0) where Q(k) = 1 - x^2*(k+1)*(k+2)/(x^2*(k+1)*(k+2) - 1/Q(k+1)).
G.f.: Q(0) where Q(k) = 1 + x*(k+1)/(x*(k+1)+1/(1- x*(k+1)/(x*(k+1) - 1/Q(k+1)))).
E.g.f.: 2 - 1/Q(0) where Q(k) = 1 + x/(4*k+1 - x/(1 - x/(4*k+3 + x/Q(k+1)))). (End)
a(n) ~ 2*n!*(2/Pi)^(n+1) if n is odd. - Vaclav Kotesovec, Jun 01 2013
a(n) = i^(n+1) * 2^n * ((-1)^n-1) * (2^(n+1)-1) * Bernoulli(n+1)/(n+1), n > 0. - Benedict W. J. Irwin, May 27 2016
a(0) = a(1) = 1; a(n) = -2 * a(n-1) + Sum_{k=0..n-1} binomial(n-1,k) * a(k) * a(n-k-1). - Ilya Gutkovskiy, Jul 05 2020

Reformatted Mar 15 1997
Definition corrected by Joerg Arndt, Apr 29 2011
Terms a(26) onward added by G. C. Greubel, Jul 21 2018

A371761 Array read by antidiagonals: The number of parades with n girls and k boys that begin with a girl and end with a boy.

Original entry on oeis.org

1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 5, 1, 0, 0, 1, 13, 13, 1, 0, 0, 1, 29, 73, 29, 1, 0, 0, 1, 61, 301, 301, 61, 1, 0, 0, 1, 125, 1081, 2069, 1081, 125, 1, 0, 0, 1, 253, 3613, 11581, 11581, 3613, 253, 1, 0, 0, 1, 509, 11593, 57749, 95401, 57749, 11593, 509, 1, 0

Offset: 0

Peter Luschny, Apr 05 2024

The name derives from a proposition by Donald Knuth, who describes the setup of a "girls and boys parade" as follows: "There are m girls {g_1, ..., g_m} and n boys {b_1, ..., b_n}, where g_i is younger than g_{i+1} and b_j is younger than b_{j+1}, but we know nothing about the relative ages of g_i and b_j. In how many ways can they all line up in a sequence such that no girl is directly preceded by an older girl and no boy is directly preceded by an older boy?" [Our notation: A <- D, n <- m, k <- n].
In A344920, the Worpitzky transform is defined as a sequence-to-sequence transformation WT := A -> B, where B(n) = Sum_{k=0..n} A163626(n, k)*A(k). (If A(n) = 1/(n + 1) then B(n) are the Bernoulli numbers (with B(1) = 1/2.)) The rows of the array are the Worpitzky transforms of the powers up to the sign (-1)^k.
The array rows are recursively generated by applying the Akiyama-Tanigawa algorithm to the powers (see the Python implementation below). In this way the array becomes the image of A004248 under the AT-transformation when applied to the rows of A004248. This makes the array closely linked to A344499, which is generated in the same way, but applied to the columns of A004248.
Conjecture: Row n + 1 is row 2^n in table A136301, where a probabilistic interpretation is given (see the link to Parsonnet's paper below).

