A120434 -id:A120434 - cp's OEIS frontend

A256890 Triangle T(n,k) = t(n-k, k); t(n,m) = f(m)t(n-1,m) + f(n)t(n,m-1), where f(x) = x + 2.

1, 2, 2, 4, 12, 4, 8, 52, 52, 8, 16, 196, 416, 196, 16, 32, 684, 2644, 2644, 684, 32, 64, 2276, 14680, 26440, 14680, 2276, 64, 128, 7340, 74652, 220280, 220280, 74652, 7340, 128, 256, 23172, 357328, 1623964, 2643360, 1623964, 357328, 23172, 256, 512, 72076, 1637860, 10978444, 27227908, 27227908, 10978444, 1637860, 72076, 512

Offset: 0

Dale Gerdemann, Apr 12 2015

Related triangles may be found by varying the function f(x). If f(x) is a linear function, it can be parameterized as f(x) = a*x + b. With different values for a and b, the following triangles are obtained:
  a\b 1.......2.......3.......4.......5.......6
  -1  A144431
  0   A007318 A038208 A038221
  1   A008292 A256890 A257180 A257606 A257607
  2   A060187 A257609 A257611 A257613 A257615
  3   A142458 A257610 A257620 A257622 A257624 A257626
  4   A142459 A257612 A257621
  5   A142460 A257614 A257623
  6   A142461 A257616 A257625
  7   A142462 A257617 A257627
  8   A167884 A257618
  9   A257608 A257619
The row sums of these, and similarly constructed number triangles, are shown in the following table:
  a\b 1.......2.......3.......4.......5.......6.......7.......8.......9
  0   A000079 A000302 A000400
  1   A000142 A001715 A001725 A049388 A049198
  2   A000165 A002866 A002866 A051580 A051582
  3   A008544 A051578 A037559 A051605 A051607 A051609
  4   A001813 A047053 A000407 A034177 A051618 A051620 A051622
  5   A047055 A008546 A008548 A034300 A034325 A051688 A051690
  6   A047657 A049308 A047058 A034689 A034724 A034788 A053101 A053103
  7   A084947 A144827 A049209 A045754 A034830 A034832 A034834 A053105
  8   A084948 A144828 A147626 A051189 A034908 A034910 A034912 A034976 A053115
  9   A084949 A144829 A147630 A049211 A045756 A035013 A035018 A035021 A035023
  10                                  A051262 A035265 A035273         A035277
  11                                  A254322
  12                                          A145448
The formula can be further generalized to: t(n,m) = f(m+s)*t(n-1,m) + f(n-s)*t(n,m-1), where f(x) = a*x + b. The following table specifies triangles with nonzero values for s (given after the slash).
  a\b  0           1           2          3
  -2   A130595/1
  -1
  0
  1    A110555/-1  A120434/-1  A144697/1  A144699/2
With the absolute value, f(x) = |x|, one obtains A038221/3, A038234/4, A038247/5, A038260/6, A038273/7, A038286/8, A038299/9 (with value for s after the slash).
If f(x) = A000045(x) (Fibonacci) and s = 1, the result is A010048 (Fibonomial).
In the notation of Carlitz and Scoville, this is the triangle of generalized Eulerian numbers A(r, s | alpha, beta) with alpha = beta = 2. Also the array A(2,1,4) in the notation of Hwang et al. (see page 31). - Peter Bala, Dec 27 2019

			Array, t(n, k), begins as:
   1,    2,      4,        8,        16,         32,          64, ...;
   2,   12,     52,      196,       684,       2276,        7340, ...;
   4,   52,    416,     2644,     14680,      74652,      357328, ...;
   8,  196,   2644,    26440,    220280,    1623964,    10978444, ...;
  16,  684,  14680,   220280,   2643360,   27227908,   251195000, ...;
  32, 2276,  74652,  1623964,  27227908,  381190712,  4677894984, ...;
  64, 7340, 357328, 10978444, 251195000, 4677894984, 74846319744, ...;
Triangle, T(n, k), begins as:
    1;
    2,     2;
    4,    12,      4;
    8,    52,     52,       8;
   16,   196,    416,     196,      16;
   32,   684,   2644,    2644,     684,      32;
   64,  2276,  14680,   26440,   14680,    2276,     64;
  128,  7340,  74652,  220280,  220280,   74652,   7340,   128;
  256, 23172, 357328, 1623964, 2643360, 1623964, 357328, 23172,   256;

Michael De Vlieger, Table of n, a(n) for n = 0..11475 (rows 0 <= n <= 150, flattened.)
L. Carlitz and R. Scoville, Generalized Eulerian numbers: combinatorial applications, J. für die reine und angewandte Mathematik, 265 (1974): 110-37. See Section 3.
Dale Gerdemann, A256890, Plot of t(m,n) mod k , YouTube, 2015.
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.

