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A144697 Triangle of 3-Eulerian numbers. 6
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 (list; table; graph; refs; listen; history; text; internal format)
OFFSET

3,3

COMMENTS

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.

REFERENCES

R. Strosser, Seminaire de theorie combinatoire, I.R.M.A., Universite de Strasbourg, 1969-1970.

LINKS

Table of n, a(n) for n=3..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.

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.

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) = 1/3!*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..inf} (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..inf} 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

EXAMPLE

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).

MAPLE

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;

CROSSREFS

Cf. A001715 (row sums), A008292, A060056, A143492, A143495, A144696, A144698, A144699.

Sequence in context: A260178 A195812 A264491 * A185419 A252501 A281000

Adjacent sequences:  A144694 A144695 A144696 * A144698 A144699 A144700

KEYWORD

easy,nonn,tabl

AUTHOR

Peter Bala, Sep 19 2008

STATUS

approved

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Last modified June 21 06:06 EDT 2018. Contains 305619 sequences. (Running on oeis4.)