1 Introduction

1.1 Stirling numbers of the second kind

Let \(\left\{ \begin{array}{ccccc} n \\ k\\ \end{array} \right\} \) denote the Stirling number of the second kind. It enumerates the number of partitions of a set with n elements consisting of k disjoint nonempty sets. It is well known that the Stirling number of the second kind satisfies the recurrence relation

$$\begin{aligned} \left\{ \begin{array}{ccccc} n \\ k\\ \end{array} \right\} =k\left\{ \begin{array}{ccccc} n-1 \\ k\\ \end{array} \right\} +\left\{ \begin{array}{ccccc} n-1 \\ k-1\\ \end{array} \right\} , \end{aligned}$$

where initial conditions \(\left\{ \begin{array}{ccccc} 0 \\ 0\\ \end{array} \right\} =1\) and \(\left\{ \begin{array}{ccccc} 0 \\ k\\ \end{array} \right\} =0\) for \(k\ge 1\) or \(k<0\). The triangular array \(\left[ \left\{ \begin{array}{ccccc} n \\ k\\ \end{array} \right\} \right] _{n,k\ge 0}\) is called the Stirling triangle of the second kind. Its row-generating function, i.e., the Bell polynomial, is defined to be \( B_n(x)=\sum _{k=0}^{n}\left\{ \begin{array}{ccccc} n \\ k\\ \end{array} \right\} x^{k}. \) There are many nice properties for the Stirling number and the Bell polynomial. For example:

  1. (i)

    The following identity is well known

    $$\begin{aligned} \left\{ \begin{array}{ccccc} n \\ k\\ \end{array} \right\} k!=\sum _{j=1}^k(-1)^{k-j}\left( {\begin{array}{c}k\\ j\end{array}}\right) j^n \end{aligned}$$
    (1.1)

    for \(n,k\ge 1\) (see [15], for instance).

  2. (ii)

    Let \(G_{n,k}=k!\left\{ \begin{array}{ccccc} n \\ k\\ \end{array} \right\} \). It counts the number of distinct ordered partitions of a set with n elements and satisfies the recurrence relation

    $$\begin{aligned} G_{n,k}=kG_{n-1,k}+kG_{n-1,k-1}. \end{aligned}$$

    Its row-generating function \(G_n(x)=\sum _{k=1}^nG_{n,k}x^k\) is called the geometric polynomial (see Tanny [39]).

  3. (iii)

    The Stirling triangle of the second kind \(\left[ \left\{ \begin{array}{ccccc} n \\ k\\ \end{array} \right\} \right] _{n,k\ge 0}\) is totally positive [9].

  4. (iv)

    The Bell polynomial \(B_n(x)\) has only real zeros and therefore is log-concave [42].

  5. (v)

    The polynomial \(B_{n+1}(x)B_{n-1}(x)-B_n^2(x)\) has no zeros in the right half plane [18].

  6. (vi)

    The exponential generating function for \(B_n(x)\) has a concise formula \( \sum _{n\ge 0}B_n(x)\frac{t^n}{n!}=e^{x(e^t-1)} \) [15].

  7. (vii)

    A Jacobi continued fraction expansion related to \(B_n(x)\) is given as

    $$\begin{aligned} \sum _{n\ge 0}B_n(x)\,t^n=\frac{1}{1-s_0t-\frac{r_1t^2}{1-s_1t-\frac{r_2t^2}{1-s_2t-\cdots }}}, \end{aligned}$$

    where \(s_n=n+x\) and \(r_{n+1}=(n+1)x\) for \(n\ge 0\) [19].

  8. (viii)

    The sequence \((B_n(q))_{n\ge 0}\) is q-log-convex, strongly q-log-convex, 3-q-log-convex and q-Stieltjes moment (see [13, 26, 44, 46, 49], for instance).

We refer the reader to [15] for more information of Stirling numbers and Bell polynomials.

1.2 Whitney numbers of the second kind

As a generalization of the partition lattice, the Dowling lattice \(Q_n(G)\) is a class of geometric lattices based on finite groups introduced by Dowling [16]. The Whitney number of the second kind, denoted by \(W_m(n, k)\), is the number of elements of corank k of \(Q_n(G)\). It satisfies the recurrence relation

$$\begin{aligned} W_m(n, k)=(mk+1)W_m(n-1, k)+W_m(n-1, k-1) \end{aligned}$$

with \(W_m(0,0)=1\). Its row-generating function \(D_n(m,x)=\sum _{k=0}^nW_m(n, k)x^k\) is called the Dowling polynomial by Benoumhani [3].

The Whitney number \(W_m(n,k)\) (resp. the Dowling polynomial) has many properties similar to those of the Stirling number \(\left\{ \begin{array}{ccccc} n \\ k\\ \end{array} \right\} \) (resp. the Bell polynomial). Let \(W^\diamond _{m}(n,k)=W_{m}(n, k)k!\). It satisfies the recurrence relation

$$\begin{aligned} W^\diamond _{m}(n, k)=(mk+1)W^\diamond _{m}(n-1, k)+kW^\diamond _{m}(n-1, k-1). \end{aligned}$$

Its row-generating function \(F_m(n,x)=\sum _{k=0}^nW^\diamond _{m}(n,k)x^k\) is called the Tanny-geometric polynomial in [3]. See [3, 4, 13, 26, 44, 46, 49] for some properties of Whitney numbers and Dowling polynomials such as explicit formulae, recurrence relations, log-concavity, real rootedness, generating functions, q-log-convexity, strong q-log-convexity, 3-q-log-convexity and q-Stieltjes moment property. We also refer the reader to [14] for more interesting properties concerning \(W_{m}(n, k)\).

1.3 A triangle of Riordan

Let \(a_{n,k}\) denote the number of set partitions of [n] in which exactly k of the blocks have been distinguished. It satisfies the recurrence relation

$$\begin{aligned} a_{n,k}=a_{n-1,k-1}+(k+1)a_{n-1,k}+(k+1)a_{n-1,k+1}, \end{aligned}$$

where initial conditions \(a_{0,0}=1\) and \(a_{n,k}=0\) unless \(0\le k\le n\) (see [34, A049020]). Its explicit formula can be written as

$$\begin{aligned} a_{n,k}=\sum _{i=0}^n\left\{ \begin{array}{ccccc} n \\ i\\ \end{array} \right\} \left( {\begin{array}{c}i\\ k\end{array}}\right) \end{aligned}$$

and \(a_{n,0}\) is exactly the famous Bell number. The triangle \([a_{n,k}]_{n,k}\) first arose in Riordan’s letter [31] and was also used to characterize Bell number by Aigner [1]. Its row-generating function can be written in terms of the next Jacobi continued fraction expansion

$$\begin{aligned} \sum _{n\ge 0}\sum _{k=0}^na_{n,k}x^kt^n=\frac{1}{1-s_0t-\frac{r_1t^2}{1-s_1t-\frac{r_2t^2}{1-s_2t-\cdots }}}, \end{aligned}$$

where \(s_n=n+1+x\) and \(r_{n+1}=(n+1)(x+1)\) for \(n\ge 0\). We refer the reader to [34, A049020] for more information of \(a_{n,k}\).