Cf. A000079, A001715, A008292, A038208, A257180, A257606, A257607, A257609.

Cf. A257610, A257612, A257614, A257616, A257617, A257618, A257619.

A256890:= func< n,k | (&+[(-1)^(k-j)*Binomial(j+3,j)*Binomial(n+4,k-j)*(j+2)^n: j in [0..k]]) >;
[A256890(n,k): k in [0..n], n in [0..10]]; // G. C. Greubel, Oct 18 2022

Table[Sum[(-1)^(k-j)*Binomial[j+3, j] Binomial[n+4, k-j] (j+2)^n, {j,0,k}], {n,0, 9}, {k,0,n}]//Flatten (* Michael De Vlieger, Dec 27 2019 *)

t(n,m) = if ((n<0) || (m<0), 0, if ((n==0) && (m==0), 1, (m+2)*t(n-1, m) + (n+2)*t(n, m-1)));
tabl(nn) = {for (n=0, nn, for (k=0, n, print1(t(n-k, k), ", ");); print(););} \\ Michel Marcus, Apr 14 2015

def A256890(n,k): return sum((-1)^(k-j)*Binomial(j+3,j)*Binomial(n+4,k-j)*(j+2)^n for j in range(k+1))
flatten([[A256890(n,k) for k in range(n+1)] for n in range(11)]) # G. C. Greubel, Oct 18 2022

T(n,k) = t(n-k, k); t(0,0) = 1, t(n,m) = 0 if n < 0 or m < 0 else t(n,m) = f(m)*t(n-1,m) + f(n)*t(n,m-1), where f(x) = x + 2.
Sum_{k=0..n} T(n, k) = A001715(n).
T(n,k) = Sum_{j = 0..k} (-1)^(k-j)*binomial(j+3,j)*binomial(n+4,k-j)*(j+2)^n. - Peter Bala, Dec 27 2019
Modified rule of Pascal: T(0,0) = 1, T(n,k) = 0 if k < 0 or k > n else T(n,k) = f(n-k) * T(n-1,k-1) + f(k) * T(n-1,k), where f(x) = x + 2. - Georg Fischer, Nov 11 2021
From G. C. Greubel, Oct 18 2022: (Start)
T(n, n-k) = T(n, k).
T(n, 0) = A000079(n). (End)

A001804 a(n) = n! * C(n,2).

Original entry on oeis.org

2, 18, 144, 1200, 10800, 105840, 1128960, 13063680, 163296000, 2195424000, 31614105600, 485707622400, 7933224499200, 137305808640000, 2510734786560000, 48373490221056000, 979563176976384000, 20801312169910272000, 462251381553561600000

Offset: 2

N. J. A. Sloane

Number of big descents in all permutations of [n+1]. A big descent in a permutation (x_1,x_2,...,x_n) is a position i such that x_i - x_(i+1) >= 2. Example: a(2)=2 because there are 2 big descents in the permutations 123, 132, 213, 23\1, 3\12, 321 of {1,2,3} (shown by a \). a(n)=Sum(k*A120434(n+1,k),k=0..n-1). - Emeric Deutsch, Oct 01 2006
a(n)/2 counts the total number of inversions in all the permutations of the set [n]; see A001809. - Peter Bala, Feb 28 2013
Equivalently, number of mappings f from a set X of n elements into itself such that f(X) has n-1 elements. - Robert FERREOL, Mar 14 2016

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 799.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 2..100
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets
Index entries for sequences related to factorial numbers

Cf. A001809, A120434.

seq(n!*binomial(n,2),n=2..20); # Emeric Deutsch, Oct 01 2006
a:=n->sum((n-j)*n!, j=1..n): seq(a(n), n=2..22); # Zerinvary Lajos, Apr 29 2007
restart: G(x):=x^2/(1-x)^3: f[0]:=G(x): for n from 1 to 18 do f[n]:=diff(f[n-1],x) od: x:=0: seq(f[n],n=2..16); # Zerinvary Lajos, Apr 01 2009

