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Generalized Euler Sums

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Generalized Euler Sums Generalized Eulerian Sums Fan Chung∗ Ron Graham∗ Dedicated to Adriano Garsia on the occasion of his 84th birthday. Abstract In this note, we derive a number of symmetrical sums involving Eu- lerian numbers and some of their generalizations. These extend earlier identities of Don Knuth and the authors, and also include several q-nomial sums inspired by recent work of Shareshian and Wachs on the joint distri- bution of various permutation statistics, such as the number of excedances, the major index and the number of fixed points of a permutation. We also produce symmetrical sums involving \restricted" Eulerian numbers which count permutations π on f1; 2; : : : ; ng with a given number of descents and which, in addition, have the value of π−1(n) specified. 1 Introduction n The classical Eulerian numbers, which we will denote by k , occur in a variety of contexts in combinatorics, number theory and computer science (see [4]). These numbers were introduced by Euler [3] in 1736 and have many interesting properties. We list some small values in Table 1. k 0 1 2 3 4 5 1 1 0 0 0 0 0 2 1 1 0 0 0 0 n 3 1 4 1 0 0 0 4 1 11 11 1 0 0 5 1 26 66 26 1 0 6 1 57 302 302 57 1 n Some Eulerian numbers k Table 1 n In particular, k enumerates the number of permutations π on [n] := f1; 2; : : : ; ng which have k descents (i.e., i ≤ n−1 with π(i) > π(i+1) ) as well as the number ∗University of California, San Diego 1 of permutations π on [n] which have k excedances (i.e., i < n with π(i) > i). Eulerian numbers satisfy the reflection property n n = ; n > 0; (1) k n − k − 1 and the recurrence n n − 1 n − 1 = (k + 1) + (n − k) : k k k − 1 They also have the explicit representation k n X n + 1 = (−1)i (k + 1 − i)n k i i=0 and have the following generating function (see [4]): X n E (t) = ti; n > 0; (2) n i i (1 − t)ex X n xn X xn E(t; x) = = 1 + ti = 1 + E (t) : (3) etx − tex i n! n n! n>0;i≥0 n>0 Eulerian numbers grow rapidly in size with increasing n, and one doesn't expect as many combinatorial identities to exist for them compared with other sequences such as binomial coefficients or Stirling numbers. Nevertheless, it was shown in [1] that the following symmetrical identity holds: X a + b k X a + b k = (4) k a − 1 k b − 1 k≥0 k≥0 0 for a; b > 0 (where unless stated otherwise, we use the convention that 0 = 0). In Section 2, we prove several natural generalizations of this identity. In Section 3, we consider a variant of the usual Eulerian numbers in which we count the number of permutations π on [n] having a given number of descents with the additional constraint that π(n)−1 is specified. In Sections 4 and 5, we deal with q-coefficients arising in the joint distribution of several permutation statistics studied by Shareshian and Wachs [7, 8]. Finally, in Section 7, we close with a number of open problems. 2 An alternating Eulerian sum identity Theorem 1. X a + bk X a + bk (−1)k = (−1)k k a k b k≥0 k≥0 for a; b > 0. 2 Proof. The generating function for our \modified” Eulerian numbers (i.e., with 0 (1−t)ex 0 = 0) is obtained by subtracting 1 from etx−tex in (3). This gives ex − etx X n xn = ti : (5) etx − tex i n! n;i≥0 Thus, 1 t X n xn 1 t ex − etx − ti = − ex etx i n! ex etx etx − tex n;i 1 1 = − : etx ex Expressing the exponentials as sums implies 0 1 X xk X tkxk X ntixn (−1)k − t (−1)k @ k! k! A i n! k≥0 k≥0 n;i X tnxn X xn = (−1)n − (−1)n ; n! n! n≥0 n≥0 X ntixn+k X nti+k+1xn+k (−1)k − (−1)k i k!n! i k!n! k;n;i k;n;i X (tn − 1)xn = (−1)n : n! n≥0 Shifting the indices of summation yields X n − kntixn X n − k ntixn (−1)k − (−1)k i k n! i − k − 1 k n! k;n;i k;n;i X (tn − 1)xn = (−1)n : n! n≥0 (6) Identifying the coefficients of xn gives X n − kn X n − k n (−1)k ti − (−1)k ti = (−1)n(tn − 1): i k i − k − 1 k k;i k;i Assuming i 6= 0; n and identifying the coefficients of ti