Some identities involving $q$-Stirling numbers of the second kind in type B

The recent interest in $q$-Stirling numbers of the second kind in type B prompted us to give a type B analogue of a classical identity connecting the $q$-Stirling numbers of the second kind and Carlitz's major $q$-Eulerian numbers, which turns out to be a $q$-analogue of an identity due to Bagno, Biagioli and Garber. We provide a combinatorial proof of this identity and an analytical proof of a more general identity for colored permutations. In addition, we prove some $q$-identities about the $q$-Stirling numbers of the second kind in types A, B and D.


Introduction
The Stirling number of the second kind, denoted S(n, k), is the number of ways to partition n distinct objects into k nonempty subsets.It satisfies the well-known triangular recurrence S(n, k) = S(n − 1, k − 1) + kS(n − 1, k) with the initial conditions S(0, k) = δ 0k , where δ ij is the Kronecker delta.Carlitz [7] introduced the type A q-Stirling numbers of the second kind S[n, k] by where [k] q := 1 + q + q 2 + • • • + q k−1 for k 1 and [0] q := 0, and S[0, k] = δ 0k .Let S n be the symmetric group on the set [n] = {1, 2, . . ., n}.An element π ∈ S n is written as π = π 1 π 2 • • • π n .The descent set of π ∈ S n is defined by and the cardinality of Des(π) is called the number of descents of π, denoted des(π).The Eulerian number A n,k is the number of π ∈ S n with k descents.There exists a wellknown identity connecting the Stirling numbers of the second kind and Eulerian numbers as follows: for all nonnegative integers 0 k n.A combinatorial proof of identity (1.2) in terms of the ordered set partitions and permutations is quite easy and well known, see [5, Theorem 1.17], for example.The q-binomial coefficients are defined for n, k ∈ N by where [n] q !:= [1] q [2] q • • • [n] q is the q-factorial of n.To give a q-analogue of identity (1.2) we need to find a suitable Mahonian statistic over permutations, that is, a statistic whose generating function over S n is [n] q !.It turns out that MacMahon's major index [17] is a good fit for our q-analogue.Recall that the major index (maj) of π ∈ S n is defined by maj(π) := i∈Des(π) i.
We define the corresponding q-analogue of Eulerian polynomial (of type A) by A n (t, q) := π∈Sn t des(π) q maj(π) = n k=0 A n,k (q)t k . (1. 3) The reader is referred to [11,18] and references therein for further q-Eulerian polynomials.
Using analytic method, Zeng and Zhang [27,Proposition 4.5] proved the following q-analogue of identity (1.2) for nonnegative integers 0 k n.In 1997, in order to give a combinatorial proof of (1.4), Steingrímsson [23] proposed several statistics on ordered set partitions and conjectured that their generating functions were given by either side of (1.4).In the following years, Zeng et al. [16,13,14] confirmed all his conjectures, and finally Remmel and Wilson [20, Section 5.1] found a combinatorial proof of (1.4) using the major index on the starred permutations.
This paper arose from the desire to give a type B analogue of (1.4).In analogy with the usual (type A) Stirling numbers of the second kind (see [26,10,3,21]), the type B Stirling numbers of the second kind S B (n, k) can be defined by S B (n, k) := S B (n − 1, k − 1) + (2k + 1)S B (n − 1, k) with the initial conditions S B (0, k) = δ 0k .
For integer i ∈ Z we denote its opposite integer −i by i.Let B n be the group of signed permutations of [n], i.e., the set of all permutations on the set [±n] := {n, . . ., 1, 1, . . ., n} such that π(i) = π(i).In what follows, we write π(i) as π i for i ∈ [±n] and use the natural order on n := {n, . . ., 1, 0, 1, . . ., n}, namely, The type B descent set of π ∈ B n [18,Section 11.5.2] is defined by with π 0 = 0, and the cardinality of Des B (π) is called the number of type B descents of π, denoted des B (π).
Let B n,k be the number of permutations in B n with k descents.By a bijection between the set of ordered type B set partitions and the set of signed permutations with separators, Bagno, Biagioli and Garber [3] combinatorially proved the following type B analogue of (1.2): for all nonnegative integers 0 k n.Recently Sagan and Swanson [21] studied the type B q-Stirling numbers of the second kind S B [n, k], which are defined by the recurrence relation with the initial conditions S B [0, k] = δ 0k , see [25,Section 1.10] and [4] for related works.
