A144698 -id:A144698 - cp's OEIS frontend

A143496 4-Stirling numbers of the second kind.

1, 4, 1, 16, 9, 1, 64, 61, 15, 1, 256, 369, 151, 22, 1, 1024, 2101, 1275, 305, 30, 1, 4096, 11529, 9751, 3410, 545, 39, 1, 16384, 61741, 70035, 33621, 7770, 896, 49, 1, 65536, 325089, 481951, 305382, 95781, 15834, 1386, 60, 1, 262144, 1690981, 3216795

Offset: 4

Peter Bala, Aug 20 2008

This is the case r = 4 of the r-Stirling numbers of the second kind. The 4-Stirling numbers of the second kind count the ways of partitioning the set {1,2,...,n} into k nonempty disjoint subsets with the restriction that the elements 1, 2, 3 and 4 belong to distinct subsets. For remarks on the general case see A143494 (r = 2). The corresponding array of 4-Stirling numbers of the first kind is A143493. The theory of r-Stirling numbers of both kinds is developed in [Broder]. For 4-Lah numbers refer to A143499.
From Wolfdieter Lang, Sep 29 2011: (Start)
T(n,k) = S(n,k,4), n >= k >= 4, in Mikhailov's first paper, eq.(28) or (A3). E.g.f. column k from (A20) with k->4, r->k. Therefore, with offset [0,0], this triangle is the Sheffer triangle (exp(4*x),exp(x)-1) with e.g.f. of column no. m >= 0: exp(4*x)*((exp(x)-1)^m)/m!. See one of the formulas given below. For Sheffer matrices see the W. Lang link under A006232 with the S. Roman reference, also found in A132393.
(End)

			Triangle begins
n\k|.....4.....5.....6.....7.....8.....9
========================================
4..|.....1
5..|.....4.....1
6..|....16.....9.....1
7..|....64....61....15.....1
8..|...256...369...151....22.....1
9..|..1024..2101..1275...305....30.....1
...
T(6,5) = 9. The set {1,2,3,4,5,6} can be partitioned into five subsets such that 1, 2, 3 and 4 belong to different subsets in 9 ways: {{1,5}{2}{3}{4}{6}}, {{1,6}{2}{3}{4}{5}}, {{2,5}{1}{3}{4}{6}}, {{2,6}{1}{3}{4}{5}}, {{3,5}{1}{2}{4}{6}}, {{3,6}{1}{2}{4}{5}}, {{4,5}{1}{2}{3}{6}}, {{4,6}{1}{2}{3}{5}} and {{5,6}{1}{2}{3}{4}}.

Andrei Z. Broder, The r-Stirling numbers, Report Number: CS-TR-82-949, Stanford University, Department of Computer Science; see also, Discrete Math. 49, 241-259 (1984).
Askar Dzhumadil'daev and Damir Yeliussizov, Path decompositions of digraphs and their applications to Weyl algebra, arXiv preprint arXiv:1408.6764v1 [math.CO], 2014. [Version 1 contained many references to the OEIS, which were removed in Version 2. - _N. J. A. Sloane_, Mar 28 2015]
Askar Dzhumadil'daev and Damir Yeliussizov, Walks, partitions, and normal ordering, Electronic Journal of Combinatorics, 22(4) (2015), #P4.10.
Paweł Hitczenko, A class of polynomial recurrences resulting in (n/log n, n/log^2 n)-asymptotic normality, arXiv:2403.03422 [math.CO], 2024. See p. 8.
V. V. Mikhailov, Ordering of some boson operator functions, J. Phys A: Math. Gen. 16 (1983) 3817-3827.
V. V. Mikhailov, Normal ordering and generalised Stirling numbers, J. Phys A: Math. Gen. 18 (1985) 231-235.
Erich Neuwirth, Recursively defined combinatorial functions: Extending Galton's board, Discrete Math. 239 No. 1-3, 33-51 (2001)
Lily L. Liu and Yi Wang, A unified approach to polynomial sequences with only real zeros, arXiv:math/0509207 [math.CO], 2005-2006.
Michael J. Schlosser and Meesue Yoo, Elliptic Rook and File Numbers, Electronic Journal of Combinatorics, 24(1) (2017), #P1.31.

Cf. A003468 (column 7), A005060 (column 5), A008277, A016103 (column 6), A045379 (row sums), A049459 (matrix inverse), A143493, A143494, A143495, A143499.

with combinat: T := (n, k) -> 1/(k-4)!*add ((-1)^(k-i)*binomial(k-4,i)*(i+4)^(n-4),i = 0..k-4): for n from 4 to 13 do seq(T(n, k), k = 4..n) end do;

t[n_, k_] := StirlingS2[n, k] - 6*StirlingS2[n-1, k] + 11*StirlingS2[n-2, k] - 6*StirlingS2[n-3, k]; Flatten[ Table[ t[n, k], {n, 4, 13}, {k, 4, n}]] (* Jean-François Alcover, Dec 02 2011 *)

