On certain combinatorial expansions of the Legendre-Stirling numbers

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. In this paper, we provide a combinatorial code for Legendre-Stirling set partitions. As an application, we obtain combinatorial expansions of the Legendre-Stirling numbers of both kinds. Moreover, we present grammatical descriptions of the Jacobi-Stirling numbers of both kinds.

Using the triangular recurrence relation n+1 k = n k + n k−1 , we get LS (n + 2, n) = 40 n + 3 6 + 32 n + 3 5 Egge [6,Theorem 3.1] showed that for k ≥ 0, the quantity LS (n + k, n) is a polynomial of degree 3k in n with leading coefficient 1 3 k k! . This paper is a continuation of [6], and it is motivated by 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 paper is organized as follows. In Section 2, by introducing a combinatorial code for Legendre-Stirling set partitions, we give a solution of Problem 1. Moreover, we get a combinatorial expansion of the Legendre-Stirling numbers of the first kind. In Section 3, we present grammatical interpretations of Jacobi-Stirling numbers of both kinds.

Legendre-Stirling set partitions
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, 2, 2, . . . , 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 < 2 = 2 < · · · < 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 2 ) The 'nonzero boxes' B 1 , B 2 , . . . , 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 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}{2, 2} < 3 >∈ LS (3,2). A classical result of Andrews and Littlejohn [3, Theorem 2] says that LS (n, k) = #LS(n, k).
We now provide a combinatorial code for Legendre-Stirling partitions (CLS -sequence for short).
Definition 2. We call Y n = (y 1 , y 2 , . . . , y n ) a CLS -sequence of length n if y 1 = X and where n x (Y k ) is the number of the symbol X in Y k = (y 1 , y 2 , . . . , y k ).
For example, (X, X, A 1,2 ) is a CLS -sequence, while (X, X, A 1,2 , B 3 ) is not since y 4 = B 3 and 3 > n x (Y 3 ) = 2. Let CLS n denote the set of CLS -sequences of length n.
The following lemma is a fundamental result.
Now we start to construct a bijection, denoted by Φ, between LS(n, k) and CLS(n, k). When 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 . Consider the following three cases: 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 correspondence between Y 5 and Φ(Y 5 ) is built up as follows: As an application of the CLS -sequences, we present the following result.
Lemma 5. Let k be a given positive integer. Then for n ≥ 1, we have Proof. It follows from Lemma 3 that Let Y n+k = y 1 y 2 · · · 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 2 + 2, · · · , t k + k. Let σ be the corresponding Legendre-Stirling partition of Y n+k . For 1 ≤ ℓ ≤ k, consider the value of y t ℓ +ℓ . Note that the number of the symbol X before y t ℓ +ℓ is t ℓ . Let σ be the corresponding Legendre-Stirling partition of y 1 y 2 · · · y t ℓ +ℓ−1 . Now we insert y t ℓ +ℓ . We distinguish two cases: (i) If y t ℓ +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 2 t ℓ 2 possibilities, since 1 ≤ i, j ≤ t ℓ and i = j.
(ii) If y t ℓ +ℓ = B s (resp. y t ℓ +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 Therefore, there are exactly 2 t ℓ 2 Legendre-Stirling partitions of M t ℓ +ℓ can be generated from σ by inserting the entry y t ℓ +ℓ . Note that 1 ≤ t j−1 ≤ t j ≤ n for 2 ≤ j ≤ k. Applying the product rule for counting, we immediately get (5).
The following simple result will be used in our discussion. Lemma 6. Let a and b be given integers. Then In particular, Proof. Note that This yields the desired result.
We can now conclude the main result of this paper from the discussion above.
Theorem 7. Let k be a given nonnegative integer. Then for n ≥ 1, the numbers LS (n + k, n) can be expanded in the binomial basis as where the coefficients γ(k, i) are all positive integers for k + 2 ≤ i ≤ 3k and satisfy the recurrence relation with the initial conditions γ(0, 0) = 1, Then the polynomials γ k (x) satisfy the recurrence relation with the initial conditions γ 0 (x) = 1, γ 1 (x) = x 3 and γ 2 (x) = x 4 + 8x 5 + 10x 6 .

