On certain combinatorial expansions of the Legendre-Stirling numbers

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1 On certain combinatorial expansions of the Legendre-Stirling numbers Shi-Mei Ma School of Mathematics and Statistics Northeastern University at Qinhuangdao Hebei, P.R. China Yeong-Nan Yeh Institute of Mathematics Academia Sinica Taipei, Taiwan Jun Ma Department of Mathematics Shanghai Jiao Tong University Shanghai, P.R. China Submitted: Aug 5, 018; Accepted: Dec,018; Published: Dec 1, 018 c The authors. Released under the CC BY-ND license (International 4.0). Abstract The Legendre-Stirling numbers of the second kind were introduced by Everitt et al. in the spectral theory of powers of the Legendre differential expressions. As a continuation of the work of Andrews and Littlejohn (Proc. Amer. Math. Soc., 137 (009), ), we provide a combinatorial code for Legendre-Stirling set partitions. As an application, we obtain expansions of the Legendre-Stirling numbers of both kinds in terms of binomial coefficients. Mathematics Subject Classifications: 05A18, 05A19 1 Introduction The study on Legendre-Stirling numbers and Jacobi-Stirling numbers has become an active area of research in the past decade. In particular, these numbers are closely related to set partitions [3], symmetric functions [1], special functions [11] and so on. Let l[y](t) = (1 t )y (t) + ty (t) be the Legendre differential operator. Then the Legendre polynomial y(t) = P n (t) is an eigenvector for the differential operator l Supported by Natural Science Foundation of Hebei Province (A ) and NSFC ( ). Supported by NSFC Supported by NSC M MY3. the electronic journal of combinatorics 5(4) (018), #P4.57 1

2 corresponding to n(n + 1), i.e., l[y](t) = n(n + 1)y(t). Following Everitt et al. [6], for n N, the Legendre-Stirling numbers LS (n, k) of the second kind appeared originally as the coefficients in the expansion of the n-th composite power of l, i.e., l n [y](t) = ( 1) k LS (n, k)((1 t ) k y (k) (t)) (k). k=0 For each k N, Everitt et al. [6, Theorem 4.1)] obtained that k 1 1 r(r + 1)x = LS (n, k)x n k, r=0 n=0 ( x < ) 1. (1) k(k + 1) According to [, Theorem 5.4], the numbers LS (n, k) have the following horizontal generating function k 1 x n = LS (n, k) (x i(1 + i)). () k=0 It follows from () that the numbers LS (n, k) satisfy the recurrence relation i=0 LS (n, k) = LS (n 1, k 1) + k(k + 1)LS (n 1, k). with the initial conditions LS (n, 0) = δ n,0 and LS (0, k) = δ 0,k, where δ i,j is the Kronecker s symbol. By using (1), Andrews et al. [, Theorem 5.] derived that the numbers LS (n, k) satisfy the vertical recurrence relation LS (n, j) = LS (k 1, j 1)(j(j + 1)) n k. k=j Following [7, Theorem 4.1], the Jacobi-Stirling number JS k n(z) of the second kind may be defined by k 1 x n = JS k n(z) (x i(z + i)). (3) k=0 It follows from (3) that the numbers JS k n(z) satisfy the recurrence relation i=0 JS k n(z) = JS k 1 n 1(z) + k(k + z)js k n 1(z), with the initial conditions JS 0 n(z) = δ n,0 and JS k 0(z) = δ 0,k (see [11] for instance). It is clear that JS k n(1) = LS (n, k). In [9], Gessel, Lin and Zeng studied generating function of the coefficients of JS n n+k(z). Note that JS n n+k(1) = LS (n + k, n). This paper is devoted to the following problem. Problem 1. Let k be a given nonnegative integer. Could the numbers LS (n + k, n) be expanded in the binomial basis? the electronic journal of combinatorics 5(4) (018), #P4.57