Table[n! Binomial[n, 2], {n, 2, 20}] (* T. D. Noe, Aug 10 2012 *)

a(n) = n!*binomial(n, 2); \\ Michel Marcus, Mar 14 2016

E.g.f.: x^2/(1-x)^3. - Geoffrey Critzer, Aug 19 2012
a(n) = 2 * A001809(n).
From Ilya Gutkovskiy, Jan 20 2017: (Start)
a(n) ~ sqrt(Pi/2)*n^(n+5/2)/exp(n).
Sum_{n>=2} 1/a(n) = 2*(3 - exp(1)) = 0.563436343081909529... (End)

A123513 Triangle read by rows: T(n,k) is the number of permutations of [n] having k small descents (n >= 1; 0 <= k <= n-1). A small descent in a permutation (x_1,x_2,...,x_n) is a position i such that x_i - x_(i+1) = 1.

Original entry on oeis.org

1, 1, 1, 3, 2, 1, 11, 9, 3, 1, 53, 44, 18, 4, 1, 309, 265, 110, 30, 5, 1, 2119, 1854, 795, 220, 45, 6, 1, 16687, 14833, 6489, 1855, 385, 63, 7, 1, 148329, 133496, 59332, 17304, 3710, 616, 84, 8, 1, 1468457, 1334961, 600732, 177996, 38934, 6678, 924, 108, 9, 1

Offset: 1

Emeric Deutsch, Oct 02 2006

This triangle is essentially A010027 (ascending pairs in permutations of [n]) with a different offset. The same triangle gives the number of permutations of [n] having k unit ascents (n >= 1; 0 <= k <= n-1). For permutations sorted by number of non-unitary (i.e., > 1) descents (also called "big" descents), see A120434. For permutations sorted by number of unitary moves (i.e., ascent or descent), see A001100. - Olivier Gérard, Oct 09 2007

With offsets n=0 (k=0) this is a binomial convolution triangle, a Sheffer triangle of the Appell type: ((exp(-x))/(1-x)^2),x). See the e.g.f. given below.

			Triangle starts:
     1;
     1,    1;
     3,    2,   1;
    11,    9,   3,   1;
    53,   44,  18,   4,  1;
   309,  265, 110,  30,  5, 1;
  2119, 1854, 795, 220, 45, 6, 1;
  ...
T(4,2)=3 because we have 14/3/2, 2/14/3 and 3/2/14 (the unit descents are shown by a /).
T(4,2)=3 because we have 14/3/2, 2/14/3 and 3/2/14 (the small descents are shown by a /).

Ch. A. Charalambides, Enumerative Combinatorics, Chapman & Hall/CRC, Boca Raton, Florida, 2002, p. 179, Table 5.4 for S_{n,k} (without row n=1 and column k=0).
F. N. David, M. G. Kendall and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 263 (Table 7.5.1).

Alois P. Heinz, Rows n = 1..150, flattened
Bhadrachalam Chitturi and Krishnaveni K S, Adjacencies in Permutations, arXiv preprint arXiv:1601.04469 [cs.DM], 2016. See Table 0.
FindStat - Combinatorial Statistic Finder, The number of adjacencies of a permutation
Sergey Kitaev and Philip B. Zhang, Distributions of mesh patterns of short lengths, arXiv:1811.07679 [math.CO], 2018.
J. Liese and J. Remmel, Q-analogues of the number of permutations with k-excedances, PU. M. A. Vol. 21 (2010), No. 2, pp. 285-320 (see E_{n,1}(x) in Table 1 p. 291).
F. Poussin, Énumération des permutations par nombre de marches, RAIRO, Informatique théorique, 13 no. 3, 1979, p. 251-255.

Cf. A000166, A000255, A000274, A000313, A001260.

Cf. A010027 (mirror image), A120434, A001100.