T(n+4,k+4) = (1/k!)*Sum_{i = 0..k} (-1)^(k-i)*C(k,i)*(i+4)^n, n,k >= 0.
T(n,k) = Stirling2(n,k) - 6*Stirling2(n-1,k) + 11*Stirling2(n-2,k) - 6*Stirling2(n-3,k) for n,k >= 4.
Recurrence relation: T(n,k) = T(n-1,k-1) + k*T(n-1,k) for n > 4 with boundary conditions: T(n,3) = T(3,n) = 0 for all n; T(4,4) = 1; T(4,k) = 0 for k > 4. Special cases: T(n,4) = 4^(n-4); T(n,5) = 5^(n-4) - 4^(n-4).
E.g.f. (k+4)-th column (with offset 4): (1/k!)*exp(4*x)*(exp(x)-1)^k.
O.g.f. k-th column: Sum_{n>=k} T(n,k)*x^n = x^k/((1-4*x)*(1-5*x)*...*(1-k*x)).
E.g.f.: exp(4*t + x*(exp(t)-1)) = Sum_{n = 0..infinity} Sum_(k = 0..n) T(n+4,k+4)*x^k*t^n/n! = Sum_{n = 0..infinity} B_n(4;x)*t^n/n! = 1 + (4+x)*t/1! + (16+9*x+x^2)*t^2/2! + ..., where the row polynomials, B_n(4;x) := Sum_{k = 0..n} T(n+4,k+4)*x^k, may be called the 4-Bell polynomials.
Dobinski-type identities: Row polynomial B_n(4;x) = exp(-x)*Sum_{i = 0..infinity} (i+4)^n*x^i/i!; Sum_{k = 0..n} k!*T(n+4,k+4)*x^k = Sum_{i = 0..infinity} (i+4)^n*x^i/(1+x)^(i+1).
The T(n,k) are the connection coefficients between the falling factorials and the shifted monomials (x+4)^(n-4). For example, 16 + 9*x + x*(x-1) = (x+4)^2; 64 + 61*x + 15*x*(x-1) + x*(x-1)*(x-2) = (x+4)^3.
This array is the matrix product P^3 * S, where P denotes Pascal's triangle, A007318 and S denotes the lower triangular array of Stirling numbers of the second kind, A008277 (apply Theorem 10 of [Neuwirth]).
The inverse array is A049459, the signed 4-Stirling numbers of the first kind.
From Peter Bala, Sep 19 2008: (Start)
Let D be the derivative operator d/dx and E the Euler operator x*d/dx. Then x^(-4)*E^n*x^4 = Sum_{k = 0..n} T(n+4,k+4)*x^k*D^k.
The row generating polynomials R_n(x) := Sum_{k=4..n} T(n,k)*x^k satisfy the recurrence R_(n+1)(x) = x*R_n(x) + x*d/dx(R_n(x)) with R_4(x) = x^4. It follows that the polynomials R_n(x) have only real zeros (apply Corollary 1.2. of [Liu and Wang]).
Relation with the 4-Eulerian numbers E_4(n,j) := A144698(n,j): T(n,k) = 4!/k!*Sum_{j = n-k..n-4} E_4(n,j)*binomial(j,n-k) for n >= k >= 4.
(End)

A144696 Triangle of 2-Eulerian numbers.

Original entry on oeis.org

1, 1, 2, 1, 7, 4, 1, 18, 33, 8, 1, 41, 171, 131, 16, 1, 88, 718, 1208, 473, 32, 1, 183, 2682, 8422, 7197, 1611, 64, 1, 374, 9327, 49780, 78095, 38454, 5281, 128, 1, 757, 30973, 264409, 689155, 621199, 190783, 16867, 256

Offset: 2

Peter Bala, Sep 19 2008

Let [n] denote the ordered set {1,2,...,n}. The symmetric group S_n consists of the injective mappings p:[n] -> [n]. Such a permutation p has an excedance at position i, 1 <= i < n, if p(i) > i. One well-known interpretation of the Eulerian numbers A(n,k) is that they count the permutations in the symmetric group S_n with k excedances. The triangle of Eulerian numbers is A008292 (but with an offset of 1 in the column numbering). We generalize this definition to restricted permutations as follows.
Let r be a nonnegative integer and let Permute(n,n-r) denote the set of injective maps p:[n-r] -> [n], which we think of as permutations of n numbers taken n-r at a time. Clearly, |Permute(n,n-r)| = n!/r!. We say that p has an excedance at position i, 1 <= i <= n-r, if p(i) > i. Then the r-Eulerian number, denoted by A(r;n,k), is defined as the number of permutations in Permute(n,n-r) having k excedances. Thus the current array of 2-Eulerian numbers gives the number of permutations in Permute(n,n-2) with k excedances. See the example section below for some numerical examples.
Clearly A(0;n,k) = A(n,k). The case r = 1 also produces the ordinary Eulerian numbers A(n,k). There is an obvious bijection from Permute(n,n) to Permute(n,n-1) that preserves the number of excedances of a permutation. Consequently, the 1-Eulerian numbers are equal to the 0-Eulerian numbers: A(1;n,k) = A(0;n,k) = A(n,k).
For other cases of r-Eulerian numbers see A144697 (r = 3), A144698 (r = 4) and A144699 (r = 5). There is also a concept of r-Stirling numbers of the first and second kinds - see A143491 and A143494. If we multiply the entries of the current array by a factor of 2 and then reverse the rows we obtain A120434.
An alternative interpretation of the current array due to [Strosser] involves the 2-excedance statistic of a permutation (see also [Foata & Schutzenberger, Chapitre 4, Section 3]). We define a permutation p in Permute(n,n-2) to have a 2-excedance at position i (1 <=i <= n-2) if p(i) >= i + 2.
Given a permutation p in Permute(n,n-2), define ~p to be the permutation in Permute(n,n-2) that takes i to n+1 - p(n-i-1). The map ~ is a bijection of Permute(n,n-2) with the property that if p has (resp. does not have) an excedance in position i then ~p does not have (resp. has) a 2-excedance at position n-i-1. Hence ~ gives a bijection between the set of permutations with k excedances and the set of permutations with (n-k) 2-excedances. Thus reading the rows of this array in reverse order gives a triangle whose entries count the permutations in Permute(n,n-2) with k 2-excedances.
Example: Represent a permutation p:[n-2] -> [n] in Permute(n,n-2) by its image vector (p(1),...,p(n-2)). In Permute(10,8) the permutation p = (1,2,4,7,10,6,5,8) does not have an excedance in the first two positions (i = 1 and 2) or in the final three positions (i = 6, 7 and 8). The permutation ~p = (3,6,5,1,4,7,9,10) has 2-excedances only in the first three positions and the final two positions.
From Peter Bala, Dec 27 2019: (Start)
This is the array A(1,1,3) in the notation of Hwang et al. (p. 25), where the authors remark that the r-Eulerian numbers were first studied by Shanlan Li (Duoji Bilei, Ch. 4), who gave the summation formulas
Sum_{i = 2..n+1} (i-1)*C(i,2) = C(n+3,4) + 2*C(n+2,4)
Sum_{i = 2..n+1} (i-1)^2*C(i,2) = C(n+4,5) + 7*C(n+3,5) + 4*C(n+2,5)
Sum_{i = 2..n+1} (i-1)^3*C(i,2) = C(n+5,6) + 18*C(n+4,6) + 33*C(n+3,6) + 8*C(n+2,6). (End)