3 A particular value of LS (n, k) is provided at the end of [3]: n + LS (n + 1, n) =. (4) 3 In [5, Eq. (19)], Egge obtained that n + n + n + n + LS (n +, n) = Using the triangular recurrence relation ( n+1 k = n ( k) + n k 1), we get n + 3 n + 3 n + 3 LS (n +, n) = (5) Egge [5, Theorem 3.1] showed that for k 0, the quantity LS (n + k, n) is a polynomial 1 of degree 3k in n with leading coefficient. 3 k k! As a continuation of [3] and [5], in Section, we give a solution of Problem 1. Moreover, we get an expansion of the Legendre-Stirling numbers of the first kind in terms of binomial coefficients. Main results The combinatorial interpretation of the Legendre-Stirling numbers LS (n, k) of the second kind was first given by Andrews and Littlejohn [3]. For n 1, let M n denote the multiset {1, 1,,,..., n, n}, in which we have one unbarred copy and one barred copy of each integer i, where 1 i n. Throughout this paper, we always assume that the elements of M n are ordered by 1 = 1 < = < < n = n. A Legendre-Stirling set partition of M n is a set partition of M n with k + 1 blocks B 0, B 1,..., B k and with the following rules: (r 1 ) The zero box B 0 is the only box that may be empty and it may not contain both copies of any number; (r ) The nonzero boxes B 1, B,..., B k are indistinguishable and each is non-empty. For any i [k], the box B i contains both copies of its smallest element and does not contain both copies of any other number. Let LS(n, k) denote the set of Legendre-Stirling set partitions of M n with one zero box and k nonzero boxes. The standard form of an element of LS(n, k) is written as σ = B 1 B B k B 0, the electronic journal of combinatorics 5(4) (018), #P4.57 3

4 where B 0 is the zero box and the minima of B i is less than that of B j when 1 i < j k. Clearly, the minima of B 1 are 1 and 1. Throughout this paper we always write σ LS(n, k) in the standard form. As usual, we let angle bracket symbol < i, j,... > and curly bracket symbol {k, k,...} denote the zero box and nonzero box, respectively. In particular, let <> denote the empty zero box. For example, {1, 1, 3}{, } < 3 > LS(3, ). A classical result of Andrews and Littlejohn [3, Theorem ] says that LS (n, k) = #LS(n, k). We now provide a combinatorial code for Legendre-Stirling partitions (CLS -sequence for short). Definition. We call Y n = (y 1, y,..., y n ) a CLS -sequence of length n if y 1 = X and y k+1 {X, A i,j, B s, B s, 1 i, j, s n x (Y k ), i j} for k = 1,,..., n 1, where n x (Y k ) is the number of the symbol X in Y k = (y 1, y,..., y k ). For example, (X, X, A 1, ) is a CLS -sequence, while (X, X, A 1,, B 3 ) is not since y 4 = B 3 and 3 > n x (Y 3 ) =. Let CLS n denote the set of CLS -sequences of length n. The following lemma is a fundamental result. Lemma 3. For n 1, we have LS (n, k) = #{Y n CLS n n x (Y n ) = k}. Proof. Let CLS(n, k) = {Y n CLS n n x (Y n ) = k}. Now we start to construct a bijection, denoted by Φ, between LS(n, k) and CLS(n, k). When n = 1, we have y 1 = X. Set Φ(Y 1 ) = {1, 1} <>. This gives a bijection from CLS(1, 1) to LS(1, 1). Let n = m. Suppose Φ is a bijection from CLS(n, k) to CLS(n, k) for all k. Consider the case n = m + 1. Let Y m+1 = (y 1, y,..., y m, y m+1 ) CLS m+1. Then Y m = (y 1, y,..., y m ) CLS(m, k) for some k. Assume Φ(Y m ) LS(m, k). Consider the following three cases: (i) If y m+1 = X, then let Φ(Y m+1 ) be obtained from Φ(Y m ) by putting the box {m + 1, m + 1} just before the zero box. In this case, Φ(Y m+1 ) LS(m + 1, k + 1). (ii) If y m+1 = A i,j, then let Φ(Y m+1 ) be obtained from Φ(Y m ) by inserting the entry m + 1 to the ith nonzero box and inserting the entry m + 1 to the jth nonzero box. In this case, Φ(Y m+1 ) LS(m + 1, k). (iii) If y m+1 = B s (resp. y m+1 = B s ), then let Φ(Y m+1 ) be obtained from Φ(Y m ) by inserting the entry m + 1 (resp. m + 1) to the sth nonzero box and inserting the entry m + 1 (resp. m + 1) to the zero box. In this case, Φ(Y m+1 ) LS(m + 1, k). After the above step, it is clear that the obtained Φ(Y m+1 ) is in standard form. By induction, we see that Φ is the desired bijection from CLS(n, k) to CLS(n, k), which also gives a constructive proof of Lemma 3. the electronic journal of combinatorics 5(4) (018), #P4.57 4