1.4 Structure of this paper

The Stirling numbers of the second kind are well known for their many nice properties (see [17, 21, 23, 47], for instance). Hence, the Stirling triangle formed by the Stirling numbers of the second kind has been widely studied. Note that both the Whitney triangle and the Riordan triangle mentioned previously have inner link with the Stirling triangle. The properties similar to those of the Stirling triangle have also been considered therefore. Motivated by the attractiveness of studying their properties, based on the fact of their inner relation, we consider a new class of triangle generalized from those three. The purpose of this paper is to provide a unified platform to study the generalized properties of the new class.

Let \(\mathbb {R}\) (resp. \(\mathbb {R}^{\ge 0}\)) denote the set of all (resp. nonnegative) real numbers. For \(\{\lambda ,a_1,a_2,b_1,b_2\}\subseteq \mathbb {R}\), define an array \(T=[T_{n,k}]_{n,k}\), which satisfies the recurrence relation:

$$\begin{aligned} T_{n,k}= & {} (b_1k+b_2)T_{n-1,k-1}+[(2\lambda b_1+a_1)k+a_2+\lambda ( b_1+b_2)] T_{n-1,k}\\&+\lambda (a_1+\lambda b_1)(k+1)T_{n-1,k+1}, \end{aligned}$$

where \(T_{0,0}=1\) and \(T_{n,k}=0\) unless \(0\le k\le n\). Let its row-generating function \(T_n(q)=\sum _{k\ge 0}T_{n,k}q^k\) for \(n\ge 0\). Obviously, we have

  • \(T_{n,k}=\left\{ \begin{array}{ccccc} n \\ k\\ \end{array} \right\} \) if \(a_1=b_2=1\) and \(a_2=b_1=\lambda =0\);

  • \(T_{n,k}=G_{n,k}\) if \(a_1=b_1=1,a_2=b_2=\lambda =0\);

  • \(T_{n,k}=W_m(n, k)\) if \(a_1=m,a_2=b_2=1\) and \(b_1=\lambda =0\);

  • \(T_{n,k}=W^\diamond _{m}(n, k)\) if \(a_1=m,a_2=b_1=1\) and \(b_2=\lambda =0\);

  • \(T_{n,k}=a_{n,k}\) if \(a_1=b_2=\lambda =1\) and \(a_2=b_1=0\);

  • \(T_{n,k}=\left( {\begin{array}{c}n\\ k\end{array}}\right) k!\) if \(b_1=a_2=1\) and \(a_1=b_2=\lambda =0\) ([34, A008279]);

  • \(T_{n,k}\) is A154602 in [34] if \(a_1=2\), \(b_2=\lambda =1\) and \(a_2=b_1=0\).

We call this array \([T_{n,k}]_{n,k}\) a Stirling–Whitney–Riordan triangle. The number \(T_{n,k}\) can be interpreted in terms of weighted Motzkin paths due to Flajolet [19]. Let \(u_k=b_1k+b_2+b_1,v_k=[(2\lambda b_1+a_1)k+a_2+\lambda ( b_1+b_2)]\) and \(w_k=\lambda (a_1+\lambda b_1)n\) for \(k\ge 0\). Then, \(T_{n,k}\) counts the number of weighted paths starting from the origin (0, 0) never falling below the x-axis and ending at (n, k) with up diagonal steps (1, 1) weighted \(u_{i-1}\), down diagonal steps \((1, -1)\) weighted \(w_{i+1}\) and horizontal steps (1, 0) weighted \(v_i\) on the line \(y=i\).

In Section 2, we prove that the Stirling–Whitney–Riordan triangle T is \(\mathbf{x} \)-totally positive with \(\mathbf{x} =(a_1,a_2,b_1,b_2,\lambda )\). In Section 3, using the method of zeros interlacing, we show that \(T_n(q)\) has only real roots and the Turán-type polynomial \(T_{n+1}(q)T_{n-1}(q)-T^2_n(q)\) is stable. In Section 4, we present explicit formulae for \(T_{n,k}\) and the exponential generating function of \(T_n(q)\). In Section 5, using addition formulae of the Stieltjes–Rogers type, we get a Jacobi continued fraction expansion of the ordinary generating function of \(T_n(q)\). Furthermore, we derive the \(\mathbf{x} \)-Stieltjes moment property and 3-\(\mathbf{x} \)-log-convexity of \(T_n(q)\) and hence show that the triangular convolution \(z_n=\sum _{k=0}^nT_{n,k}x_ky_{n-k}\) preserves Stieltjes moment property of sequences. Finally, in Section 6, for the first column \((T_{n,0})_{n\ge 0}\), we derive some properties similar to those of \((T_n(q))_{n\ge 0}.\)

2 Total positivity of the Stirling–Whitney–Riordan triangle

Let \(A=[a_{n,k}]_{n,k\ge 0}\) be a matrix of real numbers. It is called totally positive (TP for short) if all its minors are nonnegative. It is called \(TP_r\) if all minors of order \(k\le r\) are nonnegative. Let \(\mathbf{x} =(x_i)_{i\in {I}}\) be a set of indeterminates. A matrix A with entries being polynomials in \(\mathbb {R}[\mathbf{x} ]\) is x-totally positive (x-TP for short) if all its minors are polynomials with nonnegative coefficients in the indeterminates \(\mathbf{x} \) and is x-totally positive of order r (x-\(TP_r\) for short) if all its minors of order \(k\le r\) are polynomials with nonnegative coefficients in the indeterminates \(\mathbf{x} \). Total positivity of matrices is an important and powerful concept that arises often in various branches of mathematics (see the monographs [24, 30] for general details about total positivity). We also refer the reader to [7, 9,10,11, 20, 28, 36, 47, 51, 55] for total positivity in combinatorics. The following presents the total positivity of the Stirling–Whitney–Riordan triangle.

Theorem 2.1

The Stirling–Whitney–Riordan triangle T is \(\mathbf{x} \)-TP with \(\mathbf{x} =(a_1,a_2,b_1,b_2,\lambda )\).

Proof

Let \(\overline{T}\) denote the triangle obtained from T by deleting its first row. Assume that

$$\begin{aligned} J=\left[ \begin{array}{ccccc} s_0 &{} r_0 &{} &{} &{}\\ t_1 &{} s_1 &{} r_1 &{}\\ &{} t_2 &{} s_2 &{}r_2&{}\\ &{} &{} \ddots &{}\ddots &{} \ddots \\ \end{array}\right] , \end{aligned}$$

where \(r_n=b_1n+b_1+b_2\), \(s_n=(2\lambda b_1+a_1)n+a_2+\lambda ( b_1+b_2)\) and \(t_n=\lambda (a_1+\lambda b_1)n\). The recurrence relation:

$$\begin{aligned} T_{n,k}= & {} (b_1k+b_2)T_{n-1,k-1}+[(2\lambda b_1+a_1)k+a_2+\lambda ( b_1+b_2)] T_{n-1,k}\\&+\lambda (a_1+\lambda b_1)(k+1)T_{n-1,k+1} \end{aligned}$$

implies that

$$\begin{aligned} \overline{T}=TJ. \end{aligned}$$

It follows from [54, Theorem 2.1] that \(\mathbf{x} \)-total positivity of J implies that of T. In addition, J is \(\mathbf{x} \)-TP if and only if

$$\begin{aligned} J^{*}=\left[ \begin{array}{ccccc} s_0 &{} r^{*}_0 &{} &{} &{}\\ t^{*}_1 &{} s_1 &{} r^{*}_1 &{}\\ &{} t^{*}_2 &{} s_2 &{}r^{*}_2&{}\\ &{} &{} \ddots &{}\ddots &{} \ddots \\ \end{array}\right] \end{aligned}$$

is \(\mathbf{x} \)-TP, where \(r^{*}_n=\lambda (b_1n+b_1+b_2)\), \(s_n=(2\lambda b_1+a_1)n+a_2+\lambda ( b_1+b_2)\) and \(t^{*}_n=(a_1+\lambda b_1)n\). By [54, Proposition 3.3 (i)], we directly get that \(J^{*}\) is \(\mathbf{x} \)-TP with \(\mathbf{x} =(a_1,a_2,b_1,b_2,\lambda )\). In consequence, we show that the triangular matrix T is \(\mathbf{x} \)-TP with \(\mathbf{x} =(a_1,a_2,b_1,b_2,\lambda )\). The proof is complete. \(\square \)