			The triangle begins
===========================================
n\k|..0.....1.....2.....3.....4.....5.....6
===========================================
2..|..1
3..|..1.....2
4..|..1.....7.....4
5..|..1....18....33.....8
6..|..1....41...171...131....16
7..|..1....88...718..1208...473....32
8..|..1...183..2682..8422..7197..1611....64
...
Row 4 = [1,7,4]: We represent a permutation p:[n-2] -> [n] in Permute(n,n-2) by its image vector (p(1),...,p(n-2)). Here n = 4. The permutation (1,2) has no excedances; 7 permutations have a single excedance, namely, (1,3), (1,4), (2,1), (3,1), (3,2), (4,1) and (4,2); the remaining 4 permutations, (2,3), (2,4), (3,4) and (4,3) each have two excedances.

J. Riordan. An introduction to combinatorial analysis. New York, J. Wiley, 1958.
R. Strosser. Séminaire de théorie combinatoire, I.R.M.A., Université de Strasbourg, 1969-1970.
Li, Shanlan (1867). Duoji bilei (Series summation by analogies), 4 scrolls. In Zeguxizhai suanxue (Mathematics from the Studio Devoted to the Imitation of the Ancient Chinese Tradition) (Jinling ed.), Volume 4.
Li, Shanlan (2019). Catégories analogues d’accumulations discrètes (Duoji bilei), traduit et commenté par Andrea Bréard. La Bibliothèque Chinoise. Paris: Les Belles Lettres.

G. C. Greubel, Rows n = 2..52 of the triangle, flattened
J. F. Barbero G., J. Salas, and E. J. S. Villaseñor, Bivariate Generating Functions for a Class of Linear Recurrences. II. Applications, arXiv preprint arXiv:1307.5624 [math.CO], 2013-2015.
Mark Conger, A refinement of the Eulerian polynomials and the joint distribution of pi(1) and Des(pi) in S_n, arXiv:math/0508112 [math.CO], 2005.
Ming-Jian Ding and Bao-Xuan Zhu, Some results related to Hurwitz stability of combinatorial polynomials, Advances in Applied Mathematics, Volume 152, (2024), 102591. See p. 9.
Sergi Elizalde, Descents on quasi-Stirling permutations, arXiv:2002.00985 [math.CO], 2020.
D. Foata and M. Schutzenberger, Théorie Géométrique des Polynômes Eulériens, Lecture Notes in Math., no.138, Springer Verlag 1970; arXiv:math/0508232 [math.CO], 2005.
Hsien-Kuei Hwang, Hua-Huai Chern, and Guan-Huei Duh, An asymptotic distribution theory for Eulerian recurrences with applications, arXiv:1807.01412 [math.CO], 2018-2019.
Tanya Khovanova and Rich Wang, Ending States of a Special Variant of the Chip-Firing Algorithm, arXiv:2302.11067 [math.CO], 2023.
L. Liu and Y. Wang, A unified approach to polynomial sequences with only real zeros, arXiv:math/0509207 [math.CO], 2005-2006.
Shi-Mei Ma, Some combinatorial sequences associated with context-free grammars, arXiv:1208.3104 [math.CO], 2012. - From _N. J. A. Sloane_, Aug 21 2012
Carla D. Savage and Gopal Viswanathan, The 1/k-Eulerian polynomials, Elec. J. of Comb., Vol. 19, Issue 1, #P9 (2012). - From _N. J. A. Sloane_, Feb 06 2013

Cf. A008292, A120434, A143491, A143494, A143497, A144697, A144698, A144699.
Cf. A000079 (right diagonal), A001710 (row sums).
Cf. A000182 (related to alt. row sums).

m:=2; [(&+[(-1)^(k-j)*Binomial(n+1,k-j)*Binomial(j+m,m-1)*(j+1)^(n-m+1): j in [0..k]])/m: k in [0..n-m], n in [m..m+10]]; // G. C. Greubel, Jun 04 2022

with(combinat):
T:= (n,k) -> 1/2!*add((-1)^(k-j)*binomial(n+1,k-j)*(j+1)^(n-1)*(j+2), j = 0..k):
for n from 2 to 10 do
seq(T(n,k),k = 0..n-2)
end do;

T[n_, k_]:= 1/2!*Sum[(-1)^(k-j)*Binomial[n+1, k-j]*(j+1)^(n-1)*(j+2), {j, 0, k}];
Table[T[n, k], {n,2,10}, {k,0,n-2}]//Flatten (* Jean-François Alcover, Oct 15 2019 *)

m=2 # A144696
def T(n,k): return (1/m)*sum( (-1)^(k-j)*binomial(n+1,k-j)*binomial(j+m,m-1)*(j+1)^(n-m+1) for j in (0..k) )
flatten([[T(n,k) for k in (0..n-m)] for n in (m..m+10)]) # G. C. Greubel, Jun 04 2022