5 Example 4. Let Y 5 = (X, X, A,1, B, B 1 ). The correspondence between Y 5 and Φ(Y 5 ) is built up as follows: X {1, 1} <>; X {1, 1}{, } <>; A,1 {1, 1, 3}{,, 3} <>; B {1, 1, 3}{,, 3, 4} < 4 >; B 1 {1, 1, 3, 5}{,, 3, 4} < 4, 5 >. As an application of Lemma 3, we present the following lemma. Lemma 5. Let k be a given positive integer. Then for n 1, we have LS (n + k, n) = k tk + 1 tk tk t 3 t + 1 t t (6) n t k =1 Proof. It follows from Lemma 3 that t k 1 =1 t =1 LS (n + k, n) = #{Y n+k CLS n+k n x (Y n+k ) = n}. Let Y n+k = (y 1, y,..., y n+k ) be a given element in CLS n+k. Since n x (Y n+k ) = n, it is natural to assume that y i = X except i = t 1 + 1, t +,, t k + k. Let σ be the corresponding Legendre-Stirling partition of Y n+k. For 1 l k, consider the value of y tl +l. Note that the number of the symbols X before y tl +l is t l. Let σ be the corresponding Legendre-Stirling set partition of (y 1, y,..., y tl +l 1). Now we insert y tl +l. We distinguish two cases: (i) If y tl +l = A i,j, then we should insert the entry t l + l to the ith nonzero box of σ and insert t l + l to the jth nonzero box. This gives ( t l ) possibilities, since 1 i, j tl and i j. (ii) If y tl +l = B s (resp. y tl +l = B s ), then we should insert the entry t l + l (resp. t l + l) to the sth nonzero box of σ and insert t l + l (resp. t l + l) to the zero box. This gives ( t l 1) possibilities, since 1 s tl. Therefore, there are exactly ( t l ( ) + tl ) ( 1 = tl ) +1 Legendre-Stirling set partitions of M tl +l can be generated from σ by inserting the entry y tl +l. Note that 1 t j 1 t j n for j k. Applying the product rule for counting, we immediately get (6). The following simple result will be used in our discussion. Lemma 6. Let a and b be two given integers. Then x b x a + x x a b x = + (a + 1)(a b) +. a a + a + 1 a In particular, x 1 x a + x = a a + x + (a 1) + a + 1 t 1 =1 a 1 x. a the electronic journal of combinatorics 5(4) (018), #P4.57 5