3 Real rootedness and log-concavity of row-generating functions

Let \((a_k)_{k\ge 0}\) be a sequence of nonnegative numbers. The sequence \((a_k)_{k\ge 0}\) is log-concave if \(a_{i-1}a_{i+1}\le a_i^2\) for \(i\ge 1\). A basic approach to prove log-concavity is to use Newton’s inequalities: Suppose that the polynomial \(\sum _{k=0}^{n}a_kx^k\) has only real zeros. Then,

$$\begin{aligned} a_k^2\ge a_{k-1}a_{k+1}\left( 1+\frac{1}{k}\right) \left( 1+\frac{1}{n-k}\right) ,\quad k=1,2,\ldots ,n-1, \end{aligned}$$

and \((a_k)_{k\ge 0}\) is therefore log-concave (see Hardy, Littlewood and Pólya [22, p. 104]). Log-concave sequences and real-rooted polynomials often occur in combinatorics and have been extensively investigated. We refer the reader to Brenti [8], Stanely [35] and Wang and Yeh [43] for the log-concavity, Brändén [5, 6], Brenti [7], Liu and Wang [25], Wang and Yeh [42] for the real rootedness.

Following Wagner [41], a real polynomial is said to be standard if either it is identically zero or its leading coefficient is positive. Assume that both polynomials f and g only have real zeros. Let \(\{r_i\}\) and \(\{s_j\}\) be all zeros of f and g in nondecreasing order, respectively. We say that g interlaces f denoted by \(g\preceq f\) if \(\deg f=\deg g+1=n\) and

$$\begin{aligned} r_n\le s_{n-1}\le \cdots \le s_2\le r_2\le s_1\le r_1. \end{aligned}$$
(3.1)

For two interlacing polynomials, Fisk showed the following result.

Lemma 3.1

[18, Lemma 1.20] Let both f(x) and g(x) be standard real polynomials with only real zeros. Assume that \(\deg (f(x))=n\) and all real zeros of f(x) are \(s_1, \ldots , s_n\). If \(\deg (g)=n-1\) and we write

$$\begin{aligned} g(x)=\sum _{i=1}^n\frac{c_if(x)}{x-s_i}, \end{aligned}$$

then \(g\preceq f\) if and only if all \(c_i\) are positive.

A real polynomial is weakly (Hurwitz) stable if all of its zeros lie in the closed left half of the complex plane. See [27, Chapter 9] for deep surveys on the stability theory of polynomials. Let \(L_n(x)\) be the nth Legendre polynomial. Turán-type inequalities [40] state that

$$\begin{aligned} L^2_n(x)-L_{n+1}(x)L_{n-1}(x)>0\,\, \text {for}\, -1<x<1. \end{aligned}$$

In 1948, Szegö gave four different proofs of the famous Turán-type inequality on Legendre polynomials [38]. It has been proved that many important (orthogonal) polynomials and special functions satisfy some Turán-type inequalities (see [2], for instance).

The following gives the zeros properties related to \(T_{n}(x)\).

Theorem 3.2

Let \(T_n(x)\) be the row-generating function of the Stirling–Whitney–Riordan triangle T. If \(\{\lambda ,a_1,a_2,b_1,b_2\}\subseteq \mathbb {R}^{\ge 0}\) and \(a_1(b_1+b_2)> b_1a_2\)Footnote 1, then

  1. (i)

    \(T_n(x)\) has only simple real zeros in \((-\lambda -\frac{a_1}{b_1},-\lambda )\)Footnote 2 and \(T_{n-1}(x)\preceq T_n(x)\) for \(n\ge 1\). Therefore, \(T_n(x)\) is log-concave for \(n\ge 1\).

  2. (ii)

    The Turán-type polynomial \(T_{n+1}(x)T_{n-1}(x)-(T_n(x))^2\) is a weakly stable polynomial for \(n\ge 1\).

Proof

(i) It follows from the recurrence relation:

$$\begin{aligned} T_{n,k}= & {} (b_1k+b_2)T_{n-1,k-1}+[(2\lambda b_1+a_1)k+a_2+\lambda ( b_1+b_2)] T_{n-1,k}\\&+\lambda (a_1+\lambda b_1)(k+1)T_{n-1,k+1} \end{aligned}$$

that

$$\begin{aligned} T_{n}(x)=[a_2+(b_1+b_2)(x+\lambda )]T_{n-1}(x)+(x+\lambda )[a_1+b_1(x+\lambda )]T'_{n-1}(x)\nonumber \\ \end{aligned}$$
(3.2)

for \(n\ge 1\). We will prove by induction on n that \(T_n(x)\) has only simple real zeros in \((-\lambda -\frac{a_1}{b_1},-\lambda )\) and \(T_{n-1}(x)\preceq T_n(x)\) for \(n\ge 1\). Clearly, \(T_0(x)=1\).

Case 1: Assume \(b_1\ne 0\). For \(n=1\), we have

$$\begin{aligned} T_1(x)=a_2+(b_1+b_2)(x+\lambda ). \end{aligned}$$

Obviously, it follows from \(a_1(b_1+b_2)\!>\! b_1a_2\) that \(T_1(x)\) has only real zero in \((-\lambda -\frac{a_1}{b_1},-\lambda )\). Suppose for \(n\ge 2\) that \(T_{n-1}(x)\) has \(n-1\) real zeros denoted by

$$\begin{aligned} -\lambda> s_1> s_2>\ldots> s_{n-1}>-\lambda -\frac{a_1}{b_1}. \end{aligned}$$

Then, by the recurrence relation (3.2), we have

$$\begin{aligned} sign[ T_n(s_k)]=(-1)^k. \end{aligned}$$

In consequence, \(T_n(x)\) has n simple real zeros denoted by \(r_1> r_2>\ldots > r_n\) such that

$$\begin{aligned} r_1> s_1> r_2> s_2>\ldots> s_{n-1}> r_n. \end{aligned}$$
(3.3)

On the other hand, by the recurrence relation (3.2), we have

$$\begin{aligned} T_n(-\lambda )=a_2T_{n-1}(-\lambda )>0,\quad T_n\left( -\lambda -\frac{a_1}{b_1}\right) =\frac{a_2b_1-a_1(b_1+b_2)}{b_1}T_{n-1}\left( -\lambda -\frac{a_1}{b_1}\right) , \end{aligned}$$

which imply \(r_1<-\lambda \) and \(r_n>-\lambda -\frac{a_1}{b_1}\).