T(n,k) = (1/2!)*Sum_{j = 0..k} (-1)^(k-j)*binomial(n+1,k-j)*(j+1)^(n-1)*(j+2);
T(n,n-k) = (1/2!)*Sum_{j = 2..k} (-1)^(k-j)*binomial(n+1,k-j)*j^(n-1)*(j-1).
Recurrence relations:
T(n,k) = (k+1)*T(n-1,k) + (n-k)*T(n-1,k-1) with boundary conditions T(n,0) = 1 for n >= 2, T(2,k) = 0 for k >= 1. Special cases: T(n,n-2) = 2^(n-2); T(n,n-3) = A066810(n-1).
E.g.f. (with suitable offsets): (1/2)*[(1 - x)/(1 - x*exp(t - t*x))]^2 = 1/2 + x*t + (x + 2*x^2)*t^2/2! + (x + 7*x^2 + 4*x^3)*t^3/3! + ... .
The row generating polynomials R_n(x) satisfy the recurrence R_(n+1)(x) = (n*x+1)*R_n(x) + x*(1-x)*d/dx(R_n(x)) with R_2(x) = 1. It follows that the polynomials R_n(x) for n >= 3 have only real zeros (apply Corollary 1.2. of [Liu and Wang]).
The (n+1)-th row generating polynomial = (1/2!)*Sum_{k = 1..n} (k+1)!*Stirling2(n,k) *x^(k-1)*(1-x)^(n-k).
For n >= 2,
(1/2)*(x*d/dx)^(n-1) (1/(1-x)^2) = x/(1-x)^(n+1) * Sum_{k = 0..n-2} T(n,k)*x^k,
(1/2)*(x*d/dx)^(n-1) (x^2/(1-x)^2) = 1/(1-x)^(n+1) * Sum_{k = 2..n} T(n,n-k)*x^k,
1/(1-x)^(n+1)*Sum_{k = 0..n-2} T(n,k)*x^k = (1/2!) * Sum_{m = 0..inf} (m+1)^(n-1)*(m+2)*x^m,
1/(1-x)^(n+1)*Sum_{k = 2..n} T(n,n-k)*x^k = (1/2!) * Sum_{m = 2..inf} m^(n-1)*(m-1)*x^m.
Worpitzky-type identities:
Sum_{k = 0..n-2} T(n,k)*binomial(x+k,n) = (1/2!)*x^(n-1)*(x - 1);
Sum_{k = 2..n} T(n,n-k)*binomial(x+k,n) = (1/2!)*(x + 1)^(n-1)*(x + 2).
Relation with Stirling numbers (Frobenius-type identities):
T(n+1,k-1) = (1/2!) * Sum_{j = 0..k} (-1)^(k-j)*(j+1)!* binomial(n-j,k-j)*Stirling2(n,j) for n,k >= 1;
T(n+1,k-1) = (1/2!) * Sum_{j = 0..n-k} (-1)^(n-k-j)*(j+1)!* binomial(n-j,k)*S(2;n+2,j+2) for n,k >= 1 and
T(n+2,k) = (1/2!) * Sum_{j = 0..n-k} (-1)^(n-k-j)*(j+2)!* binomial(n-j,k)*S(2;n+2,j+2) for n,k >= 0, where S(2;n,k) denotes the 2-Stirling numbers A143494(n,k).
The row polynomials of this array are related to the Eulerian polynomials. For example, 1/2*x*d/dx [x*(x + 4*x^2 + x^3)/(1-x)^4] = x^2*(1 + 7*x + 4*x^2)/(1-x)^5 and 1/2*x*d/dx [x*(x + 11*x^2 + 11*x^3 + x^4)/(1-x)^5] = x^2*(1 + 18*x + 33*x^2 + 8*x^3)/(1-x)^6.
Row sums A001710. Alternating row sums [1, -1, -2, 8, 16, -136, -272, 3968, 7936, ... ] are alternately (signed) tangent numbers and half tangent numbers - see A000182.
Sum_{k = 0..n-2} 2^k*T(n,k) = A069321(n-1). Sum_{k = 0..n-2} 2^(n-k)*T(n,k) = 4*A083410(n-1).
For n >=2, the shifted row polynomial t*R(n,t) = (1/2)*D^(n-1)(f(x,t)) evaluated at x = 0, where D is the operator (1-t)*(1+x)*d/dx and f(x,t) = (1+x*t/(t-1))^(-2). - Peter Bala, Apr 22 2012

A144697 Triangle of 3-Eulerian numbers.

Original entry on oeis.org

1, 1, 3, 1, 10, 9, 1, 25, 67, 27, 1, 56, 326, 376, 81, 1, 119, 1314, 3134, 1909, 243, 1, 246, 4775, 20420, 25215, 9094, 729, 1, 501, 16293, 115105, 248595, 180639, 41479, 2187, 1, 1012, 53388, 590764, 2048710, 2575404, 1193548, 183412, 6561