			Array starts:
[0] 1, 0,   0,     0,      0,       0,        0,         0,          0, ...
[1] 0, 1,   1,     1,      1,       1,        1,         1,          1, ...
[2] 0, 1,   5,    13,     29,      61,      125,       253,        509, ...
[3] 0, 1,  13,    73,    301,    1081,     3613,     11593,      36301, ...
[4] 0, 1,  29,   301,   2069,   11581,    57749,    268381,    1191989, ...
[5] 0, 1,  61,  1081,  11581,   95401,   673261,   4306681,   25794781, ...
[6] 0, 1, 125,  3613,  57749,  673261,  6487445,  55213453,  431525429, ...
[7] 0, 1, 253, 11593, 268381, 4306681, 55213453, 610093513, 6077248381, ...
.
Seen as triangle T(n, k) = A(n - k, k):
  [0] 1;
  [1] 0, 0;
  [2] 0, 1,  0;
  [3] 0, 1,  1,   0;
  [4] 0, 1,  5,   1,   0;
  [5] 0, 1, 13,  13,   1,  0;
  [6] 0, 1, 29,  73,  29,  1, 0;
  [7] 0, 1, 61, 301, 301, 61, 1, 0;
.
A(n, k) as sum of powers:
  A(2, k) =  -3+   2*2^k;
  A(3, k) =   7-  12*2^k+    6*3^k;
  A(4, k) = -15+  50*2^k-   60*3^k+   24*4^k;
  A(5, k) =  31- 180*2^k+  390*3^k-  360*4^k+  120*5^k;
  A(6, k) = -63+ 602*2^k- 2100*3^k+ 3360*4^k- 2520*5^k+  720*6^k;
  A(7, k) = 127-1932*2^k+10206*3^k-25200*4^k+31920*5^k-20160*6^k+5040*7^k;

Peter Luschny, Table of n, a(n) for antidiagonals 0..140.
Beáta Bényi, A Bijection for the Boolean Numbers of Ferrers Graphs, Graphs and Combinatorics (2022) Vol. 38, No. 10.
Beáta Bényi and Péter Hajnal, Combinatorial properties of poly-Bernoulli relatives, arXiv preprint arXiv:1602.08684 [math.CO], 2016. See D_{n, k}.
Don Knuth, Parades and poly-Bernoulli bijections, Mar 31 2024. See (16.2).
Brian Parsonnet, Probability of Derangements.

Variant: A272644.
Rows include: A344920 (row 2, signed), A006230 (row 3).
Row sums of triangle (n>=2): A297195, alternating row sums: A226158.
Diagonal of array: A048144.
Cf. A004248, A075263, A028246, A173018, A136301, A371762, A136126 (similar), A099594, A344499.

egf := 1/(exp(w) + exp(z) - exp(w + z)): serw := n -> series(egf, w, n + 1):
# Returns row n (>= 0) with length len (> 0):
R := n -> len -> local k;
seq(k!*coeff(series(n!*coeff(serw(n), w, n), z, len), z, k), k = 0..len - 1):
seq(lprint(R(n)(9)), n = 0..7);
# Explicit with Stirling2 :
A := (n, k) -> local j; add(j!^2*Stirling2(n, j)*Stirling2(k, j), j = 0..min(n, k)): seq(lprint(seq(A(n, k), k = 0..8)), n = 0..7);
# Using the unsigned Worpitzky transform.
WT := (a, len) -> local n, k;
seq(add((-1)^(n - k)*k!*Stirling2(n + 1, k + 1)*a(k), k=0..n), n = 0..len-1):
Arow := n -> WT(x -> x^n, 8): seq(lprint(Arow(n)), n = 0..8);
# Two recurrences:
A := proc(n, k) option remember; local j; if k = 0 then return k^n fi;
add(binomial(n, j)*(A(n-j, k-1) + A(n-j+1, k-1)), j = 1..n) end:
A := proc(n, k) option remember; local j; if n = 0 then 0^k else
add(binomial(k + `if`(j=0,0,1), j+1)*A(n-1, k-j), j = 0..k) fi end:

(* Using the unsigned Worpitzky transform. *)
Unprotect[Power]; Power[0, 0] = 1;
W[n_, k_] := (-1)^(n - k) k! StirlingS2[n + 1, k + 1];
WT[a_, len_] := Table[Sum[W[n, k] a[k], {k, 0, n}], {n, 0, len-1}];
A371761row[n_, len_] := WT[#^n &, len];
Table[A371761row[n, 9], {n, 0, 8}] // MatrixForm
(* Row n >= 1 by linear recurrence: *)
RowByLRec[n_, len_] := LinearRecurrence[Table[-StirlingS1[n+1, n+1-k], {k, 1, n}],
A371761row[n, n+1], len]; Table[RowByLRec[n, 9], {n, 1, 8}] // MatrixForm

from functools import cache
from math import comb as binomial
@cache
def A(n, k):
    if n == 0: return int(k == 0)
    return sum(binomial(k + int(j > 0), j + 1) * A(n - 1, k - j)
           for j in range(k + 1))
for n in range(8): print([A(n, k) for k in range(8)])