Case 2: Assume \(b_1=0\). For \(n=1\), we have

$$\begin{aligned} T_1(x)=a_2+b_2(x+\lambda ). \end{aligned}$$

Obviously, it follows from \(a_1b_2> b_1a_2\) that \(T_1(x)\) has only real zero in \((-\infty ,-\lambda )\). Suppose for \(n\ge 2\) that \(T_{n-1}(x)\) has \(n-1\) real zeros denoted by

$$\begin{aligned} -\lambda> s_1> s_2>\ldots > s_{n-1}. \end{aligned}$$

Then, by the recurrence relation (3.2), we have

$$\begin{aligned} sign[ T_n(s_k)]=(-1)^k. \end{aligned}$$

This implies that \(T_n(x)\) has n simple real zeros denoted by \(r_1> r_2>\ldots > r_n\) such that

$$\begin{aligned} r_1> s_1> r_2> s_2>\ldots> s_{n-1}> r_n. \end{aligned}$$
(3.4)

In addition, by the recurrence relation (3.2), we have

$$\begin{aligned} T_n(-\lambda )=a_2T_{n-1}(-\lambda )>0. \end{aligned}$$

Thus, \(r_1<-\lambda \). This completes the proof of (i).

(ii) By (3.2), we deduce that

$$\begin{aligned}&T_{n+1}(x)T_{n-1}(x)-(T_n(x))^2\nonumber \\&\quad =[a_2+(b_1+b_2)(x+\lambda )]T_n(x)T_{n-1}(x)+(x+\lambda )\nonumber \\&\qquad [a_1+b_1(x+\lambda )]T'_{n}(x)T_{n-1}(x)-\nonumber \\&\qquad [a_2+(b_1+b_2)(x+\lambda )]T_n(x)T_{n-1}(x)-(x+\lambda )[a_1+b_1(x+\lambda )]T'_{n-1}(x)T_{n}(x) \nonumber \\&\quad =(x+\lambda )[a_1+b_1(x+\lambda )]\left[ T'_{n}(x)T_{n-1}(x)-T'_{n-1}(x)T_{n}(x)\right] \nonumber \\&\quad =-(x+\lambda )[a_1+b_1(x+\lambda )][T_n(x)]^2\left( \frac{T_{n-1}(x)}{T_{n}(x)}\right) '. \end{aligned}$$
(3.5)

By (i), assume that \(T_n(x)\) has n negative zeros as \( r_1> r_2>\ldots > r_n\). It follows from \(T_{n-1}(x)\preceq T_n(x)\) in (i) and Lemma 3.1 that we get

$$\begin{aligned} \frac{T_{n-1}(x)}{T_n(x)}=\sum _{k=1}^n\frac{t_k}{x-r_k}, \end{aligned}$$
(3.6)

where \(t_k> 0\) for \(1\le k\le n\). Combining (3.5) and (3.6) gives the iterated Turán-type polynomial

$$\begin{aligned} T_{n+1}(x)T_{n-1}(x)-(T_n(x))^2= & {} (x+\lambda )[a_1+b_1(x+\lambda )][T_n(x)]^2\sum _{k=1}^n\frac{t_k}{(x-r_k)^2}. \end{aligned}$$

Thus, in order get that \(T_{n+1}(x)T_{n-1}(x)-(T_n(x))^2\) is sable, it suffices to prove that

$$\begin{aligned} \sum _{k=1}^n\frac{t_k}{(x-r_k)^2}\ne 0 \end{aligned}$$

for x in the right half plane. Obviously, \(\sum _{k=1}^n\frac{t_k}{(x-r_k)^2}>0\) for \(x\ge 0\). In addition, for \(x=u+vi\) with \(u>0\) and \(v\ne 0\),

$$\begin{aligned}&Im\left( \sum _{k=1}^n\frac{t_k}{(x-r_k)^2}\right) =-2v\sum _{k=1}^n\frac{t_k(u-r_k)}{(u-r_k)^2+v^2}\ne 0 \end{aligned}$$

since \(\frac{t_k(u-r_k)}{(u-r_k)^2+v^2}>0\) for each \(k\in [1,n]\). In consequence, we get that

$$\begin{aligned} \sum _{k=1}^n\frac{t_k}{(x-r_k)^2} \end{aligned}$$

has no zeros in the right half plane. So does the polynomial \(T_{n+1}(x)T_{n-1}(x)-(T_n(x))^2\). We complete the proof of (ii). \(\square \)

4 Exponential generating function and explicit formula

In this section, we will present the exponential generating function of \(T_n(q)\) and use it to derive an explicit formula for the Stirling–Whitney–Riordan triangle T as follows.

Theorem 4.1

Let \(T_n(q)\) be the row-generating function of T with \(a_1^2+b_1^2\ne 0\).

  1. (i)

    The exponential generating function of \(T_n(q)\) is given asFootnote 3

    $$\begin{aligned} \sum _{n\ge 0}T_n(q)\frac{t^n}{n!}= & {} e^{a_2t}\left[ 1+\frac{b_1(q+\lambda )(1-e^{a_1t})}{a_1}\right] ^{-(1+\frac{b_2}{b_1})}. \end{aligned}$$
  2. (ii)

    An explicit formula for \(T_{n,k}\) can be written as

    $$\begin{aligned} T_{n,k}= & {} {\left\{ \begin{array}{ll} \sum _{i\ge k}\frac{\prod _{j=1}^i(b_2+b_1j)}{a^i_1}\times \left( {\begin{array}{c}i\\ k\end{array}}\right) {\lambda }^{i-k}\times \frac{1}{i!}\sum _{j=0}^i\left( {\begin{array}{c}i\\ j\end{array}}\right) (-1)^{i-j}(a_2+a_1j)^n, &{}\text {for}\quad a_1\ne 0\\ \sum _{i\ge k}\prod _{j=1}^i(b_2+b_1j)\times \left( {\begin{array}{c}n\\ i\end{array}}\right) \left( {\begin{array}{c}i\\ k\end{array}}\right) {\lambda }^{i-k}a^{n-i}_2,&\text {for}\quad a_1=0. \end{array}\right. } \end{aligned}$$

Proof

Let the exponential generating function

$$\begin{aligned} \mathcal {T}(q,t)=\sum _{n,k\ge 0}T_{n,k}q^k\frac{t^n}{n!}=\sum _{n\ge 0}T_n(q)\frac{t^n}{n!}. \end{aligned}$$

Then, by the recurrence relation:

$$\begin{aligned} T_{n,k}= & {} (b_1k+b_2)T_{n-1,k-1}+[(2\lambda b_1+a_1)k+a_2\\&+\lambda (b_1+b_2)] T_{n-1,k}+\lambda (a_1+\lambda b_1)(k+1)T_{n-1,k+1}, \end{aligned}$$

we have the next partial differential equation

$$\begin{aligned} \mathcal {T}_t(q,t)-[b_1(q+\lambda )+a_1](q+\lambda )\mathcal {T}_q(q,t)=[a_2+(b_1+b_2)(q+\lambda )]\mathcal {T}(q,t) \end{aligned}$$

with the initial condition \(\mathcal {T}(q,0)=1\). It is routine to check that

$$\begin{aligned} \mathcal {T}(q,t)=e^{a_2t}\left[ 1+\frac{b_1(q+\lambda )(1-e^{a_1t})}{a_1}\right] ^{-\left( 1+\frac{b_2}{b_1}\right) } \end{aligned}$$
(4.1)

is a solution of the above partial differential with the initial condition.