Remark 1.1.Chow-Gessel [8, Eq. ( 18) and Proposition 4.2] defined a kind of type B q-Stirling numbers of the second kind S n,k (q) by the following recurrence relation with the initial conditions S n,0 (q) = 1 for n 0. It is routine to verify that the above two types B q-Stirling numbers of the second kind are related as follows Adin and Roichman [1] defined the flag-major index of π ∈ B n as follows fmaj(π) := where neg(π) is the number of negative elements in π, i.e., |{i ∈ [n] : π i < 0}|.Then, as a q-analogue of Eulerian polynomial of type B, Chow and Gessel [8] studied the enumerative polynomials of statistic (des B , fmaj) over B n , In this paper, using Sagan and Swanson's q-Stirling numbers of the second kind in type B [21] and Chow and Gessel's q-Eulerian numbers of type B, we prove a q-analogue of Bagno et al.'s identity (1.5).The following is our first main result.
Theorem 1.2.For 0 k n we have We shall provide a combinatorial proof for Theorem 1.2 in Section 2. In Section 3, we define a q-Stirling numbers of the second kind in type D and give q-analogues of some known identities connecting the Stirling numbers of the second kind in types A, B and D. Next, we prove algebraically a general identity (see Theorem 4.2) between the r-colored q-Stirling numbers of the second kind and q-Eulerian numbers of colored permutations in Section 4. Note that the proof of Theorem 4.2 yields another proof of Theorem 1.2.

Combinatorial proof of Theorem 1.2
In this section, we give a combinatorial proof of (1.10) by generalizing Remmel and Wilson's proof of identity (1.4) in [20].Our strategy is to study the polynomial and interpret the coefficient of z k combinatorially in two different ways.
Remark 2.1.Let r : π → π r be the reversing operator on B n defined by π r i = π n+1−i and c : π → π c the type B completion operator on B n defined by Clearly, if i is a descent (resp., an ascent) position in π ∈ B n and the product of π i and π i+1 is positive, then n − i is an ascent (resp., a descent) position in π; if i is a descent (resp., an ascent) position in π ∈ B n and the product of π i and π i+1 is negative, then n − i is a descent (resp., an ascent) position in π.
In fact, the mapping ψ is a bijection between all permutations in B n with k descents and all permutations in B n with n − k descents by the following result.
Lemma 2.2.The mapping ψ is a bijection on B n such that for any π ∈ B n , we have des Proof.It is convenient to associate a permutation in B n with a character string in {+, −} n by replacing each positive (resp., negative) element with + (resp., −).For example, the string for permutation 1 5 3 4 6 2 is + + − + +−.Let π ∈ B n with des B (π) = k.We consider the following four cases in terms of the signs of π 1 and π n .
(i) π 1 > 0 and π n < 0, we have ℓ = r + 1.It is easy to see that i k+1 −j k is the number of positive elements between the kth ascent position and the (k + 1)th descent position from left to right.Note that elements between the kth ascent position and the kth descent position from left to right.Then we have r k=1 (i k − j k ) = n − m.Combining the above two cases completes the proof of (a).
The following q-symmetry of B n,k (q) is crucial for our combinatorial proof of identity (1.10).
Proposition 2.4.For each fixed nonnegative integer n and the polynomial B n,k (q) defined in (1.9), we have for 0 k n.
Let π = ψ(π), we consider the proof in terms of the signs of π 1 and π n .We only give the proof for this case π 1 > 0 and π n > 0 and omit similar discussions for other three cases for the brevity.
If π 1 > 0 and π n > 0, by the definition of the mapping ψ, then the set of descents in π is the disjoint union and π has n − m negative elements.Then, By Case (i) in the proof of Proposition 2.2, we have where the second equality uses the fact that the sum of all descent and ascent indexes is This is the desired result.

Ordered set partitions of type B
Recall that n = {n, . . ., 1, 0, 1, . . ., n}.There are at least two equivalent definitions of type B set partition.We say that a set partition of n is a type B partition if it satisfies the following properties the electronic journal of combinatorics 30 (2023), #P00 (1) there exactly is one zero block T such that 0 ∈ T and −T = T ; (2) if T appears as a block then −T is also a block.
It is known [3,21] that S B (n, k) is the number of type B partitions of n with 2k + 1 blocks.An ordered signed partition of n is a sequence (T 0 , T 1 , T 2 , . . ., T 2k ) of disjoint subsets (blocks) T i of n satisfying (1) 0 ∈ T 0 and T 0 = T 0 , and where T = {t : t ∈ T }.The blocks T 2i and T 2i−1 are called paired.Clearly the number of all ordered signed partitions of n with 2k + 1 blocks is 2 k k!S B (n, k).