It follows from Lemma 5 that
By using Lemma 6, it is routine to verify that the coefficients γ(k, i) satisfy the recurrence relation (7), and so (6) is established for general k. Multiplying both sides of (7) by x i and summing for all i, we immediately get (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 where the coefficients γ(k, i) are defined by (7).
It follows from (8) that Since γ(1, 3) = 1 and γ 1 (−1) = −1, it is easy to verify that for k ≥ 1, we have It should be noted that the number γ(k, 3k) is the number of partitions of {1, 2, . . . , 3k} into blocks of size 3 (see [18, A025035]), and the number (k+1)!k! 2 k is the product of first k positive triangular numbers (see [18, A006472]). 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 which yields [6, Theorem 3.1].

Grammatical interpretations of Jacobi-Stirling numbers of both kinds
In this section, a context-free grammar is in the sense of Chen [4]: for an alphabet A, let According to [8,Theorem 4.1], the Jacobi-Stirling number JS k n (z) of the second kind is defined by It follows from (11) that the numbers JS k n (z) satisfy the recurrence relation with the initial conditions JS 0 n (z) = δ n,0 and JS k 0 (z) = δ 0,k . It is clear that JS k n (1) = LS (n, k). with the initial conditions Jc 0 n (z) = δ n,0 and Jc k 0 (z) = δ k,0 . In particular, Jc k n (1) = Lc (n, k). Properties and combinatorial interpretations of the Jacobi-Stirling numbers of both kinds were extensively studied in [1,10,11,12,15,16,17]. The Jacobi-Stirling numbers share many similar properties to those of the Stirling numbers. A question arises immediately: are there grammatical descriptions of the Jacobi-Stirling numbers of both kinds? In this section, we give the answer.
As a variant of the CLS -sequence, we now introduce a marked scheme for Legendre-Stirling partitions. Given a Legendre-Stirling partition σ = B 1 B 2 · · · B k B 0 ∈ LS(n, k), where B 0 is the zero box of σ. We mark the box vector (B 1 , B 2 , . . . , B k ) by the label a k . We mark 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 partition that generated from σ by inserting n + 1 and n + 1. If n + 1 and n + 1 are in the same box, then 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 → 2b. (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 any 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 2 , a 3 , . . . , b, c} be a set of alphabet. Using the above marked scheme, it is natural to consider the following grammars: where k ≥ 1.
Theorem 9. Let G k be the grammars defined by (12). Then we have Proof. Note that D 1 (a 0 ) = a 1 c and D 2 D 1 (a 0 ) = a 2 bc 2 + (1 + z)a 1 c. Thus the result holds for n = 1, 2. For m ≥ 2, we define P k m (z) by We proceed by induction. Consider the case n = m + 1. Since Therefore, we obtain P k m+1 (z) = P k−1 m (z) + k(k + z)P k m (z). Since the numbers P k n (z) and JS k n (z) satisfy the same recurrence relation and initial conditions, so they agree.
Combining the marked scheme for Legendre-Stirling partitions and Theorem 9, 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 exactly i barred entries in the zero box, which gives a proof of [10, Theorem 2].
We end this section by giving the following result.
Theorem 10. Let A = {a, b 0 , b 1 , . . .} be a set of alphabet. Let G k be the grammars defined by where k ≥ 1. Then we have D n D n−1 · · · D 1 (ab 0 ) = a n k=1 Jc k n (z)b k .

Concluding remarks
Note that the Jacobi-Stirling numbers are polynomial refinements of the Legendre-Stirling numbers. It would be interesting to explore combinatorial expansions of Jacobi-Stirling numbers of both kinds.
Let γ k (x) be the polynomials defined by (8). We end our paper by proposing the following.