(ii) Its proof will be divided into the following three cases.

Case 1: \(a_1b_1\ne 0\). We have

$$\begin{aligned} \sum _{n,k\ge 0} T_{n,k}q^k\frac{t^n}{n!}= & {} e^{a_2t}\left[ 1+\frac{b_1(q+\lambda )(1-e^{a_1t})}{a_1}\right] ^{-(1+\frac{b_2}{b_1})}\\= & {} e^{a_2t}\left[ 1-\frac{b_1(q+\lambda )(e^{a_1t}-1)}{a_1}\right] ^{-(1+\frac{b_2}{b_1})}\\= & {} e^{a_2t}\sum _{i\ge 0}\left( {\begin{array}{c}\frac{b_2}{b_1}+i\\ i\end{array}}\right) \left[ \frac{b_1(q+\lambda )(e^{a_1t}-1)}{a_1}\right] ^i\\= & {} e^{a_2t}\sum _{i\ge 0}\left( {\begin{array}{c}\frac{b_2}{b_1}+i\\ i\end{array}}\right) \left( \frac{b_1(q+\lambda )}{a_1}\right) ^i\sum _{j=0}^i\left( {\begin{array}{c}i\\ j\end{array}}\right) (-1)^{i-j}e^{a_1jt}\\= & {} \sum _{i\ge 0}\left( {\begin{array}{c}\frac{b_2}{b_1}+i\\ i\end{array}}\right) \left( \frac{b_1(q+\lambda )}{a_1}\right) ^i\sum _{j=0}^i\left( {\begin{array}{c}i\\ j\end{array}}\right) (-1)^{i-j}e^{(a_2+a_1j)t}\\= & {} \sum _{i\ge 0}\frac{\prod _{j=1}^i(b_2+b_1j)}{i!}\times \frac{(q+\lambda )^i}{a^i_1}\sum _{j=0}^i\left( {\begin{array}{c}i\\ j\end{array}}\right) (-1)^{i-j}\sum _{n\ge 0}\frac{(a_2+a_1j)^nt^n}{n!}, \end{aligned}$$

which implies

$$\begin{aligned} T_{n,k}=\sum _{i\ge k}\frac{\prod _{j=1}^i(b_2+b_1j)}{a^i_1}\times \left( {\begin{array}{c}i\\ k\end{array}}\right) {\lambda }^{i-k}\times \frac{1}{i!}\sum _{j=0}^i\left( {\begin{array}{c}i\\ j\end{array}}\right) (-1)^{i-j}(a_2+a_1j)^n. \end{aligned}$$

Case 2: \(b_1=0\) and \(a_1\ne 0\). We have

$$\begin{aligned} \sum _{n,k\ge 0} T_{n,k}q^k\frac{t^n}{n!}= & {} e^{a_2t+\left[ \frac{b_2(q+\lambda )(e^{a_1t}-1)}{a_1}\right] }\\= & {} e^{a_2t}\sum _{i\ge 0}\left[ \frac{b_2(q+\lambda )(e^{a_1t}-1)}{a_1}\right] ^i\times \frac{1}{i!}\\= & {} e^{a_2t}\sum _{i\ge 0}\left( \frac{b_2(q+\lambda )}{a_1}\right) ^i\times \frac{1}{i!}\sum _{j=0}^i\left( {\begin{array}{c}i\\ j\end{array}}\right) (-1)^{i-j}e^{a_1jt}\\= & {} \sum _{i\ge 0}\left( \frac{b_2}{a_1}\right) ^i(q+\lambda )^i\times \frac{1}{i!}\sum _{j=0}^i\left( {\begin{array}{c}i\\ j\end{array}}\right) (-1)^{i-j}e^{(a_2+a_1j)t}\\= & {} \sum _{i\ge 0}\left( \frac{b_2}{a_1}\right) ^i(q+\lambda )^i\times \frac{1}{i!}\sum _{j=0}^i\left( {\begin{array}{c}i\\ j\end{array}}\right) (-1)^{i-j}\sum _{n\ge 0}\frac{(a_2+a_1j)^nt^n}{n!}. \end{aligned}$$

Obviously,

$$\begin{aligned} T_{n,k}=\sum _{i\ge k}\frac{b^i_2}{a^i_1}\left( {\begin{array}{c}i\\ k\end{array}}\right) {\lambda }^{i-k}\times \frac{1}{i!}\sum _{j=0}^i\left( {\begin{array}{c}i\\ j\end{array}}\right) (-1)^{i-j}(a_2+a_1j)^n. \end{aligned}$$

Case 3: \(a_1=0\) and \(b_1\ne 0\). We have

$$\begin{aligned} \sum _{n,k\ge 0} T_{n,k}q^k\frac{t^n}{n!}= & {} e^{a_2t}\left[ 1-b_1(q+\lambda )t\right] ^{-(1+\frac{b_2}{b_1})}\\= & {} e^{a_2t}\sum _{i\ge 0}\left( {\begin{array}{c}\frac{b_2}{b_1}+i\\ i\end{array}}\right) \left[ b_1(q+\lambda )t\right] ^i\\= & {} e^{a_2t}\sum _{i\ge 0}\prod _{j=1}^i(b_2+b_1j)\times (q+\lambda )^i\frac{t^i}{i!}. \end{aligned}$$

Thus, we get

$$\begin{aligned} T_{n,k}=\sum _{i\ge k}\prod _{j=1}^i(b_2+b_1j)\times \left( {\begin{array}{c}n\\ i\end{array}}\right) \left( {\begin{array}{c}i\\ k\end{array}}\right) {\lambda }^{i-k}a^{n-i}_2. \end{aligned}$$

This completes the proof. \(\square \)

5 Stieltjes moment property and continued fractions

In this section, we will present a continued fraction expansion of \(\sum _{n\ge 0}T_n(q)t^n\) and demonstrate a Stieltjes moment property for \(T_n(q)\). Continued fraction is an important tool in combinatorics, which is closely related to many aspects, e.g., combinatorial lattice paths, combinatorial interpretations, combinatorial identities, combinatorial positivity, determinants of sequences, and so on. We refer the reader to Flajolet [19] for more information concerning continued fraction expansions related to many important combinatorial objects.

Continued fraction plays an important role in studying Hankel-total positivity and Stieltjes moment sequences. Given a sequence \(\alpha =(a_k)_{k\ge 0}\), define its Hankel matrix \(H(\alpha )\) by

$$\begin{aligned} H(\alpha )=[a_{i+j}]_{i,j\ge 0}= \left[ \begin{array}{ccccc} a_0 &{} a_1 &{} a_2 &{} a_3 &{} \cdots \\ a_1 &{} a_2 &{} a_3 &{} a_4 &{} \cdots \\ a_2 &{} a_3 &{} a_4 &{} a_5 &{}\cdots \\ a_3 &{} a_4 &{} a_5 &{} a_6 &{} \cdots \\ \vdots &{}\vdots &{} \vdots &{} \vdots &{} \ddots \\ \end{array}\right] . \end{aligned}$$

We say that \(\alpha \) is a Stieltjes moment (SM for short) sequence if it has the form

$$\begin{aligned} a_k=\int _0^{+\infty }x^kd\mu (x), \end{aligned}$$
(5.1)

where \(\mu \) is a nonnegative measure on \([0,+\infty )\) (see [30, Theorem 4.4], for instance). The Stieltjes moment problem is one of classical moment problems and arises naturally in many branches of mathematics [33, 45]. It is well known that the following are equivalent:

  1. (i)

    \(\alpha \) is a Stieltjes moment sequence.