On the other hand, as in [20], we can consider an ordered set partition with sign as a descent-starred signed permutation, i.e, for any π ∈ B n , the space following element π i , satisfying π i > π i+1 for some 0 i n − 1, is starred or unstarred.That is to say, instead of using brackets to signify separations between blocks, the spaces between elements sharing a block can be marked with stars and all blocks are written in decreasing order.Note that we require that the block including element 0 always stands first on the list.
and the polynomial q fmaj((π,S)) . (2.5) the electronic journal of combinatorics 30 (2023), #P00 By the definition of the statistic fmaj((π, S)), we attach the ith descent position of π (from right to left) with the weight 1 if this descent position is unstarred and the weight z/q 2i−1 if this descent position is starred.Therefore, the following identity holds (2.6) For convenience, we recall two known q-identities (see [2,Theorem 3.3]) N j q (−1) j z j q j(j−1)/2 ; (2.7) We first establish the following result for polynomials B n,n−ℓ (q) and B fmaj n,n−k (q) defined by (1.9) and (2.5).
Proposition 2.5.For 0 k n we have Proof.Let (π, S) ∈ B > n,n−k , then there are n − k starred descents in (π, S), this means that the number of ascents is in {0} ∪ [k].Suppose that the signed permutation π has ℓ ascents, where ℓ ∈ {0} ∪ [k], then the signed permutation π can be any permutation in B n with n − ℓ descents.Therefore, the sum of q-counting about the flag-major statistic for all possible signed permutations with n − ℓ descents is the polynomial B n,n−ℓ (q).
In addition, for a signed permutation π with n − ℓ descents, we can choose n − k descents from n − ℓ descents in π and mark them with stars.By the definition of the statistic fmaj((π, S)) and identities (2.6) and (2.7), we have we complete the proof.
To derive a recurrence relation for the polynomials B fmaj n,n−k (q), we introduce some notations.For other unstarred positions, we label the rightmost position in our descent-starred the electronic journal of combinatorics 30 (2023), #P00 signed permutation with 0, and then label its unlabelled descent positions from right to left with 1,2,. . . .Next, all other unlabelled positions from left to right are labelled with increasing labels starting from the next number.We call the above labelling as fmajlabelling.For example, if (π, S) = 4 * 3 * 1 7 * 2 6 8 * 5, then the fmaj-labelling for (π, S) is For α = n or n, we define the mapping by sending (i, (π, S)) to the descent-starred signed permutation obtained from (π, S) by ( 1) inserting α at the fmaj-labelling i, and then (2) moving each star on the right of α one descent to its left.
Clearly, the rightmost descent will be unstarred when the letter n is not inserted after π n−1 .Thus, we have the following relation between these labels and insertion mappings.
Lemma 2.6.For 0 k n − 1 we have Proof.We will discuss the change of the statistic fmaj((π, S)) in terms of the insertion position of n or n.Suppose that the space labelled i under the fmaj-labelling of (π, S) is the space immediately following π p .Moreover, we suppose that there are a starred descents and b unstarred descents to the left of π p and c unstarred descents and d starred descents to the right of π p+1 in (π, S).For (a), inserting n into the space labelled i.Let (τ, T ) = φ | n,k (i, (π, S)).If i = 0, that is to say we insert n at the end, then the insertion of n does not affect fmaj((π, S)), thus fmaj((τ, T )) = fmaj((π, S)).For i = 0, there will exist two cases in terms of the values of π p and π p+1 .
Summarising the above cases we have completed the proof.
As mentioned before, the mapping φ | α,k preserves the number of stars in the mapping process.Similarly, we need to define some mappings that increase the number of stars by one as follows: ) which send (i, (π, S)) to the descent-starred signed permutation obtained from (π, S) by (1) inserting n (resp., n) at the fmaj-labelling i, then (2) moving each star on the right of n (resp., n) one descent to its left, and then (3) placing a star at the rightmost descent of the resulting descent-starred signed permutation.
In analogy with the discussion in the proof of Lemma 2.6, let α = n or n and (τ, T ) = φ * α,k (i, (π, S)).The first step and second one from the mapping φ * α,k have same effect with φ | α,k to the statistics fmaj(π) and j∈S (2| Des B (π) ∩ {j, . . ., n − 2}| − 1).The last step from the mapping φ * α,k , placing a star at the rightmost of resulting descent-starred signed permutation, which increases the sum j∈S (2| Des B (π) ∩ {j, . . ., n − 2}| − 1) by one.Therefore, we have the following results, of which the proof is omitted for the brevity.Lemma 2.7.For 1 k n we have By definitions (2.9) and (2.10), the mappings φ | α,k and φ * α,k (α = n or n) have their images I 0 ∪ I 1 and I 2 , respectively, where Obviously, the disjoint union of those three sets is B > n,k .Now, we are ready to prove the following recurrence relation for the polynomial B fmaj n,k (q) defined in (2.5).
