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A144698 Triangle of 4-Eulerian numbers. 5
1, 1, 4, 1, 13, 16, 1, 32, 113, 64, 1, 71, 531, 821, 256, 1, 150, 2090, 6470, 5385, 1024, 1, 309, 7470, 40510, 65745, 33069, 4096, 1, 628, 25191, 221800, 612295, 592884, 194017, 16384, 1, 1267, 81853, 1113919, 4835875, 7843369, 4915423, 1101157 (list; table; graph; refs; listen; history; text; internal format)
OFFSET

4,3

COMMENTS

This is the case r = 4 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 4-Eulerian numbers are the number of permutations in Permute(n,n-4) with k excedances. For other cases see A008292 (r = 0 and r = 1), A144696 (r = 2), A144697 (r = 3) and A144699 (r = 5).

An alternative interpretation of the current array due to [Strosser] involves the 4-excedance statistic of a permutation (see also [Foata & Schutzenberger, Chapter 4, Section 3]). We define a permutation p in Permute(n,n-4) to have a 4-excedance at position i (1 <= i <= n-4) if p(i) >= i + 4.

Given a permutation p in Permute(n,n-4), define ~p to be the permutation in Permute(n,n-4) that takes i to n+1 - p(n-i-3). The map ~ is a bijection of Permute(n,n-4) with the property that if p has (resp. does not have) an excedance in position i then ~p does not have (resp. has) a 4-excedance at position n-i-3. Hence ~ gives a bijection between the set of permutations with k excedances and the set of permutations with (n-k) 4-excedances. Thus reading the rows of this array in reverse order gives a triangle whose entries are the number of permutations in Permute(n,n-4) with k 4-excedances.

Example: Represent a permutation p:[n-4] -> [n] in Permute(n,n-4) by its image vector (p(1),...,p(n-4)). In Permute(10,6) the permutation p = (1,2,4,10,3,6) does not have an excedance in the first two positions (i = 1 and 2) or in the final two positions (i = 5 and 6). The permutation ~p = (5,8,1,7,9,10) has 4-excedances only in the first two positions and the final two positions.

REFERENCES

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

LINKS

Table of n, a(n) for n=4..47.

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.

D. Foata, M. Schutzenberger, Théorie Géometrique des Polynômes Eulériens, arXiv:math/0508232 [math.CO], 2005; Lecture Notes in Math., no. 138, Springer Verlag, 1970.

L. Liu, 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

FORMULA

T(n,k) = 1/4!*Sum_{j = 0..k} (-1)^(k-j)*binomial(n+1,k-j)*(j+1)^(n-3)*(j+2)*(j+3)*(j+4).

T(n,n-k) = 1/4!*Sum_{j = 4..k} (-1)^(k-j)*binomial(n+1,k-j)*j^(n-3)*(j-1)*(j-2)*(j-3).

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 >= 4, T(4,k) = 0 for k >= 1. Special cases: T(n,n-4) = 4^(n-4); T(n,n-5) = 5^(n-3) - 4^(n-3) - (n-3)*4^(n-4).

E.g.f. (with suitable offsets): 1/4*[(1 - x)/(1 - x*exp(t - t*x))]^4 = 1/4 + x*t + (x + 4*x^2)*t^2/2! + (x + 13*x^2 + 16*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_4(x) = 1. It follows that the polynomials R_n(x) for n >= 5 have only real zeros (apply Corollary 1.2. of [Liu and Wang]).

The (n+3)-th row generating polynomial = 1/4!*Sum_{k = 1..n} (k+3)!*Stirling2(n,k)*x^(k-1)*(1-x)^(n-k).

For n >= 4,

1/4*(x*d/dx)^(n-3) (1/(1-x)^4) = x/(1-x)^(n+1) * Sum_{k = 0..n-4} T(n,k)*x^k,

1/4*(x*d/dx)^(n-3) (x^4/(1-x)^4) = 1/(1-x)^(n+1) * Sum_{k = 4..n} T(n,n-k)*x^k,

1/(1-x)^(n+1) * Sum {k = 0..n-4} T(n,k)*x^k = 1/4! * Sum_{m = 0..inf} (m+1)^(n-3)*(m+2)*(m+3)*(m+4)*x^m,

1/(1-x)^(n+1) * Sum {k = 4..n} T(n,n-k)*x^k = 1/4! * Sum_{m = 4..inf} m^(n-3)*(m-1)*(m-2)*(m-3)*x^m,

Worpitzky-type identities:

Sum_{k = 0..n-4} T(n,k)*binomial(x+k,n) = 1/4!*x^(n-3)*(x-1)*(x-2)*(x-3).

Sum_{k = 4..n} T(n,n-k)* binomial(x+k,n) = 1/4!*(x+1)^(n-3)*(x+2)*(x+3)*(x+4).

Relation with Stirling numbers (Frobenius-type identities):

T(n+3,k-1) = 1/4! * Sum_{j = 0..k} (-1)^(k-j)* (j+3)!* binomial(n-j,k-j)*Stirling2(n,j) for n,k >= 1;

T(n+3,k-1) = 1/4! * Sum_{j = 0..n-k} (-1)^(n-k-j)*(j+3)!* binomial(n-j,k)*S(4;n+4,j+4) for n,k >= 1 and

T(n+4,k) = 1/4! * Sum_{j = 0..n-k} (-1)^(n-k-j)*(j+4)!* binomial(n-j,k)*S(4;n+4,j+4) for n,k >= 0, where S(4;n,k) denotes the 4-Stirling numbers of the second kind A143496(n,k).

For n >=4, the shifted row polynomial t*R(n,t) = 1/4*D^(n-3)(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))^(-4). - Peter Bala, Apr 22 2012

EXAMPLE

Triangle begins

  ===+=============================================

  n\k|  0      1      2      3      4      5      6

  ===+=============================================

   4 |  1

   5 |  1      4

   6 |  1     13     16

   7 |  1     32    113     64

   8 |  1     71    531    821    256

   9 |  1    150   2090   6470   5385   1024

  10 |  1    309   7470  40510  65745  33069   4096

  ...

T(6,1) = 13: We represent a permutation p:[n-4] -> [n] in Permute(n,n-4) by its image vector (p(1),...,p(n-4)). The 13 permutations in Permute(6,2) having 1 excedance are (1,3), (1,4), (1,5), (1,6), (3,2), (4,2), (5,2), (6,2), (2,1), (3,1), (4,1), (5,1) and (6,1).

MAPLE

with(combinat):

T:= (n, k) -> 1/4!*add((-1)^(k-j)*binomial(n+1, k-j)*(j+1)^(n-3)*(j+2)*(j+3)*(j+4), j = 0..k):

for n from 4 to 12 do

seq(T(n, k), k = 0..n-4)

end do;

CROSSREFS

Cf. A001720 (row sums), A008292, A143493, A143496, A143499, A144696, A144697, A144699.

Sequence in context: A193843 A116414 A215502 * A115154 A292270 A051928

Adjacent sequences:  A144695 A144696 A144697 * A144699 A144700 A144701

KEYWORD

easy,nonn,tabl

AUTHOR

Peter Bala, Sep 19 2008

STATUS

approved

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Last modified September 19 01:08 EDT 2018. Contains 315153 sequences. (Running on oeis4.)