  2. (ii)

    Its Hankel matrix \(H(\alpha )\) is TP.

  3. (iii)

    Its generating function has the Stieltjes continued fraction expansion

    $$\begin{aligned} \sum _{n\ge 0}a_nz^n=\frac{1}{1-\frac{\beta _0z}{1-\frac{\beta _1z}{1-\frac{\beta _2z}{1-\cdots }}}} \end{aligned}$$

    with \(\beta _i\ge 0\) for \(i\ge 0\).

  4. (iv)

    Positivity characterization: \(\sum _{n=0}^Nc_na_n\ge 0\) for every polynomial \( \sum _{n=0}^Nc_nq^n\ge 0 \) on \([0,\infty )\).

Let \(\mathbf{x} =(x_i)_{i\in {I}}\) be a set of indeterminates. A polynomial sequence \((\alpha _n(\mathbf{x} ))_{n\ge 0}\) in \(\mathbb {R}[\mathbf{x} ]\) is called a \(\mathbf{x} \) -Stieltjes moment (\(\mathbf{x} \)-SM for short) sequence if its associated infinite Hankel matrix is \(\mathbf{x} \)-TP, see Zhu [51, 54], for instance. When \((\alpha _n(\mathbf{x} ))_{n\ge 0}\) is a sequence of real numbers, \(\mathbf{x} \)-SM sequence reduces to the classical Stieltjes moment sequence. For \(\mathbf{x} \)-SM sequences, the following criterion was proved in [51, 54].

Lemma 5.1

[54] Let \(\{s_n(\mathbf{x} ),r_{n}(\mathbf{x} ),H_n(\mathbf{x} )\}\subseteq \mathbb {R}^{\ge 0}[\mathbf{x} ]\) for \(n\in \mathbb {N}\) and

$$\begin{aligned} \sum _{n\ge 0} H_n(\mathbf{x} )t^n=\frac{1}{1-s_0(\mathbf{x} )t-\frac{r_1(\mathbf{x} )t^2}{1-s_1(\mathbf{x} )t-\frac{r_2(\mathbf{x} )t^2}{1-\cdots }}}. \end{aligned}$$

If there exists \(\{\lambda _n(\mathbf{x} ),u_n(\mathbf{x} ),v_n(\mathbf{x} )\}\subseteq \mathbb {R}^{\ge 0}[\mathbf{x} ]\) such that \(s_n=\lambda _n+u_n+v_n\) and \(r_{n+1}=u_{n+1}v_{n}\) for \(n\ge 0\), then polynomials \(H_n(\mathbf{x} )\) form a \(\mathbf{x} \)-SM sequence for \(n\ge 0\).

In order to compute continued fraction, we need the following addition formulae of the Stieltjes–Rogers type.

Lemma 5.2

[32, 37] For a sequence \((\alpha _n)_{n\ge 0}\), define the function

$$\begin{aligned} h(x)= & {} \sum _{n\ge 0}\alpha _n\frac{x^n}{n!}. \end{aligned}$$

If there exist two sequences \((e_{n})_{n\ge 0}\) and \((w_n)_{n\ge 0}\) such that the generating function

$$\begin{aligned} h(x+y)=\sum _{n\ge 0}w_kf_k(x)f_k(y), \end{aligned}$$

where

$$\begin{aligned} f_k(x)=\frac{x^k}{k!}+e_{k+1}\frac{x^{k+1}}{(k+1)!}+O(x^{k+2}), \end{aligned}$$

then we have

$$\begin{aligned} \sum _{n\ge 0}\alpha _n t^n=\frac{1}{1-s_0t-\frac{r_1t^2}{1-s_1t-\frac{r_2t^2}{1-s_2t-\frac{r_3t^2}{1-s_3t-\cdots }}}}, \end{aligned}$$

where \(s_n=e_{n+1}-e_n\) and \(r_{n+1}=w_{n+1}/w_n\) for \(n\ge 0\).

If a polynomial sequence \((A_n(q))_{n\ge 0}\) in an indeterminate q is q-SM, then its triangular convolution preserves the SM property in terms of the next result.

Lemma 5.3

[44] For \(n\in \mathbb {N},\) let \(A_n(q)=\sum _{k=0}^{n}A_{n,k}q^k\) be the nth row-generating function of a matrix \([A_{n,k}]_{n,k\ge 0}\). Assume that \((A_n(q))_{n\ge 0}\) is a SM sequence for any fixed \(q\ge 0\). If both \((x_n)_{n\ge 0}\) and \((y_n)_{n\ge 0}\) are SM sequences, then so is \((z_n)_{n\ge 0}\) defined by

$$\begin{aligned} z_n=\sum _{k=0}^{n}A_{n,k}x_ky_{n-k}. \end{aligned}$$
(5.2)

For the \(\mathbf{x} \)-SM property, one necessary condition is \(\mathbf{x} \)-log-convexity. For a polynomial sequence \((f_n(\mathbf{x} ))_{n\ge 0}\), it is \(\mathbf{x} \) -log-convex if

$$\begin{aligned} f_{n+1}(\mathbf{x} )f_{n-1}(\mathbf{x} )- f_n(\mathbf{x} )^2 \end{aligned}$$

is a polynomial with nonnegative coefficients for \(n\ge 1\). Define the operator \(\mathcal {L}\) which maps a polynomial sequence \((f_n(\mathbf{x} ))_{n\ge 0}\) to another polynomial sequence \((g_i(\mathbf{x} ))_{i\ge 1}\) given by

$$\begin{aligned} g_i(\mathbf{x} ):=f_{i-1}(\mathbf{x} )f_{i+1}(\mathbf{x} )-f_i(\mathbf{x} )^2. \end{aligned}$$

Then, the \(\mathbf{x} \)-log-convexity of \((f_n(\mathbf{x} ))_{n\ge 0}\) is equivalent to the \(\mathbf{x} \)-positivity of \(\mathcal {L}\{f_i(\mathbf{x} )\}\), i.e., the coefficients of \(g_i(\mathbf{x} )\) are nonnegative for all \(i\ge 1\). Generally, we say that \((f_i(\mathbf{x} ))_{i\ge 0}\) is k-\(\mathbf{x} \) -log-convex if the coefficients of \(\mathcal {L}^m\{f_i(\mathbf{x} )\}\) are nonnegative for all \(m\le k\), where \(\mathcal {L}^m=\mathcal {L}(\mathcal {L}^{m-1})\).

Lemma 5.4

[54] If the Hankel matrix \([A_{i+j}(\mathbf{x} )]_{i,j\ge 0}\) is \(\mathbf{x} \)-TP\(_4\), then the sequence \((A_n(\mathbf{x} ))_{n\ge 0}\) is 3-\(\mathbf{x} \)-log-convex.

Obviously, if \((A_n(\mathbf{x} ))_{n\ge 0}\) is a \(\mathbf{x} \)-SM sequence, then \([A_{i+j}(\mathbf{x} )]_{i,j}\) is \(\mathbf{x} \)-Hankel-TP\(_4\). Thus, it is 3-\(\mathbf{x} \)-log-convex by Lemma 5.4.