Offset: 3

Peter Bala, Sep 19 2008

This is the case r = 3 of the r-Eulerian numbers, denoted by A(r;n,k), defined as follows. Let [n] denote the ordered set {1,2,...,n} and let r be a nonnegative integer. Let Permute(n,n-r) denote the set of injective maps p:[n-r] -> [n], which we think of as permutations of n numbers taken n-r at a time. Clearly, |Permute(n,n-r)| = n!/r!. We say that the permutation p has an excedance at position i, 1 <= i <= n-r, if p(i) > i. Then the r-Eulerian number A(r;n,k) is defined as the number of permutations in Permute(n,n-r) with k excedances. Thus the 3-Eulerian numbers count the permutations in Permute(n,n-3) with k excedances (see the example section below for a numerical example).
For other cases see A008292 (r = 0 and r = 1), A144696 (r = 2), A144698 (r = 4) and A144699 (r = 5).
An alternative interpretation of the current array due to [Strosser] involves the 3-excedance statistic of a permutation (see also [Foata & Schutzenberger, Chapitre 4, Section 3]). We define a permutation p in Permute(n,n-3) to have a 3-excedance at position i (1 <= i <= n-3) if p(i) >= i + 3.
Given a permutation p in Permute(n,n-3), define ~p to be the permutation in Permute(n,n-3) that takes i to n+1 - p(n-i-2). The map ~ is a bijection of Permute(n,n-3) with the property that if p has (resp. does not have) an excedance in position i then ~p does not have (resp. has) a 3-excedance at position n-i-2. Hence ~ gives a bijection between the set of permutations with k excedances and the set of permutations with (n-k) 3-excedances. Thus reading the rows of this array in reverse order gives a triangle whose entries count the permutations in Permute(n,n-3) with k 3-excedances.
Example: Represent a permutation p:[n-3] -> [n] in Permute(n,n-3) by its image vector (p(1),...,p(n-3)). In Permute(10,7) the permutation p = (1,2,4,10,3,6,5) does not have an excedance in the first two positions (i = 1 and 2) or in the final three positions (i = 5, 6 and 7). The permutation ~p = (6,5,8,1,7,9,10) has 3-excedances only in the first three positions and the final two positions.

			Triangle begins
=================================================
n\k|..0......1......2......3......4......5......6
=================================================
3..|..1
4..|..1......3
5..|..1.....10......9
6..|..1.....25.....67.....27
7..|..1.....56....326....376.....81
8..|..1....119...1314...3134...1909....243
9..|..1....246...4775..20420..25215...9094....729
...
T(5,1) = 10: We represent a permutation p:[n-3] -> [n] in Permute(n,n-3) by its image vector (p(1),...,p(n-3)). The 10 permutations in Permute(5,2) having 1 excedance are (1,3), (1,4), (1,5), (3,2), (4,2), (5,2), (2,1), (3,1), (4,1) and (5,1).

R. Strosser, Séminaire de théorie combinatoire, I.R.M.A., Universite de Strasbourg, 1969-1970.

G. C. Greubel, Rows n = 3..53 of the triangle, flattened
J. F. Barbero G., J. Salas, and E. J. S. Villaseñor, Bivariate Generating Functions for a Class of Linear Recurrences. II. Applications, arXiv preprint arXiv:1307.5624 [math.CO], 2013-2015.
Mark Conger, A refinement of the Eulerian polynomials and the joint distribution of pi(1) and Des(pi) in S_n, arXiv:math/0508112 [math.CO], 2005.
Ming-Jian Ding and Bao-Xuan Zhu, Some results related to Hurwitz stability of combinatorial polynomials, Advances in Applied Mathematics, Volume 152, (2024), 102591. See p. 9.
Sergi Elizalde, Descents on quasi-Stirling permutations, arXiv:2002.00985 [math.CO], 2020.
D. Foata and M. Schutzenberger, Théorie Géométrique des Polynômes Eulériens, arXiv:math/0508232 [math.CO], 2005; Lecture Notes in Math., no. 138, Springer Verlag, 1970.
L. Liu and Y. Wang, A unified approach to polynomial sequences with only real zeros, arXiv:math/0509207 [math.CO], 2005-2006.
Shi-Mei Ma, Some combinatorial sequences associated with context-free grammars, arXiv:1208.3104v2 [math.CO], 2012. - From _N. J. A. Sloane_, Aug 21 2012

Cf. A008292, A060056, A143492, A143495, A144696, A144698, A144699.

Cf. A001715 (row sums), A000244 (right diagonal).

m:=3; [(&+[(-1)^(k-j)*Binomial(n+1,k-j)*Binomial(j+m,m-1)*(j+1)^(n-m+1): j in [0..k]])/m: k in [0..n-m], n in [m..13]]; // G. C. Greubel, Jun 04 2022

with(combinat):
T:= (n,k) -> 1/3!*add((-1)^(k-j)*binomial(n+1,k-j)*(j+1)^(n-2)*(j+2)*(j+3),j = 0..k):
for n from 3 to 11 do
seq(T(n,k),k = 0..n-3)
end do;

T[n_, k_] /; 0 < k <= n-3 := T[n, k] = (k+1) T[n-1, k] + (n-k) T[n-1, k-1];
T[, 0] = 1; T[, _] = 0;
Table[T[n, k], {n, 3, 11}, {k, 0, n-3}] // Flatten (* Jean-François Alcover, Nov 11 2019 *)

m=3 # A144697
def T(n,k): return (1/m)*sum( (-1)^(k-j)*binomial(n+1,k-j)*binomial(j+m,m-1)*(j+1)^(n-m+1) for j in (0..k) )
flatten([[T(n,k) for k in (0..n-m)] for n in (m..13)]) # G. C. Greubel, Jun 04 2022