6 Proof. Note that a + (x a)(x a 1) (a + )(a + 1) This yields the desired result. + (a + 1)(a b) x a a ( a b x b = We can now conclude the main result of this paper from the discussion above. Theorem 7. Let k be a given nonnegative integer. For n 1, the numbers LS (n + k, n) can be expanded in the binomial basis as LS (n + k, n) = k 3k i=k+ ). n + k + 1 γ(k, i), (7) i where the coefficients γ(k, i) are all positive integers for k + i 3k and satisfy the recurrence relation i k 1 i 1 γ(k +1, i) = γ(k, i 1)+(i 1)(i k )γ(k, i )+ γ(k, i 3), (8) with the initial conditions γ(0, 0) = 1, γ(0, i) = γ(i, 0) = 0 for i 0. Let γ k (x) = 3k i=k+ γ(k, i)xi. Then the polynomials γ k (x) satisfy the recurrence relation k(k + 1) γ k+1 (x) = kx + x xγ k (x) (k + (k )x x )x γ k(x) + (1 + x) x 3 γ k(x), (9) with the initial conditions γ 0 (x) = 1, γ 1 (x) = x 3 and γ (x) = x 4 + 8x x 6. Proof. We prove (7) by induction on k. It is clear that LS (n, n) = 1 = n+1 0. When k = 1, by using the Chu Shih-Chieh s identity we obtain n + 1 = k + 1 i=k i, k t1 + 1 n + =, 3 t 1 =1 and so (4) is established. When k =, it follows from Lemma 5 that LS (n +, n) = 4 = 4 t + 1 t t1 + 1 t 1 =1 t + 1 t +. 3 t =1 t =1 the electronic journal of combinatorics 5(4) (018), #P4.57 6

7 Setting x = t + and a = 3 in Lemma 6, we get ( ) t + t + t + LS (n +, n) = t =1 ( ) n + 3 n + 3 n + 3 = , which yields (5). Along the same lines, it is not hard to verify that ( ) t3 + 1 t3 + 3 t3 + 3 t3 + 3 LS (n + 3, n) = t 3 =1 ( ) n + 4 n + 4 n + 4 n + 4 n + 4 = Hence the formula (7) holds for k = 0, 1,, 3, so we proceed to the inductive step. For k 3, assume that It follows from Lemma 5 that LS (n + k, n) = k LS (n + k + 1, n) = k+1 t k+1 =1 3k i=k+ ( tk n + k + 1 γ(k, i). i ) 3k i=k+ tk+1 + k + 1 γ(k, i) i By using Lemma 6 and the Chu Shih-Chieh s identity, it is routine to verify that the coefficients γ(k, i) satisfy the recurrence relation (8), and so (7) is established for general k. Multiplying both sides of (8) by x i and summing for all i, we immediately get (9). In [], Andrews et al. introduced the (unsigned) Legendre-Stirling numbers Lc (n, k) of the first kind, which may be defined by the recurrence relation Lc (n, k) = Lc (n 1, k 1) + n(n 1)Lc (n 1, k), with the initial conditions Lc (n, 0) = δ n,0 and Lc (0, n) = δ 0,n. Let f k (n) = LS (n + k, n). According to Egge [5, Eq. (3)], we have Lc (n 1, n k 1) = ( 1) k f k ( n) (10) for k 0. For m, k N, we define m ( m)( m 1) ( m k + 1) =. k k! Combining (7) and (10), we immediately get the following result. the electronic journal of combinatorics 5(4) (018), #P4.57 7