If \(\mathbf{x} \) is an indeterminate q, then it has been proved that many famous polynomials have the q-log-convexity, e.g., the Bell polynomials, the classical Eulerian polynomials, the Narayana polynomials of types A and B, Jacobi–Stirling polynomials, and so on (see Liu and Wang [26], Chen et al. [12], Zhu [46,47,48,49, 53], for instance). These polynomials also have 3-q-log-convexity (see Zhu [50, 52]).

The main result of this section is the following.

Theorem 5.5

Let \(T_n(q)\) be the row-generating function of T. Then, we have the next results.

  1. (i)

    The ordinary generating function of \(T_n(q)\) has a Jacobi continued fraction expression

    $$\begin{aligned} \sum _{n\ge 0}T_n(q)t^n=\frac{1}{1-s_0t-\frac{r_1t^2}{1-s_1t-\frac{r_2t^2}{1-s_2t-\cdots }}}, \end{aligned}$$

    where \(s_n=a_2+a_1n+[b_1(2n+1)+b_2]\lambda \) and \(r_{n+1}=[b_1(n+1)+b_2]\lambda (b_1\lambda +a_1)(n+1)\) for \(n\ge 0\).

  2. (ii)

    The sequence \((T_{n}(q))_{n\ge 0}\) is \(\mathbf{x} \)-SM and 3-\(\mathbf{x} \)-log-convex with \(\mathbf{x} =(a_1,a_2,b_1,b_2,\lambda ,q)\).

  3. (iii)

    The convolution \(z_n=\sum _{k\ge 0}T_{n,k}x_ky_{n-k}\) preserves SM property if \(\{\lambda ,a_1,a_2,b_1,b_2\}\subseteq \mathbb {R}^{\ge 0}\).

  4. (iv)

    We have Hankel determinants

    $$\begin{aligned} \det _{0\le i,j\le n-1}(T_{i+j}(q))= & {} r_1^{n-1}r_2^{n-2}\ldots r_{n-2}^2r_{n-1} \end{aligned}$$

    and

    $$\begin{aligned} \det _{0\le i,j\le n-1}(T_{i+j+1}(q))= & {} r_1^{n-1}r_2^{n-2}\cdots r_{n-2}^2r_{n-1}Q_n, \end{aligned}$$

    where \((Q_n)_{n\ge 0}\) is defined by \(Q_{n+1}=s_nQ_n-r_nQ_{n-1}\) with \(Q_0=1\) and \(Q_1=s_0\).

Proof

Let the exponential generating function

$$\begin{aligned} G(q,t)=\sum _{n\ge 0} T_n(q)\frac{t^n}{n!}. \end{aligned}$$

Then,

$$\begin{aligned} G(q,t)=e^{a_2t}\left[ 1+\frac{b_1(q+\lambda )(1-e^{a_1t})}{a_1}\right] ^{-(1+\frac{b_2}{b_1})}. \end{aligned}$$

In the following, we only need to consider the case \(a_1b_1\ne 0\). (For the case \(a_1b_1=0\), it is the corresponding limits in terms of continuity.) Let \(\gamma =\frac{b_1}{a_1}\), \(\beta =\frac{b_2}{b_1}\) and \(p=q+\lambda \). Assume that

$$\begin{aligned} h(x+y)=e^{(x+y)a_2}\left[ 1-\gamma p(e^{a_1(x+y)}-1)\right] ^{-(1+\beta )}. \end{aligned}$$

Then,

$$\begin{aligned} h(x+y)= & {} e^{(x+y)a_2}\left\{ [1-\gamma p(e^{a_1x}-1)][1-\gamma p(e^{a_1y}-1)]\right. \\&\left. -\gamma p(1+\gamma p)(e^{a_1y}-1)(e^{a_1x}-1)\right\} ^{-(1+\beta )}\\= & {} \sum _{k\ge 0}k!a_1^{2k}\langle 1+\beta \rangle _k(\gamma p)^k(1+\gamma p)^kf_k(x)f_k(y), \end{aligned}$$

where \(\langle 1+\beta \rangle _k=(1+\beta )(2+\beta )\cdots (k+\beta )\) and

$$\begin{aligned} f_k(x)= & {} \frac{1}{k!a_1^k}e^{xa_2}(e^{a_1x}-1)^k\left[ 1-\gamma p(e^{a_1x}-1)\right] ^{-(1+\beta +k)}\\= & {} \frac{1}{k!a_1^k}\left( 1+a_2 x+\frac{(a_2 x)^2}{2}+\cdots \right) \left( a_1x+\frac{a^2_1x^2}{2}+\cdots \right) ^k\left[ 1+\gamma p(1+\beta )a_1x+\cdots \right] \\= & {} \frac{x^k}{k!}+\frac{x^{k+1}}{(k+1)!}\left[ a_2+\frac{a_1k}{2}+(1+\beta +k)\gamma pa_1\right] (k+1)+O(x^{k+2}). \end{aligned}$$

By Lemma 5.2, we have

$$\begin{aligned}&w_k=k!a_1^{2k}\langle 1+\beta \rangle _k(\gamma p)^k(1+\gamma p)^k, \,\\&\quad e_{k+1}=\left[ a_2+\frac{a_1k}{2}+(1+\beta +k)\gamma pa_1\right] (k+1). \end{aligned}$$

Thus, we get

$$\begin{aligned} s_k= & {} e_{k+1}-e_k=a_2+a_1k+(2k+1+\beta )\gamma pa_1, \, \\ r_{k+1}= & {} \frac{w_{k+1}}{w_k}=a_1^2(k+1+\beta )\gamma p(\gamma p+1)(k+1) \end{aligned}$$

for \(k\ge 0\). So

$$\begin{aligned} \sum _{n\ge 0}T_n(q)t^n=\frac{1}{1-s_0t-\frac{r_1t^2}{1-s_1t-\frac{r_2t^2}{1-s_2t-\cdots }}}. \end{aligned}$$

where \(s_n=a_2+a_1n+[b_1(2n+1)+b_2](q+\lambda )\) and \(r_{n+1}=[b_1(n+1)+b_2](q+\lambda )[b_1(q+\lambda )+a_1](n+1)\) for \(n\ge 0\).