gives X nn − k X n n − k (−1)k − (−1)k = 0: k i k i − k − 1 k k 3 Finally, using the reflection properties of the Eulerian numbers in (1), replacing k by n − k, and using the reflection properties of the binomial coefficients, we obtain X nk X n k (−1)k = (−1)k ; k i k n − i k k for 0 < i < n: Therefore, if we set n = a + b; i = a; then we have X a + bk X a + bk (−1)k = (−1)k (7) k a k b k≥0 k≥0 for a; b > 0. This proves the theorem. Note that (7) can be considered as a companion to (4). We can rewrite (4) in the form X nn − k X nn − k = ; (8) k i − k k i − 1 k k for 0 < i < n. As pointed out in [1], this is the first in an infinite sequence of (increasingly complex) sums of this type. The next two are: X nn − k X nn − k 2k − 2k k i − k k i − 2 k k n n = − ; for 1 < i < n: (9) i i − 1 X nn − k X nn − k 3k − 3k k i − k k i − 3 k k n n = 2i + 2n−i+1 i i − 1 n n −2i−1 − 2n−i+2 i − 1 i − 2 for 2 < i < n: (10) The companion sum (7) is also the first of an infinite sequence of similar sums, 4 the next few being: X nk X n k (−1)k2n−k − (−1)k2n−k k i k n − i + 1 k k n n = − ; for 1 ≤ i ≤ n: (11) i i − 1 X nk X n k (−1)k3n−k − (−1)k3n−k k i k n − i + 2 k k n n = 2n−i − 2n−i+1 i i − 1 n n +2i−1 − 2i−2 i − 1 i − 2 for 1 ≤ i ≤ n: (12) The proofs of these follow the same lines as that of (7) and are omitted. 3 Generalized Eulerian identities for restricted descent polynomials In this section we consider a restricted version of Eulerian numbers. We define the restricted Eulerian number b(n; k; i), for 1 ≤ k ≤ n, 0 ≤ i < n, to be the number of π 2 Sn with des(π) = i and with π(k) = n. This restriction is similar to one used by the authors for investigating the joint statistics of permutations having a given number of descents and a given bound on drop size [2]. (A permutation π has a drop at i if π(i) < i and the drop size is i − π(i)). From the definition of b(n; k; i), it follows that X n b(n; k; i) = ; i k k − 1 b(k; k; i) = ; i b(n; k; i) = 0; if n < k or i ≥ n or i < 0: The quantities b(n; k; i) satisfy the following reflection property which will be needed later. Lemma 1. For 1 < k < n;, we have b(n; k; i) = b(n; k; n − i − 1): (13) Proof. The proof follows by considering the bijection which maps 0 0 π 2 Sn with π(k) = n to π 2 Sn with π (k) = n defined by π0 = (π(k − 1); : : : ; π(1); n; π(n); : : : ; π(k + 1)). 5 We consider the following descent polynomial Bn;k(t) defined by: X i Bn;k(t) = b(n; k; i)t : i≥0 We will need the generating function for this polynomial. Lemma 2. The descent polynomial Bn;k(t) has the following generating func- tion: X xn−1 xk−1 (etx − tetx) B (t; x) = B (t) = E (t) : (14) k n;k (n − 1)! k−1 (k − 1)!(etx − tex) n≥k Proof. We first consider Bn;k(t) for the case that k < n. Let Sm;l denote the set of permutations π 2 Sm which have l descents. For a fixed (k − 1)-subset of f1; : : : ; ng, there is a straightforward bijection from the set of permutations π 2 Sn with π(k) = n and having i descents to the following set: [ Sk−1;j × Sn−k;i−j−1 0≤j<i Thus, for k < n, we have X i Bk;n(t) = b(n; k; i)t i n − 1 X k − 1 X n − k = t tj1 tj2 : k − 1 j1 j2 j1 j2 xn−1 By multiplying both sides by (n−1)! and summing over all n > k, we have for fixed k, n−1 n−1 X X x t X En−k(t)x b(n; k; i)ti = E (t) (n − 1)! (k − 1)! k−1 (n − k)! n>k i n>k k−1 n−k tx Ek−1(t) X En−k(t)x = (k − 1)! (n − k)! n>k txk−1E (t) (1 − t)ex = k−1 − 1 (k − 1)! etx − tex txk−1E (t) ex − etx = k−1 : (k − 1)! etx − tex 6 Adding to this sum a term for the case n = k, we obtain X X xn−1 B (t; x) = b(n; k; i)ti k (n − 1)! n≥k i k−1 x tx k−1 tx Ek−1(t) e − e X x = + b(k; k; i)ti (k − 1)! etx − tex (k − 1)! i txk−1E (t) ex − etx xk−1E (t) = k−1 + k−1 (k − 1)! etx − tex (k − 1)! xk−1E (t) t(ex − etx) = k−1 + 1 (k − 1)! etx − tex xk−1E (t) etx − tetx = k−1 : (k − 1)! etx − tex Now we are ready to prove the following generalized equality for our re- stricted descent polynomials.
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