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A144697
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Triangle of 3-Eulerian numbers.
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7
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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
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OFFSET
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3,3
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COMMENTS
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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).
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.
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REFERENCES
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R. Strosser, Séminaire de théorie combinatoire, I.R.M.A., Universite de Strasbourg, 1969-1970.
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LINKS
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FORMULA
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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
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EXAMPLE
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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).
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MAPLE
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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;
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MATHEMATICA
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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;
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PROG
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(Magma) 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
(SageMath)
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
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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