Let \(v_{n}=(nb_1+b_2+b_1)(q+\lambda )\) and \(u_{n}=n[a_1+b_1(q+\lambda )]\) for \(n\ge 0\). It is obvious that \(s_n=a_2+u_{n}+v_{n}\) and \(t_n=v_{n}u_{n+1}\) for \(n\ge 0\). It follows from Lemma 5.1 that \((T_n(q))_{n\ge 0}\) is a \(\mathbf{x} \)-SM sequence with \(\mathbf{x} =(a_1,a_2,b_1,b_2,\lambda ,q)\). Then \([T_{i+j}(q)]_{i,j}\) is \(\mathbf{x} \)-TP\(_4\). Thus, by Lemma 5.4, we immediately have \((T_n(q))_{n\ge 0}\) is 3-\(\mathbf{x} \)-log-convex. In addition, it follows from Lemma 5.3 that the Stirling–Whitney–Riordan triangle convolution

$$\begin{aligned} z_n=\sum _{k=0}^{n}T_{n,k} x_ky_{n-k},\quad n=0,1,2,\ldots \end{aligned}$$

preserves the SM property. Finally, for (iv), it follows from the next general criterion (see [29], for instance): If the generating function of \((u_i)_{i\ge 0}\) can be expressed by

$$\begin{aligned} \sum \limits _{i=0}^{\infty }u_ix^i=\frac{\displaystyle u_0}{\displaystyle 1- s_0x-\frac{\displaystyle t_1x^2}{\displaystyle 1-s_1x-\frac{\displaystyle t_2x^2}{\displaystyle 1-s_2x-\ldots }}}, \end{aligned}$$

then

$$\begin{aligned} \det _{0\le i,j\le n-1}(u_{i+j})= & {} u_0^nt_1^{n-1}t_2^{n-2}\ldots t_{n-2}^2t_{n-1} \end{aligned}$$

and

$$\begin{aligned} \det _{0\le i,j\le n-1}(u_{i+j+1})= & {} u_0^nt_1^{n-1}t_2^{n-2}\cdots t_{n-2}^2t_{n-1}q_n, \end{aligned}$$

where \((q_n)_{n\ge 0}\) is defined by \(q_{n+1}=s_nq_n-t_nq_{n-1}\) with \(q_0=1\) and \(q_1=s_0\). \(\square \)

Remark 5.6

Note that for a Jacobi continued fraction expansion, it has a general combinatorial interpretation in terms of weighted Motzkin paths due to Flajolet [19]. For the Jacobi continued fraction expansion

$$\begin{aligned} \sum _{n\ge 0}T_n(q)\,t^n=\frac{1}{1-s_0t-\frac{r_1t^2}{1-s_1t-\frac{r_2t^2}{1-s_2t-\frac{r_3t^2}{1-s_3t-\cdots }}}} \end{aligned}$$

with \(s_n=a_2+a_1n+[b_1(2n+1)+b_2](q+\lambda )\) and \(r_{n+1}=(n+1)[b_1(n+1)+b_2](q+\lambda )[b_1(q+\lambda )+a_1]\) for \(n\ge 0\), we can interpret \(T_n(q)\) as follows: Weighted Motzkin paths start from the origin (0, 0) never falling below the x-axis and end at (n, 0) with up diagonal steps (1, 1) weighted 1, down diagonal steps \((1, -1)\) weighted \(r_{i+1}\) and horizontal steps (1, 0) weighted \(s_i\) on the line \(y=i\). Then, \(T_n(q)\) counts the number of these weighted paths ending at (n, 0). Thus, if let \(\mathscr {M}_n\) denote the set of the weighted Motzkin paths of length n and \(w(\beta )=(w(\beta _1),w(\beta _2),\ldots ,w(\beta _n))\) be a weighted Motzkin path of length n with \(w(\beta _i)\in \{1,s_n,r_{n+1}\}_{n\ge 0}\), then we have

$$\begin{aligned} T_n(q)=\sum _{\beta \in \mathscr {M}_n}\prod _{i=1}^nw(\beta _i). \end{aligned}$$

6 Properties of the first column

The first column \((T_{n,0})_{n\ge 0}\) of the Stirling–Whitney–Riordan triangle T has properties similar to those of \((T_n(q))_{n\ge 0}\). In this section, we will present some properties for \((T_{n,0})_{n\ge 0}\).

Theorem 6.1

Let \((T_{n,0})_{n\ge 0}\) be the first column of T.

  1. (i)

    The ordinary generating function of \(T_{n,0}\) has a Jacobi continued fraction expression

    $$\begin{aligned} \sum _{n\ge 0}T_{n,0}t^n=\frac{1}{1-s_0t-\frac{r_1t^2}{1-s_1t-\frac{r_2t^2}{1-s_2t-\cdots }}}, \end{aligned}$$

    where \(s_n=a_2+a_1n+[b_1(2n+1)+b_2]\lambda \) and \(r_{n+1}=[b_1(n+1)+b_2]\lambda (b_1 \lambda +a_1)(n+1)\) for \(n\ge 0\).

  2. (ii)

    The sequence \((T_{n,0})_{n\ge 0}\) is \(\mathbf{x} \)-SM and 3-\(\mathbf{x} \)-LCX with \(\mathbf{x} =(a_1,a_2,b_1,b_2,\lambda )\).

  3. (iii)

    The exponential generating function of \(T_{n,0}\) is given as

    $$\begin{aligned} \sum _{n\ge 0}T_{n,0}\frac{t^n}{n!}= & {} e^{a_2t}\left[ 1+\frac{b_1\lambda (1-e^{a_1t})}{a_1}\right] ^{-(1+\frac{b_2}{b_1})}. \end{aligned}$$
  4. (iv)

    \(T_{n,0}\) is a polynomial in \(\lambda \) and has only real zeros.

  5. (v)

    The Turán-type polynomial \(T_{n+1,0}T_{n-1,0}-T^2_{n,0}\) is a weakly stable polynomial in \(\lambda \) for \(n\ge 1\).

Proof

(i) Assume that \(s_n=a_2+a_1n+[b_1(2n+1)+b_2]\lambda \), \(r_{n}=b_1(n+1)+b_2\) and \(t_n=\lambda (b_1\lambda +a_1)n\) for \(n\ge 0\). Let \(h_k(z)=\sum _{n\ge k}T_{n,k}z^n\) for \(k\ge 0\). It follows from the recurrence relation:

$$\begin{aligned} T_{n,k}=r_{k-1}T_{n-1,k-1}+s_k T_{n-1,k}+t_{k+1}T_{n-1,k+1} \end{aligned}$$

that we have

$$\begin{aligned} h_0(z)= & {} 1+s_0zh_0(z)+t_1zh_1(z), \\ h_k(z)= & {} r_{k-1}zh_{k-1}(z)+s_{k}zh_k(z)+t_{k+1}zh_{k+1}(z) \end{aligned}$$

for \(k\ge 1\), which imply

$$\begin{aligned} \frac{h_0(z)}{1}= & {} \frac{1}{1-s_0z-t_1z\frac{h_1(z)}{h_0(z)}}, \\ \frac{h_1(z)}{h_0(z)}= & {} \frac{r_0z}{1-s_1z-t_2z\frac{h_2(z)}{h_1(z)}},\\&\vdots&\\ \frac{h_k(z)}{h_{k-1}(z)}= & {} \frac{r_{k-1}z}{1-s_{k}z-t_{k+1}z\frac{h_{k+1}(z)}{h_k(z)}}. \end{aligned}$$

Thus, we get

$$\begin{aligned} \sum \limits _{n=0}^{\infty }T_{n,0} z^n=h_0(z)=\frac{\displaystyle 1}{\displaystyle 1- s_0z-\frac{\displaystyle r_0t_1z^2}{\displaystyle 1- s_1z-\frac{\displaystyle r_1t_2z^2}{\displaystyle 1- s_2z-\ldots }}}. \end{aligned}$$

If let \(\mathscr {T}_n(\lambda ):=T_{n,0}\), then by (i) and Theorem 5.5 (i), we immediately get

$$\begin{aligned} \mathscr {T}_n(\lambda +q)=T_n(q) \end{aligned}$$

for \(n\ge 0\). Hence, we easily get (ii) and (iii) by Theorem 5.5 (ii) and Theorem 4.1 (i), respectively. We also have (iv) and (v) by Theorem 3.2. This completes the proof. \(\square \)