G:=exp(-x+t*x)/(1-x)^2: Gser:=simplify(series(G,x=0,15)): for n from 0 to 10 do P[n+1]:=sort(n!*coeff(Gser,x,n)) od: for n from 1 to 11 do seq(coeff(P[n],t,k),k=0..n-1) od; # yields sequence in triangular form

Needs["Combinatorica`"];
Table[Map[Count[#,1]&,Map[Differences,Permutations[n]]]//Distribution,{n,1,10}]//Grid
(* Geoffrey Critzer, Dec 15 2012 *)
T[n_, k_] := (n-1)! SeriesCoefficient[Exp[-x + t x]/(1-x)^2, {x, 0, n-1}, {t, 0, k}];
Table[T[n, k], {n, 1, 10}, {k, 0, n-1}] // Flatten (* Jean-François Alcover, Sep 25 2019 *)
T[1,1]:=1;T[0,1]:=0;T[n_,1]:=T[n,1]=(n-1)T[n-1,1]+(n-2)T[n-2,1];T[n_,k_]:=T[n,k]=T[n-1, k-1](n-1)/(k-1);Flatten@Table[T[n,k],{n,1,10},{k,1,n}] (* Oliver Seipel, Dec 01 2024 *)

T(n,1) = A000255(n-1).
T(n,2) = A000166(n-1) (the derangement numbers).
T(n,3) = A000274(n).
T(n,4) = A000313(n).
T(n,5) = A001260(n);
G.f.: exp(-x+tx)/(1-x)^2 (if offset is 0), i.e., T(n,k)=(n-1)!*[x^(n-1) t^k]exp(-x+tx)/(1-x)^2.
T(n,k) = binomial(n-1,k)*A000255(n-1), n-1 >= k >= 0, else 0.

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

A136717 Triangle T(n,k), 1 <= k <= n, read by rows: T(n,k) is the number of permutations in the symmetric group S_n having k multiplicative 3-excedances. Equivalently, the number of permutations of the set {3,6,9,...,3n} with k excedances.

Original entry on oeis.org

1, 0, 2, 0, 2, 4, 0, 0, 12, 12, 0, 0, 0, 72, 48, 0, 0, 0, 72, 456, 192, 0, 0, 0, 0, 960, 3120, 960, 0, 0, 0, 0, 0, 10800, 23760, 5760, 0, 0, 0, 0, 0, 10800, 133920, 183600, 34560, 0, 0, 0, 0, 0, 0, 241920, 1572480, 1572480, 241920

Offset: 1

Peter Bala, Jan 23 2008

A permutation (p(1),p(2),...,p(n)) in the symmetric group S_n has a multiplicative 3-excedance at position i, 1 <= i <= n, if 3*p(i) > i. The (n,k)-th entry in this array gives the number of permutations in S_n with k multiplicative 3-excedances.
Compare with A008292, the triangle of Eulerian numbers, which enumerates permutations by the usual excedance number and with A136715, which enumerates permutations by multiplicative 2-excedances.
Let e(p)= |{i | 1 <= i < = n, 3*p(i) > i}| denote the number of multiplicative 3-excedances in the permutation p. This 3-excedance statistic e(p) on the symmetric group S_n is related to a descent statistic as follows. Define a permutation p in S_n to have a multiplicative 3-descent at position i, 1 <= i <= n-1, if p(i) is divisible by 3 and p(i) > p(i+1). For example, the permutation (4,1,6,5,3,2) in S_6 has two multiplicative 3-descents (at position 3 and position 5). Array A136718 records the number of permutations of S_n with k multiplicative 3-descents.
Let d(p) = |{i | 1 <= i <= n-1, p(i) is divisible by 3 & p(i) > p(i+1)}| count the multiplicative 3-descents in the permutation p. Comparison of the recursion relations for the entries of this table with the recursion relations for the entries of A136718 shows that e(p) and d(p) are related by sum {p in S_n} x^e(p) = x^ceiling(2*n/3)* sum {p in S_n} x^d(p). Thus the shifted multiplicative 3-excedance statistic e(p) - ceiling(2*n/3) and the multiplicative 3-descent statistic d(p) are equidistributed on the symmetric group S_n.
(Note: There is also an additive r-excedance statistic on the symmetric group, due to Riordan, where the condition r*p(i) > i is replaced by p(i) >= i+r. See A120434 for the r = 2 case.)
An alternative interpretation of this array is as follows: Let T_n denote the set {3,6,9,...,3n} and let now p denote a bijection p:T_n -> T_n. We say the permutation p has an excedance at position i, 1 <= i <= n, if p(3i) > i. For example, if we represent p in one line notation by the vector (p(3),p(6),...,p(3n)), then the permutation (9,18,3,12,15,6) of T_6 has four excedances in total (at positions 1, 2, 4 and 5). This array gives the number of permutations of the set T_n with k excedances. This is the viewpoint taken in [Jansson].
A137593 = A000012 * this triangular matrix. A137594 = A007318 * this triangular matrix. - Gary W. Adamson, Jan 28 2008