T(n,k) = (1/3!)*Sum_{j = 0..k} (-1)^(k-j)*binomial(n+1,k-j)*(j+1)^(n-2)*(j+2)*(j+3);
T(n,n-k) = (1/3!)*Sum_{j = 3..k} (-1)^(k-j)*binomial(n+1,k-j)*j^(n-2)*(j-1)*(j-2).
Recurrence relation:
T(n,k) = (k+1)*T(n-1,k) + (n-k)*T(n-1,k-1) with boundary conditions T(n,0) = 1 for n >= 3, T(3,k) = 0 for k >= 1. Special cases: T(n,n-3) = 3^(n-3); T(n,n-4) = A086443 (n-2).
E.g.f. (with suitable offsets): (1/3)*((1 - x)/(1 - x*exp(t - t*x)))^3 = 1/3 + x*t + (x + 3*x^2)*t^2/2! + (x + 10*x^2 + 9*x^3)*t^3/3! + ... .
The row generating polynomials R_n(x) satisfy the recurrence R_(n+1)(x) = (n*x+1)*R_n(x) + x*(1-x)*d/dx(R_n(x)) with R_3(x) = 1. It follows that the polynomials R_n(x) for n >= 4 have only real zeros (apply Corollary 1.2. of [Liu and Wang]).
The (n+2)-th row generating polynomial = (1/3!)*Sum_{k = 1..n} (k+2)!*Stirling2(n,k)*x^(k-1)*(1-x)^(n-k).
For n >= 3,
(1/3)*(x*d/dx)^(n-2) (1/(1-x)^3) = (x/(1-x)^(n+1)) * Sum_{k = 0..n-3} T(n,k)*x^k,
(1/3)*(x*d/dx)^(n-2) (x^3/(1-x)^3) = (1/(1-x)^(n+1)) * Sum_{k = 3..n} T(n,n-k)*x^k,
(1/(1-x)^(n+1)) * Sum_{k = 0..n-3} T(n,k)*x^k = (1/3!) * Sum_{m >= 0} (m+1)^(n-2)*(m+2)*(m+3)*x^m,
(1/(1-x)^(n+1)) * Sum_{k = 3..n} T(n,n-k)*x^k = (1/3!) * Sum_{m >= 3} m^(n-2)*(m-1)*(m-2)*x^m.
Worpitzky-type identities:
Sum_{k = 0..n-3} T(n,k)* binomial(x+k,n) = (1/3!)*x^(n-2)*(x-1)*(x-2);
Sum_{k = 3..n} T(n,n-k)* binomial(x+k,n) = (1/3!)*(x+1)^(n-2)*(x+2)*(x+3).
Relation with Stirling numbers (Frobenius-type identities):
T(n+2,k-1) = (1/3!) * Sum_{j = 0..k} (-1)^(k-j)* (j+2)!* binomial(n-j,k-j)*Stirling2(n,j) for n,k >= 1;
T(n+2,k-1) = (1/3!) * Sum_{j = 0..n-k} (-1)^(n-k-j)* (j+2)!* binomial(n-j,k)*S(3;n+3,j+3) for n,k >= 1 and
T(n+3,k) = (1/3!) * Sum_{j = 0..n-k} (-1)^(n-k-j)*(j+3)!* binomial(n-j,k)*S(3;n+3,j+3) for n,k >= 0, where S(3;n,k) denotes the 3-Stirling numbers A143495(n,k).
The row polynomials of this array are related to the 2-Eulerian polynomials (see A144696). For example, (1/3)*x*d/dx (x^3*(1 + 7*x + 4*x^2)/(1-x)^5) = x^3*(1 + 10*x + 9*x^2)/(1-x)^6 and (1/3)*x*d/dx (x^3*(1 + 18*x + 33*x^2 + 8*x^3)/(1-x)^6) = x^3*(1 + 25*x + 67*x^2 + 27*x^3)/(1-x)^7.
For n >=3, the shifted row polynomial t*R(n,t) = (1/3)*D^(n-2)(f(x,t)) evaluated at x = 0, where D is the operator (1-t)*(1+x)*d/dx and f(x,t) = (1+x*t/(t-1))^(-3). - Peter Bala, Apr 22 2012

A144699 Triangle of 5-Eulerian numbers.

Original entry on oeis.org

1, 1, 5, 1, 16, 25, 1, 39, 171, 125, 1, 86, 786, 1526, 625, 1, 181, 3046, 11606, 12281, 3125, 1, 372, 10767, 70792, 142647, 92436, 15625, 1, 755, 36021, 380071, 1279571, 1553145, 663991, 78125, 1, 1522, 116368, 1880494, 9818494, 19555438, 15519952, 4608946, 390625

Offset: 5

Peter Bala, Sep 19 2008

This is the case r = 5 of the r-Eulerian numbers, denoted by A(r;n,k), defined as follows. Let [n] denote the ordered set {1,2,...,n} and let r be a nonnegative integer. Let Permute(n,n-r) denote the set of injective maps p:[n-r] -> [n], which we think of as permutations of n numbers taken n-r at a time. Clearly, |Permute(n,n-r)| = n!/r!. We say that the permutation p has an excedance at position i, 1 <= i <= n-r, if p(i) > i. Then the r-Eulerian number A(r;n,k) is defined as the number of permutations in Permute(n,n-r) with k excedances. Thus the 5-Eulerian numbers count the permutations in Permute(n,n-5) with k excedances. For other cases see A008292 (r = 0 and r = 1), A144696 (r = 2), A144697 (r = 3) and A144698 (r = 4).
An alternative interpretation of the current array due to [Strosser] involves the 5-excedance statistic of a permutation (see also [Foata & Schutzenberger, Chapitre 4, Section 3]). We define a permutation p in Permute(n,n-5) to have a 5-excedance at position i (1 <=i <= n-5) if p(i) >= i + 5.
Given a permutation p in Permute(n,n-5), define ~p to be the permutation in Permute(n,n-5) that takes i to n+1 - p(n-i-4). The map ~ is a bijection of Permute(n,n-5) with the property that if p has (resp. does not have) an excedance in position i then ~p does not have (resp. has) a 5-excedance at position n-i-4. Hence ~ gives a bijection between the set of permutations with k excedances and the set of permutations with (n-k) 5-excedances. Thus reading the rows of this array in reverse order gives a triangle whose entries count the permutations in Permute(n,n-5) with k 5-excedances.
Example: Represent a permutation p:[n-5] -> [n] in Permute(n,n-5) by its image vector (p(1),...,p(n-5)). In Permute(12,7) the permutation p = (1,2,4,12,3,6,5) does not have an excedance in the first two positions (i = 1 and 2) or in the final three positions (i = 5, 6 and 7). The permutation ~p = (8,7,10,1,9,11,12) has 5-excedances only in the first three positions and the final two positions.