Proposition 2.8.For n 1 we have the recurrence relation where B fmaj n,k (q) is 1 when k = n and is 0 when k < 0 or k > n.
3 q-Stirling numbers of the second kind in type D Recently, Bagno et al. [3] studied some identities about the type D Stirling numbers of the second kind S D (n, k).As far as we know, there is no q-Stirling numbers of the second kind in type D in the literature.In this section, we first define a q-Stirling numbers of the second kind in type D and prove q-analogues of two known results about the Stirling numbers of the second kind in types A, B and D, see Proposition 3.6.Then, we establish a q-identity connecting the q-falling factorials of type D and the q-Stirling numbers of the second kind in type D, see Proposition 3.8.
3.1 Two q-identities about the q-Stirling numbers of the second kind Using the definitions and notations of ordered signed partition in Subsection 2.2, we say that the set For 0 k n, the following two identities about the Stirling numbers of the second kind in types A, B and D were implicitly given in [26,Corollary 12], [8,Eq. (19)] and [24,Proposition 3]: In this subsection, we define a kind of type D q-Stirling numbers of the second kind S D [n, k], and give q-analogues of identities (3.1) and (3.2).
Definition 3.1.For any S ⊂ Z\{0} let S = {i : i ∈ S}.A standard signed partition (SSP for short) of S is a sequence π = (S 1 , S 2 , . . ., S k ) of disjoint nonempty subsets of S ∪ S such that (1) {S 1 , . . ., S k , S 1 , . . ., S k } is a partition of S ∪ S; ( the electronic journal of combinatorics 30 (2023), #P00 The sets S 1 , S 2 , . . ., S k are the blocks of π (so π has k blocks).A partial standard signed partition (PSSP for short) of S is a standard signed partition of a subset of S.
Let B(S, k) (resp., B ⊆ (S, k)) be the set of all SSP (resp., PSSP) of S with k blocks.Let D ⊆ ([n], k) denote the set of all PSSP of [n] that excludes all SSP of [n]\{i} with k blocks for i ∈ [n], namely, is a type D signed partition of n .Due to the choice of T i and T i , both PSSP π = (T 1 , . . ., T i , . . ., T k ) and π ′ = (T 1 , . . ., T i , . . ., T k ) correspond to the type D signed partition Π, which implies the desired result.
the electronic journal of combinatorics 30 (2023), #P00 By recurrence (1.6) of S B [n, k], it suffices to show that S B (n, k, q) satisfies with the initial conditions S B (0, k, q) = δ 0k for q = 0.The case n = 0 is trivial.Suppose that n > 0 and π Thus This finishes the proof.
Definition 3.5.We define the q-Stirling numbers of the second kind in type D by The following results are q-analogues of identities (3.1) and (3.2), which also show that S D [n, k] is a polynomial in q.Let S[n, k] q 2 denote S[n, k] with q replaced by q 2 , i.e., S[n, k] q 2 := S[n, k] q←q 2 .Proposition 3.6.Let S D [n, k] be defined by (3.5).Then the identities hold for 0 k n. the electronic journal of combinatorics 30 (2023), #P00 3.2 Falling factorials and q-Stirling numbers of the second kind in type D For the Stirling numbers of the second kind S(n, k), a well-known identity involving the connection between the standard basis of the polynomial ring R n [t] and the basis consisting of falling factorials is that, for n ∈ N and t ∈ C , we have where (t ) and (t) 0 := 1.
A classical combinatorial interpretation for (3.9) pointed out that t n is the number of all mappings from the set [n] to the set [t] (t ∈ N + ) and S(n, k)(t) k is the number of surjections that map the set [n] to all k-subsets of [t], see [22,Eq.(1.96)] for more details.Similarly, for the Stirling numbers of the second kind in types B and D, Bagno et al. [3,Theorems 5.1 and 5.4] used a geometric method to obtain the following identities: where (t ) and (t) B 0 := 1, and where (t) D k is defined as Naturally, those q-analogues for identities (3.9) and (3.10) were also given as where (t) k,q = t(t ) and (t) 0,q := 1 (see Carlitz [7, Eq. (3.1)]), and where (t ) and (t) B 0,q := 1 (see Define a q-falling factorial of type D by We have a q-analogue of identity (3.11) as follows.