			T(3,3) = 4; the four permutations in S_3 with three multiplicative 3-excedances are (1,2,3), (1,3,2), (2,1,3) and (3,1,2). The remaining two permutations (2,3,1) and (3,2,1) each have two multiplicative 3-excedances.
Equivalently, the four permutations of the set {3,6,9} with 3 excedances are (3,6,9), (3,9,6), (6,3,9) and (9,3,6). The remaining two permutations (6,9,3) and (9,6,3) each have 2 excedances.
Triangle starts
n\k|..1....2....3....4....5....6
---+----------------------------
1..|..1
2..|..0....2
3..|..0....2....4
4..|..0....0...12...12
5..|..0....0....0...72...48
6..|..0....0....0...72..456..192

Fredrik Jansson, Variations on the excedance statistic in permutations, Thesis, Stockholm University, 2006.
S. Kitaev and J. Remmel, Classifying descents according to equivalence mod k, arXiv:math/0604455 [math.CO], 2006.

Cf. A000142 (row sums), A008292, A136718, A136715.

Cf. A137593, A137594.

Recurrence relations (apply proposition 2.2 of [Jansson]):
T(3n,k) = (k+1-2n)*T(3n-1,k) + (5n-k)*T(3n-1,k-1) for n >= 1;
T(3n+1,k) = (k-2n)*T(3n,k) + (5n+2-k)*T(3n,k-1) for n >= 0;
T(3n+2,k) = (k-1-2n)*T(3n+1,k) + (5n+4-k)*T(3n+1,k-1) for n >= 0.
Boundary conditions: T(0,k) = 0 all k; T(n,0) = 0 all n; T(1,1) = 1.
Define the shifted row polynomials R(n,x) by
R(n,x) := x^(1+floor(n/3)-n)* sum {k = n-floor(n/3)..n} T(n,k)*x^k.
The first few values are R(1,x) = x, R(2,x) = 2x, R(3,x) = 2x+4x^2 and R(4,x) = 12x+12x^2.
The recurrence relations yield the identities:
x*d/dx(1/x*R(3n,x)/(1-x)^(3n+1)) = R(3n+1,x)/(1-x)^(3n+2);
x*d/dx(1/x*R(3n+1,x)/(1-x)^(3n+2)) = R(3n+2,x)/(1-x)^(3n+3);
x*d/dx(R(3n+2,x)/(1-x)^(3n+3)) = R(3n+3,x)/(1-x)^(3n+4).
An easy induction argument now gives the Taylor series expansions:
R(3n,x)/(1-x)^(3n+1) = sum {m = 1..inf} m^2*(m+1)*(m+2)^2*(m+3)*...* (m+2n-2)^2*(m+2n-1)*x^m;
R(3n+1,x)/(1-x)^(3n+2) = sum {m = 1..inf} m*((m+1)^2*(m+2)*(m+3)^2*(m+4) *...*(m+2n-1)^2*(m+2n))*x^m.
R(3n+2,x)/(1-x)^(3n+3) = sum {m = 1..inf} m*((m+1)*(m+2)^2*(m+3)*(m+4)^2 *...*(m+2n-1)*(m+2n)^2)*(m+2n+1)*x^m.
For example, for row 6 (n = 2) we have the expansion (72x+456x^2+192x^3)/(1-x)^7 = 72x + 960x^2 + 5400x^3 + ... = (1^2*2*3^2*4)*x + (2^2*3*4^2*5)*x^2 + (3^2*4*5^2*6)*x^3 + ... .

A136715 Triangle T(n,k), 1 <= k <= n, read by rows: T(n,k) is the number of permutations of the set {2,4,6,...,2n} with k excedances. Equivalently, T(n,k) is the number of permutations in the symmetric group S_n having k multiplicative 2-excedances.