			Triangle begins
=======================================================
n\k|..0.......1......2......3.........4.......5.......6
=======================================================
5..|..1
6..|..1.......5
7..|..1......16......25
8..|..1......39.....171.....125
9..|..1......86.....786....1526.....625
10.|..1.....181....3046...11606...12281....3125
11.|..1.....372...10767...70792..142647...92436...15625
...
T(7,1) = 16: We represent a permutation p:[n-5] -> [n] in Permute(n,n-5) by its image vector (p(1),...,p(n-5)). The 16 permutations in Permute(7,2) having 1 excedance are (1,3), (1,4), (1,5), (1,6), (1,7), (3,2), (4,2), (5,2), (6,2), (7,2), (2,1), (3,1), (4,1), (5,1), (6,1) and (7,1).

R. Strosser, Séminaire de théorie combinatoire, I.R.M.A., Université de Strasbourg, 1969-1970.

G. C. Greubel, Rows n = 5..55 of the triangle, flattened
J. F. Barbero G., J. Salas, and E. J. S. Villaseñor, Bivariate Generating Functions for a Class of Linear Recurrences. II. Applications, arXiv preprint arXiv:1307.5624 [math.CO], 2013-2015.
Ming-Jian Ding and Bao-Xuan Zhu, Some results related to Hurwitz stability of combinatorial polynomials, Advances in Applied Mathematics, Volume 152, (2024), 102591. See p. 9.
Sergi Elizalde, Descents on quasi-Stirling permutations, arXiv:2002.00985 [math.CO], 2020.
D. Foata and M. Schutzenberger, Théorie Géométrique des Polynômes Eulériens, arXiv:math/0508232 [math.CO], 2005; Lecture Notes in Math., no. 138, Springer Verlag, 1970.
L. Liu and Y. Wang, A unified approach to polynomial sequences with only real zeros, arXiv:math/0509207 [math.CO], 2005-2006.
Shi-Mei Ma, Some combinatorial sequences associated with context-free grammars, arXiv:1208.3104v2 [math.CO], 2012. - From _N. J. A. Sloane_, Aug 21 2012

Cf. A001725 (row sums), A008292, A143494, A144696, A144697, A144698.

m:=5; [(&+[(-1)^(k-j)*Binomial(n+1,k-j)*Binomial(j+m,m-1)*(j+1)^(n-m+1): j in [0..k]])/m: k in [0..n-m], n in [m..m+10]]; // G. C. Greubel, Jun 08 2022

with(combinat):
T:= (n,k) -> 1/5!*add((-1)^(k-j)*binomial(n+1,k-j)*(j+1)^(n-4)*(j+2)*(j+3)*(j+4)*(j+5),j = 0..k):
for n from 5 to 13 do
seq(T(n,k),k = 0..n-5)
end do;

T[n_, k_] /; 0 < k <= n-5 := T[n, k] = (k+1) T[n-1, k] + (n-k) T[n-1, k-1];
T[, 0] = 1; T[, _] = 0;
Table[T[n, k], {n, 5, 13}, {k, 0, n-5}] // Flatten (* Jean-François Alcover, Nov 11 2019 *)

m=5 # A144699
def T(n,k): return (1/m)*sum( (-1)^(k-j)*binomial(n+1,k-j)*binomial(j+m,m-1)*(j+1)^(n-m+1) for j in (0..k) )
flatten([[T(n,k) for k in (0..n-m)] for n in (m..m+10)]) # G. C. Greubel, Jun 08 2022