Proof.From equation (3.7) we derive the identity Thus, multiplying both sides of (3.14) by (t) D k,q and summing over 0 k n, we have First, for the left-hand side of (3.15), we have where the second equality and last one use the facts S B [n, n] = 1 and (t) B n,q = (t) D n,q − [n] q q n−1 (t) D n−1,q , and identity (3.13), respectively.In addition, for the second summation in the right-hand side of (3.15), we have where the second equality uses identity (3.12).Combining (3.15), (3.16) and (3.17), we complete the proof.

Generalization to colored permutations
In this section, instead of proving Theorem 1.2 by an algebraic proof, we shall prove a more general identity.Define the r-colored q-Stirling numbers of the second kind S r [n, k] by the recurrence relation with the initial conditions S r [0, k] = δ 0k .It is not difficult to verify (see [19,Theorem 1] for a more general result) that where (t) r k,q = (t ) and (t) r 0,q := 1.Using Rook theory, Remmel and Wachs gave a combinatorial interpretation of identity (4.2) in [19,Theorem 7].
Substituting t by [rm + 1] q in (4.2) yields which, by (2.8), is equivalent to the generating function identity, The colored permutations group of n letters with r colors can be looked as the wreath product group Define the following total order relation on the elements of Z r ≀ S n : i+1 , where π z 0 0 = 0. Let Des r (π) denote the descent set of π ∈ Z r ≀ S n and des r (π) the number of descents of π, i.e., | Des r (π)|.The r-colored Eulerian number A r n,k is the number of all colored permutations in Z r ≀ S n with k descents.For each π ∈ Z r ≀ S n , as in [1], define the r-flag-major index of π by fmaj r (π) := r the electronic journal of combinatorics 30 (2023), #P00 A q-analogue of the r-colored Eulerian polynomial A r n (t, q) is defined by When r takes 1 and 2, (4.5) reduces to (1.3) and (1.9), respectively.The following Carlitz's identity for Z r ≀ S n was proved in [6, Proposition 8.1] and [9, Theorem 9] Combining (4.3) and (4.6) we obtain the following identity.
Proposition 4.1.For the polynomials S r [n, k] in (4.1) and A r n (t, q) in (4.5), the q-Frobenius formula holds .
The following result is a q-analogue of Theorem 6.6 in [3] about an identity between the r-colored Stirling numbers of the second kind S r (n, k) (the sequence defined by (4.1) when q = 1, see also[3, Section 6.1]) and r-colored Eulerian numbers A r n,k .Theorem 4.2.For the r-colored q-Stirling numbers of the second kind S r [n, k] in (4.1) and q-Eulerian numbers A r n,k (q) in (4.5), we have the identity q r( k+1 2 )+(1−r)k [r] k q [k] q r !S r [n, k] = k ℓ=0 q rk(k−ℓ) A r n,ℓ (q) n − ℓ k − ℓ q r (4.7) for 0 k n.
Proof.Summing for both sides of (4.7) multiplying by t k / k i=0 (1 − tq ri ) over all k, it is clear that (4.7) is equivalent to for 0 ℓ n.That is to say, the index ℓ does not affect the summation in the right-hand side of (4.8).Substituting q r → q and applying (2.8) to extract the coefficients of t m on both sides of (4.8) we obtain which is a q-analogue of Chu-Vandermonde summation [2, Eq.
the electronic journal of combinatorics 30 (2023), #P00 In addition, by (1.4) and (4.8), we have the following q-Frobenius formula [12, Eq. (4.1)] related to q-Stirling numbers of the second kind and q-Eulerian polynomials of type A: tA n (t, q) n i=0 (1 . (4.10) Following the above discussion, it is clear that identity (4.10) is a special case of Proposition 4.1 for r = 1.
Remark 4.3.Similar to the combinatorial proofs of (1.4) and (1.10), it is natural to ask for a combinatorial proof of identity (4.7).One difficulty for such a proof is that a counterpart of the q-symmetry (2.2) is missing for A r n,k (q).Note that the r-colored Eulerian polynomials A r n (t, 1) are not symmetric for r 3. We leave it as an open problem to give a combinatorial proof of identity (4.7).
where #T denotes the cardinality of a finite set T , in other words, the block T 0 contains at least two positive elements or only contains 0. Let S D (n, k) be the number of all type D signed partitions of n with 2k + 1 blocks, see an equivalent definition of S D (n, k) in[3].The numbers S D (n, k) are called the Stirling numbers of the second kind in type D.