Original entry on oeis.org

1, 1, 1, 0, 4, 2, 0, 4, 16, 4, 0, 0, 36, 72, 12, 0, 0, 36, 324, 324, 36, 0, 0, 0, 576, 2592, 1728, 144, 0, 0, 0, 576, 9216, 20736, 9216, 576, 0, 0, 0, 0, 14400, 115200, 172800, 57600, 2880, 0, 0, 0, 0, 14400, 360000, 1440000, 1440000, 360000, 14400

Offset: 1

Peter Bala, Jan 18 2008

Let E_n denote the set {2,4,6,...,2n} and let p denote a bijection p:E_n -> E_n. We say the permutation p has an excedance at position i, 1 <= i <= n, if p(2i) > i. For example, if we represent p in one line notation by the vector (p(2),p(4),...,p(2n)), then the permutation (6,2,8,4,10) of E_5 has three excedances in total (at positions 1, 3 and 5). This array gives the number of permutations of the set E_n with k excedances. This is the viewpoint taken in [Jansson]. See A136716 for the corresponding array when the set E_n is replaced by the set O_n := {1,3,5,...,2n-1}.
Alternatively, we can work with permutations (p(1),p(2),...,p(n)) in the symmetric group S_n and define p to have a multiplicative 2-excedance at position i, 1 <= i <= n, if 2*p(i) > i. Then the (n,k)-th entry of this array gives the number of permutations in S_n with k multiplicative 2-excedances. Compare with A008292, the triangle of Eulerian numbers, which enumerates permutations by the usual excedance number and A136717 which enumerates permutations by multiplicative 3-excedances.
Let e(p)= |{i | 1 <= i < = n, 2*p(i) > i}| denote the number of multiplicative 2-excedances in the permutation p of S_n. This 2-excedance statistic e(p) on the symmetric group S_n is related to a descent statistic as follows.
Define a permutation p in S_n to have a descent from even at position i, 1 <= i <= n-1, if p(i) is even and p(i) > p(i+1). For example, the permutation (2,1,3,5,6,4) in S_6 has two descents from even (at position 1 and position 5). Array A134434 records the number of permutations of S_n with k descents from even.
Let d(p) = |{i | 1 <= i <= n-1, p(i) is even & p(i) > p(i+1)}| count the descents from even in the permutation p. Comparison of the formulas for the entries of this table with the formulas for the entries of A134434 shows that e(p) and d(p) are related by sum {p in S_n} x^e(p) = x^ceiling(n/2)* sum {p in S_n} x^d(p). Thus the shifted multiplicative 2-excedance statistic e(p) - ceiling(n/2) and the descent statistic d(p) are equidistributed on the symmetric group S_n.

			T(4,2) = 4; the four permutations in S_4 with two multiplicative 2-excedances are (3,4,1,2), (4,3,1,2), (3,1,4,2) and (4,1,3,2). Alternatively, the four permutations (6,8,2,4), (8,6,2,4), (6,2,8,4) and (8,2,6,4) of the set E_4 each have 2 excedances.
Triangle starts
n\k|..1....2....3....4....5....6
--------------------------------
1..|..1
2..|..1....1
3..|..0....4....2
4..|..0....4...16....4
5..|..0....0...36...72...12
6..|..0....0...36..324..324...36

Fredrik Jansson, Variations on the excedance statistic in permutations

Cf. A000142 (row sums), A008292, A008459, A103371, A120434, A134434, A136716, A136717.