T(n,k) = (1/5!)*Sum_{j = 0..k} (-1)^(k-j)*binomial(n+1,k-j)*(j+1)^(n-4)*(j+2)*(j+3)*(j+4)*(j+5).
T(n,n-k) = (1/5!)*Sum_{j = 5..k} (-1)^(k-j)*binomial(n+1,k-j)*j^(n-4)*(j-1)*(j-2)*(j-3)*(j-4).
Recurrence relation:
T(n,k) = (k+1)*T(n-1,k) + (n-k)*T(n-1,k-1) with boundary conditions T(n,0) = 1 for n >= 5, T(5,k) = 0 for k >= 1.
E.g.f. (with suitable offsets): (1/5)*((1 - x)/(1 - x*exp(t - t*x)))^5 = 1/5 + x*t + (x + 5*x^2)*t^2/2! + (x + 16*x^2 + 25*x^3)*t^3/3! + ... .
The row generating polynomials R_n(x) satisfy the recurrence R_(n+1)(x) = (n*x+1)*R_n(x) + x*(1-x)*d/dx(R_n(x)) with R_5(x) = 1. It follows that the polynomials R_n(x) for n >= 6 have only real zeros (apply Corollary 1.2. of [Liu and Wang]).
The (n+4)-th row generating polynomial = (1/5!)*Sum_{k = 1..n} (k+4)!*Stirling2(n,k)*x^(k-1)*(1-x)^(n-k).
For n >= 5,
(1/5)*(x*d/dx)^(n-4) (1/(1-x)^5) = (x/(1-x)^(n+1)) * Sum_{k = 0..n-5} T(n,k)*x^k,
(1/5)*(x*d/dx)^(n-4) (x^5/(1-x)^5) = (1/(1-x)^(n+1)) * Sum_{k = 5..n} T(n,n-k)*x^k,
(1/(1-x)^(n+1)) * Sum_{k = 0..n-5} T(n,k)*x^k = (1/5!) * Sum_{m = 0..infinity} (m+1)^(n-4)*(m+2)*(m+3)*(m+4)*(m+5)*x^m,
(1/(1-x)^(n+1)) * Sum_{k = 5..n} T(n,n-k)*x^k = (1/5!) * Sum_{m = 5..infinity} m^(n-4)*(m-1)*(m-2)*(m-3)*(m-4)*x^m.
Worpitzky-type identities:
Sum_{k = 0..n-5} T(n,k)*binomial(x+k,n) = (1/5!)*x^(n-4)*(x-1)*(x-2)*(x-3)*(x-4).
Sum_{k = 5..n} T(n,n-k)* binomial(x+k,n) = (1/5!)*(x+1)^(n-4)*(x+2)*(x+3)*(x+4)*(x+5).
Relation with Stirling numbers (Frobenius-type identities):
T(n+4,k-1) = (1/5!) * Sum_{j = 0..k} (-1)^(k-j)* (j+4)!* binomial(n-j,k-j)*Stirling2(n,j) for n,k >= 1;
T(n+4,k-1) = (1/5!) * Sum_{j = 0..n-k} (-1)^(n-k-j)*(j+4)!* binomial(n-j,k)*S(5;n+5,j+5) for n,k >= 0 and
T(n+5,k) = (1/5!) * Sum_{j = 0..n-k} (-1)^(n-k-j)*(j+5)!* binomial(n-j,k)*S(5;n+5,j+5) for n,k >= 0, where S(5;n,k) denotes the 5-Stirling numbers of the second kind, which are given by the formula S(5;n+5,j+5) = 1/j!*Sum_{i = 0..j} (-1)^(j-i)*binomial(j,i)*(i+5)^n for n,j >= 0.

A152249 Triangle of 4 - restricted Eulerian numbers as polynomials used in exponential data smoothing: m(p,k,x)=((-1)^k(1 - x)^(p + k)/(k!(p - 1)!))Sum[(p - 1 + j)!j^kx^j/(j!), {j, 0, Infinity}]/x;n=6; t(m,l)=coefficients((-1)^mm!M[n, m, x])/n.

Original entry on oeis.org

1, 1, 6, 1, 19, 36, 1, 46, 241, 216, 1, 101, 1091, 2551, 1296, 1, 212, 4182, 18932, 24337, 7776, 1, 435, 14666, 113366, 273141, 217015, 46656, 1, 882, 48783, 600124, 2385999, 3487218, 1845697, 279936, 1, 1777, 156933, 2937109, 17931235, 42397299

Offset: 1

Roger L. Bagula, Nov 30 2008

Row sums are: {1, 7, 56, 504, 5040, 55440, 665280, 8648640, 121080960, 1816214400,...}. The sequences A008292, A144696,A144697,A144698,A144699 and this one, form a matrix of polynomials that are used in data smoothing calculations.

			{1},
{1, 6},
{1, 19, 36},
{1, 46, 241, 216},
{1, 101, 1091, 2551, 1296},
{1, 212, 4182, 18932, 24337, 7776},
{1, 435, 14666, 113366, 273141, 217015, 46656},
{1, 882, 48783, 600124, 2385999, 3487218, 1845697, 279936},
{1, 1777, 156933, 2937109, 17931235, 42397299, 40817623, 15159367, 1679616},
{1, 3568, 493900, 13631632, 121964374, 433696144, 667299052, 447815920, 121232113,10077696}

Douglas C. Montgomery, Lynwood A, Johnson, Forecasting and Time Series Analysis,McGraw-Hill, New York,1976,page 64.

A008292, A144696, A144697, A144698, A144699

M[p_, k_, x_] = ((-1)^k*(1 - x)^(p + k)/(k!(p - 1)!))*Sum[(p - 1 + j)!*j^k*x^j/(j!), {j, 0, Infinity}]/x;
Table[Table[CoefficientList[FullSimplify[ExpandAll[(-1)^m*m!*M[n, m, x]]]/n, x], {m, 1, 10}], {n, 1, 10}];
Table[Flatten[Table[CoefficientList[FullSimplify[ExpandAll[(-1)^m*m!*M[n, m, x]]]/n, x], {m, 1, 10}]], {n, 1, 10}]

m(p,k,x)=((-1)^k*(1 - x)^(p + k)/(k!(p - 1)!))*Sum[(p - 1 + j)!*j^k*x^j/(j!), {j, 0, Infinity}]/x;n=6;

t(m,l)=coefficients((-1)^m*m!*M[n, m, x])/n

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A143496 4-Stirling numbers of the second kind.

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A144696 Triangle of 2-Eulerian numbers.

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A144697 Triangle of 3-Eulerian numbers.

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A144699 Triangle of 5-Eulerian numbers.

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A152249 Triangle of 4 - restricted Eulerian numbers as polynomials used in exponential data smoothing: m(p,k,x)=((-1)^k*(1 - x)^(p + k)/(k!(p - 1)!))*Sum[(p - 1 + j)!*j^k*x^j/(j!), {j, 0, Infinity}]/x;n=6; t(m,l)=coefficients((-1)^m*m!*M[n, m, x])/n.

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A152249 Triangle of 4 - restricted Eulerian numbers as polynomials used in exponential data smoothing: m(p,k,x)=((-1)^k(1 - x)^(p + k)/(k!(p - 1)!))Sum[(p - 1 + j)!j^kx^j/(j!), {j, 0, Infinity}]/x;n=6; t(m,l)=coefficients((-1)^mm!M[n, m, x])/n.