# The Akiyama-Tanigawa algorithm for powers generates the rows.
def ATPowList(n, len):
    A = [0] * len
    R = [0] * len
    for k in range(len):
        R[k] = k**n   # Changing this to R[k] = (n + 1)**k generates A344499.
        for j in range(k, 0, -1):
            R[j - 1] = j * (R[j] - R[j - 1])
        A[k] = R[0]
    return A
for n in range(8): print([n], ATPowList(n, 9))

def A371761(n, k): return sum((-1)^(j - k) * factorial(j) * stirling_number2(k + 1, j + 1) * j^n for j in range(k + 1))
for n in range(9): print([A371761(n, k) for k in range(8)])

A(n, k) = k! * [z^k] (n! * [w^n] 1/(exp(w) + exp(z) - exp(w + z))).
A(n, k) = k! * [w^k] (Sum_{j=0..n} A075263(n, n - j) * exp(j*w)).
A(n, k) = Sum_{j=0..k} (-1)^(j-k) * Stirling2(k + 1, j + 1) * j! * j^n.
A(n, k) = Sum_{j=0..min(n,k)} (j!)^2 * Stirling2(n, j) * Stirling2(k, j).
A(n, k) = Sum_{j=0..n} (-1)^(n-j)*A028246(n, j) * j^k; this is explicit:
A(n, k) = Sum_{j=0..n} Sum_{m=0..n} binomial(n-m, n-j) * Eulerian1(n, m) * j^k *(-1)^(n-j), where Eulerian1 = A173018.
A(n, k) = Sum_{j=0..k} binomial(k + [j>0], j+1)*A(n-1, k-j) for n > 0.
A(n, k) = Sum_{j=1..n} binomial(n, j)*(A(n-j, k-1) + A(n-j+1, k-1)) for n,k >= 1.
Row n (>=1) satisfies a linear recurrence:
A(n, k) = -Sum_{j=1..n} Stirling1(n + 1, n + 1 - j)*A(n, k - j) if k > n.
A(n, k) = [x^k] (Sum_{j=0..n} A371762(n, j)*x^j) / (Sum_{j=0..n} Stirling1(n + 1, n + 1 - j)*x^j).
A(n, k) = A(k, n). (From the symmetry of the bivariate exponential g.f.)
Let T(n, k) = A(n - k, k) and G(n) = Sum_{k=0..n} (-1)^k*T(n, k) the alternating row sums of the triangle. Then G(n) = (n + 2)*Euler(n + 1, 1) and as shifted Genocchi numbers G(n) = -2*(n + 2)*PolyLog(-n - 1, -1) = -A226158(n + 2).

A130850 Triangle read by rows, 0 <= k <= n, T(n,k) = Sum_{j=0..n} A(n,j)*binomial(n-j,k) where A(n,j) are the Eulerian numbers A173018.

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 6, 12, 7, 1, 24, 60, 50, 15, 1, 120, 360, 390, 180, 31, 1, 720, 2520, 3360, 2100, 602, 63, 1, 5040, 20160, 31920, 25200, 10206, 1932, 127, 1, 40320, 181440, 332640, 317520, 166824, 46620, 6050, 255, 1, 362880, 1814400, 3780000, 4233600, 2739240, 1020600, 204630, 18660, 511, 1