Recurrence relations:
T(2n,k) = (k+1-n)*T(2n-1,k) + (3n-k)*T(2n-1,k-1) for n >= 1;
T(2n+1,k) = (k-n)*T(2n,k) + (3n+2-k)*T(2n,k-1) for n >= 0. Boundary conditions: T(0,k) = 0 all k; T(n,0) = 0 all n; T(1,1) = 1.
The recurrence relations have the explicit solution T(2n,n+k) = [n!* C(n,k)]^2 and T(2n+1,n+k+1) = 1/(k+1)*[(n+1)!*C(n,k)]^2 = n!*(n+1)!*C(n,k)*C(n+1,k+1); or as a single formula, T(n,k) = floor(n/2)! * floor((n+1)/2)! * C(floor(n/2),k-floor((n+1)/2)) * C(floor((n+1)/2),k-floor(n/2)). Also T(2n,n+k) = n!^2 * A008459 (n,k); T(2n+1,n+k+1) = n!*(n+1)!* A103371 (n,k).
For the even numbered rows, define the shifted row polynomials F(2n,x) := x^(1-n)* sum {k = n..2n} T(2n,k)*x^k = n!^2 * x * (1 + C(n,1)^2*x + C(n,2)^2*x^2 + ... + C(n,n)^2*x^n). For the odd numbered rows, define the shifted row polynomials F(2n+1,x) := x^(-n)* sum {k = n+1..2n+1} T(2n+1,k)*x^k = n!*(n+1)!* ((n+1)*N(n+1,1)*x + n*N(n+1,2)*x^2 +(n-1)* N(n+1,3)*x^3 + ... + N(n+1,n+1)*x^(n+1)), where N(n,k) denotes the Narayana numbers. The first few values are F(1,x) = x, F(2,x) =x+x^2, F(3,x) = 4x+2x^2 and F(4,x) = 4x+16x^2+4x^2.
The recurrence relations yield the identities x*d/dx(F(2n-1,x)/(1-x)^(2n)) = F(2n,x)/(1-x)^(2n+1) and x*d/dx(1/x*F(2n,x)/(1-x)^(2n+1)) = F(2n+1,x)/(1-x)^(2n+2), for n = 1,2,3,... . An easy induction argument now gives the Taylor series expansions: F(2n,x)/(1-x)^(2n+1) = sum {m = 1..inf} (m*(m+1)*...*(m+n-1))^2*x^m; F(2n+1,x)/(1-x)^(2n+2) = sum {m = 1..inf} m*((m+1)*(m+2)*...*(m+n))^2*x^m.
For example, when n = 3 we have for row 6 the expansion (36x + 324x^2 + 324x^3 + 36x^4)/(1-x)^7 = 36x + 576x^2 + 3600x^3 + ... = (1.2.3)^2*x + (2.3.4)^2*x^2 + (3.4.5)^2*x^3 + ... and for row 7 the expansion (576x + 2592x^2 + 1728x^3 + 144x^4)/(1-x)^8 = 576x + 7200x^2 + 43200x^3 + ... = 1*(2.3.4)^2*x + 2*(3.4.5)^2*x^2 + 3*(4.5.6)^2*x^3 + ... .
Relation with the Jacobi polynomials P_n(a,b,x): F(2n,x) = n!^2*x*(1-x)^n *P_n(0,0,(1+x)/(1-x)), F(2n+1,x) = n!*(n+1)!*x*(1-x)^n *P_n(1,0,(1+x)/(1-x)).
Worpitzky-type identities:
Sum {k = n..2n} T(2n,k)*C(x+k,2n) = ((x+1)*(x+2)*...*(x+n))^2;
sum {k = n+1..2n+1} T(2n+1,k)*C(x+k,2n+1) = ((x+1)*(x+2)*...*(x+n))^2*(x+n+1);
and for the odd numbered rows read in reverse order, sum {k = n+1..2n+1} T(2n+1,3n+2-k)*C(x+k,2n+1) = (x+1)*((x+2)*(x+3)*...*(x+n+1))^2.

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A256890 Triangle T(n,k) = t(n-k, k); t(n,m) = f(m)*t(n-1,m) + f(n)*t(n,m-1), where f(x) = x + 2.

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A001804 a(n) = n! * C(n,2).

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A123513 Triangle read by rows: T(n,k) is the number of permutations of [n] having k small descents (n >= 1; 0 <= k <= n-1). A small descent in a permutation (x_1,x_2,...,x_n) is a position i such that x_i - x_(i+1) = 1.

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A144696 Triangle of 2-Eulerian numbers.

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A136717 Triangle T(n,k), 1 <= k <= n, read by rows: T(n,k) is the number of permutations in the symmetric group S_n having k multiplicative 3-excedances. Equivalently, the number of permutations of the set {3,6,9,...,3n} with k excedances.

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A136715 Triangle T(n,k), 1 <= k <= n, read by rows: T(n,k) is the number of permutations of the set {2,4,6,...,2n} with k excedances. Equivalently, T(n,k) is the number of permutations in the symmetric group S_n having k multiplicative 2-excedances.

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A199335 Triangle T(n,k), read by rows, given by (0,1,0,2,0,3,0,4,0,5,0,6,0,7,0,8,0,9,...) DELTA (2,0,3,0,4,0,5,0,6,0,7,0,8,0,9,...), where DELTA is the operator defined in A084938.

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A256890 Triangle T(n,k) = t(n-k, k); t(n,m) = f(m)t(n-1,m) + f(n)t(n,m-1), where f(x) = x + 2.