8 Corollary 8. Let k be a given nonnegative integer. For n 1, the numbers Lc (n 1, n k 1) can be expanded in the binomial basis as Lc (n 1, n k 1) = ( 1) k k where the coefficients γ(k, i) are defined by (8). 3k i=k+ n + k + 1 γ(k, i), (11) i It follows from (9) that k(k + 1) (k + )(k + 1) γ(k + 1, k + 3) = k(k + ) + γ(k, k + ), ( ) 3k(3k 1) γ(k + 1, 3k + 3) = 1 + 6k + γ(k, 3k), k(k + 1) γ k+1 ( 1) = + k + 1 γ k ( 1). Since γ(1, 3) = 1 and γ 1 ( 1) = 1, it is easy to verify that for k 1, we have γ(k, k + ) = 1, γ(k, 3k) = (3k)! k!(3!), γ k (k + 1)!k! k( 1) = ( 1). k k It should be noted that the number γ(k, 3k) is the number of partitions of {1,,..., 3k} into blocks of size 3, and the number (k+1)!k! is the product of first k positive triangular k numbers. Moreover, if the number LS (n + k, n) is viewed as a polynomial in n, then its degree is 3k, which is implied by the quantity n+k+1 3k. Furthermore, the leading coefficient of LS (n + k, n) is given by k γ(k, 3k) 1 = (3k)! k (3k)! 1 = 1, which yields [5, Theorem k!(3!) k (3k)! k!3 k 3.1]. 3 Concluding remarks In this paper, by introducing the CLS -sequence, we present a combinatorial expansion of LS (n + k, n). It should be noted that the CLS -sequence has several other variants. For an alphabet A, let Q[[A]] be the rational commutative ring of formal Laurent series in monomials formed from letters in A. Following Chen [4], a context-free grammar over A is a function G : A Q[[A]] that replace a letter in A by a formal function over A. The formal derivative D = D G : Q[[A]] Q[[A]] is defined as follows: for x A, we have D(x) = G(x); for a monomial u in Q[[A]], D(u) is defined so that D is a derivation, and for a general element q Q[[A]], D(q) is defined by linearity. The reader is referred to [10] for recent results on context-free grammars. As a variant of the CLS -sequence, we now introduce a marked scheme for Legendre- Stirling set partitions. Given a set partition σ = B 1 B B k B 0 LS(n, k), where B 0 is the zero box of σ. We mark the box vector (B 1, B,..., B k ) by the label a k. We mark the electronic journal of combinatorics 5(4) (018), #P4.57 8

9 any box pair (B i, B j ) by a label b and mark any box pair (B s, B 0 ) by a label c, where 1 i < j k and 1 s k. Let σ denote the Legendre-Stirling set partition that generated from σ by inserting n + 1 and n + 1. If n + 1 and n + 1 are in the same box, then σ = B 1 B B k B k+1 B 0, where B k+1 = {n + 1, n + 1}. This case corresponds to the operator a k a k+1 b k c. If n + 1 and n + 1 are in different boxes, then we distinguish two cases: (i) Given a box pair (B i, B j ), where 1 i < j k. We can put n + 1 (resp. n + 1) into the box B i and put n + 1 (resp. n + 1) into the box B j. This case corresponds to the operator b b. (ii) Given a box pair (B i, B 0 ), where 1 i k. We can put n + 1 (resp. n + 1) into the box B i and put n + 1 (resp. n + 1) into the zero box B 0. Moreover, we mark each barred entry in the zero box B 0 by a label z. This case corresponds to the operator c (1 + z)c. Let A = {a 0, a 1, a, a 3,..., b, c} be a set of alphabet. scheme, we consider the grammars Following the above marked G k = {a 0 a 1 c, a 1 a bc,..., a k 1 a k b k 1 c, b b, c (1 + z)c}, where k 1. It is a routine check to verify that D n D n 1 D 1 (a 0 ) = JS k n(z)a k b (k ) c k. k=1 Therefore, it is clear that for n k, the number JS k n(z) is a polynomial of degree n k in z, and the coefficient z i of JS k n(z) is the number of Legendre-Stirling partitions in LS(n, k) with i barred entries in zero box, which gives a grammatical proof of [8, Theorem ]. We end our paper by proposing the following. Conjecture 9. For any k 1, the polynomial γ k (x) has only real zeros. Set k k γ k (x) = γ(k, 3k)x k+ (x r i ), γ k+1 (x) = γ(k + 1, 3k + 3)x k+3 (x s i ), i=1 where r k < r k 3 < < r < r 1 and s k < s k 1 < s k < < s < s 1. Then s k < r k < s k 1 < r k 3 < s k < < r k < s k+1 < s k < r k 1 < < s < r 1 < s 1, in which the zeros s k+1 and s k of γ k+1 (x) are continuous appearance, and the other zeros of γ k+1 (x) separate the zeros of γ k (x). Acknowledgements The authors thank the referee for his valuable suggestions. i=1 the electronic journal of combinatorics 5(4) (018), #P4.57 9

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