Offset: 0

Philippe Deléham, Aug 20 2007

Old name was: Triangle T(n,k), 0<=k<=n, read by rows given by [1,1,2,2,3,3,4,4,5,5,...] DELTA [1,0,2,0,3,0,4,0,5,0,6,0,...] where DELTA is the operator defined in A084938.
Vandervelde (2018) refers to this as the Worpitzky number triangle - N. J. A. Sloane, Mar 27 2018 [Named after the German mathematician Julius Daniel Theodor Worpitzky (1835-1895). - Amiram Eldar, Jun 24 2021]
Triangle given by A123125*A007318 (as infinite lower triangular matrices), A123125 = Euler's triangle, A007318 = Pascal's triangle; A007318*A123125 gives A046802.
Taylor coefficients of Eulerian polynomials centered at 1. - Louis Zulli, Nov 28 2015
A signed refinement is A263634. - Tom Copeland, Nov 14 2016
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 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 this entry (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

			Triangle begins:
1
1      1
2      3       1
6      12      7       1
24     60      50      15      1
120    360     390     180     31      1
720    2520    3360    2100    602     63      1
5040   20160   31920   25200   10206   1932    127    1
40320  181440  332640  317520  166824  46620   6050   255   1
362880 1814400 3780000 4233600 2739240 1020600 204630 18660 511 1
...

G. C. Greubel, Table of n, a(n) for n = 0..1034 [a(438) and a(901) corrected by _Georg Fischer_, Nov 10 2021]
Nguyen-Huu-Bong, Some Combinatorial Properties of Summation Operators, J. Comb. Theory, Ser. A, Vo. 11, No. 3 (1971), pp. 213-221. See Table on page 214.
John K. Sikora, On Calculating the Coefficients of a Polynomial Generated Sequence Using the Worpitzky Number Triangles, arXiv:1806.00887 [math.NT], 2018.
Sam Vandervelde, The Worpitzky Numbers Revisited, Amer. Math. Monthly, Vol. 125, No. 3 (2018), pp. 198-206.

Essentially reverse of A028246.
Variants are A130850, A075263, A163626.
Cf. A000142, A001710, A005460, A005461, A005462, A005463, A005464, A263634.
Cf. A019538, A028246, A046802, A090582, A119879, A123125, A248727.

Table[(n-k)!*StirlingS2[n+1, n-k+1], {n, 0, 10}, {k, 0, n}] (* G. C. Greubel, Nov 15 2015 *)

t(n, k) = (n-k)!*stirling(n+1, n-k+1, 2);
tabl(nn) = for (n=0, 10, for (k=0, n, print1(t(n,k),", ")); print()); \\ Michel Marcus, Nov 16 2015

from sage.combinat.combinat import eulerian_number
def A130850(n, k):
    return add(eulerian_number(n, j)*binomial(n-j, k) for j in (0..n))
for n in (0..7): [A130850(n, k) for k in (0..n)] # Peter Luschny, May 21 2013

T(n,k) = (-1)^k*A075263(n,k).
T(n,k) = (n-k)!*A008278(n+1,k+1).
T(n,n-1) = 2^n - 1 for n > 0. - Derek Orr, Dec 31 2015
E.g.f.: x/(e^(-x*t)*(1+x)-1). - Tom Copeland, Nov 14 2016
Sum_{k=1..floor(n/2)} T(n,2k) = Sum_{k=0..floor(n/2)} T(n,2k+1) = A000670(n). - Jacob Sprittulla, Oct 03 2021

New name from Peter Luschny, May 21 2013

cp's OEIS Frontend

A106338 Triangle T, read by rows, equal to the matrix inverse of the triangle defined by [T^-1](n,k) = A075263(n,k)/n!, for n>=k>=0.

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A123125 Triangle of Eulerian numbers T(n,k), 0 <= k <= n, read by rows.

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A028246 Triangular array a(n,k) = (1/k)*Sum_{i=0..k} (-1)^(k-i)*binomial(k,i)*i^n; n >= 1, 1 <= k <= n, read by rows.

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A009006 Expansion of e.g.f.: 1 + tan(x).

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A371761 Array read by antidiagonals: The number of parades with n girls and k boys that begin with a girl and end with a boy.

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A028246 Triangular array a(n,k) = (1/k)Sum_{i=0..k} (-1)^(k-i)binomial(k,i)*i^n; n >= 1, 1 <= k <= n, read by rows.