Search: id:A144696 Results 1-1 of 1 results found. %I A144696 %S A144696 1,1,2,1,7,4,1,18,33,8,1,41,171,131,16,1,88,718,1208,473,32,1,183,2682, %T A144696 8422,7197,1611,64,1,374,9327,49780,78095,38454,5281,128,1,757,30973, %U A144696 264409,689155,621199,190783,16867,256 %N A144696 Triangle of 2-restricted Eulerian numbers. %C A144696 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 number of 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 generalise this definition to restricted permutations as follows. %C A144696 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-restricted 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-restricted Eulerian numbers gives the number of permutations in Permute(n,n-2) with k excedances. See the example section below for some numerical examples. %C A144696 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-restricted Eulerian numbers are equal to the 0-restricted Eulerian numbers: A(1;n,k) = A(0;n,k) = A(n,k). %C A144696 For other cases of restricted Eulerian numbers see A144697 (r = 3), A144698 (r = 4) and A144699 (r = 5). There is also a concept of r-restricted 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. %C A144696 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. %C A144696 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. %C A144696 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. %D A144696 J. Riordan. - An introduction to combinatorial analysis. - New York, J. Wiley, 1958. %D A144696 R. Strosser. - Seminaire de theorie combinatoire, I.R.M.A., Universite de Strasbourg, 1969-1970. %H A144696 Mark Conger. - A refinement of the Eulerian polynomials and the joint distribution of pi(1) and Des(pi) in S_n. %H A144696 D. Foata, M. Schutzenberger. - Theorie Geometrique des Polynomes Euleriens, Lecture Notes in Math., no.138, Springer Verlag 1970. %H A144696 L. Liu, Y. Wang, - A unified approach to polynomial sequences with only real zeros %F A144696 T(n,k) = 1/2!*sum {j = 0..k} (-1)^(k-j)*binomial(n+1,k-j)*(j+1)^(n-1)*(j+2); %F A144696 T(n,n-k) = 1/2!*sum {j = 2..k} (-1)^(k-j)*binomial(n+1,k-j)*j^(n-1)*(j-1). %F A144696 Recurrence relations: %F A144696 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). %F A144696 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). %F A144696 For n >= 2, %F A144696 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, %F A144696 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, %F A144696 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, %F A144696 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. %F A144696 Worpitzky-type identities: %F A144696 sum {k = 0..n-2} T(n,k)*binomial(x+k,n) = 1/2!*x^(n-1)*(x - 1); %F A144696 sum {k = 2..n} T(n,n-k)*binomial(x+k,n) = 1/2!*(x + 1)^(n-1)*(x + 2). %F A144696 Relation with Stirling numbers (Frobenius-type identities): %F A144696 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; %F A144696 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 %F A144696 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-restricted Stirling numbers A143494(n,k). %F A144696 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. %F A144696 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). %e A144696 The triangle begins %e A144696 =========================================== %e A144696 n\k|..0.....1.....2.....3.....4.....5.....6 %e A144696 =========================================== %e A144696 2..|..1 %e A144696 3..|..1.....2 %e A144696 4..|..1.....7.....4 %e A144696 5..|..1....18....33.....8 %e A144696 6..|..1....41...171...131....16 %e A144696 7..|..1....88...718..1208...473....32 %e A144696 8..|..1...183..2682..8422..7197..1611....64 %e A144696 ... %e A144696 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. %p A144696 with(combinat): %p A144696 T:= (n,k) -> 1/2!*add((-1)^(k-j)*binomial(n+1,k-j)*(j+1)^(n-1)*(j+2), j = 0..k): %p A144696 for n from 2 to 10 do %p A144696 seq(T(n,k),k = 0..n-2) %p A144696 end do; %Y A144696 A000182 (related to alt. row sums), A008292, A001710 (row sums), A120434, A143491, A143494, A143497, A144697, A144698, A144699. %Y A144696 Sequence in context: A120903 A021050 A115629 this_sequence A072248 A092276 A011274 %Y A144696 Adjacent sequences: A144693 A144694 A144695 this_sequence A144697 A144698 A144699 %K A144696 easy,nonn,tabl %O A144696 2,3 %A A144696 Peter Bala (pbala(AT)toucansurf.com), Sep 19 2008 Search completed in 0.002 seconds