2. Strong $q$-log-concavity of $D(q)$

Let $f(q)$ and $g(q)$ be two real polynomials in $q$. We say that $f(q)$ is $q$-non-negative if $f(q)$ has non-negative coefficients. Denote $f(q)\geq _q g(q)$ if the difference $f(q)-g(q)$ is $q$-non-negative. For a polynomial sequence $(f_n(q))_{n\geq 0}$, it is called $q$-log-concave (or $q$-log-convex) if

\[ f_n(q)^{2} \geq_q f_{n+1}(q)f_{n-1}(q)\ \left(\text{or}\ f_n(q)^{2} \leq_q f_{n+1}(q)f_{n-1}(q)\right) \]

for $n \geq 1$. It is called strongly $q$-log-concave (or strongly $q$-log-convex) if

\[ f_n(q)f_m(q) \geq_q f_{n+1}(q)f_{m-1}(q)\ \left(\text{or}\ f_n(q)f_m(q) \leq_q f_{n+1}(q)f_{m-1}(q)\right) \]

for $n \geq m \geq 1$. Clearly, the strong $q$-log-concavity (strong $q$-log-convexity) of polynomial sequences implies the $q$-log-concavity ($q$-log-convexity), which further implies the log-concavity (log-convexity) for any fixed $q \geq 0$, however, not vice versa. The (strong) $q$-log-concavity has been extensively studied; see [Reference Butler5, Reference Leroux8, Reference Sagan13].

It is known that $D(0)$ is the Pascal triangle $P$. Su and Wang [Reference Su and Wang15] proved the log-concavity of the sequence located in a transversal line of $P$ or a line parallel to the boundary of $P$. Yu [Reference Yu24] pointed out that such properties also hold in $D(1)$ which is the Delannoy triangle.

The central coefficients $d_{2n,n}(q)$ of $D(q)$ are $q$-central Delannoy numbers

\[ D_n(q)=\sum_{i\geq 0}\binom{n}{i}\binom{2n-i}{n}q^{i}. \]

Liu and Wang [Reference Liu and Wang9] proved that the sequence of $q$-central Delannoy numbers $(D_n(q))_{n\geq 0}$ is $q$-log-convex. Zhu [Reference Zhu26, Reference Zhu27] later proved that $(D_n(q))_{n\geq 0}$ is strongly $q$-log-convex. Wang and Zhu [Reference Wang and Zhu23] gave a stronger result that $(D_n(q))_{n\geq 0}$ forms a $q$-Stieltjes moment sequence, i.e., all minors of the corresponding Hankel matrix $[D_{i+j}(q)]$ are $q$-non-negative.

In this section, we aim to study the strong $q$-log-concavity of a polynomial sequence located in a ray or a transversal line of $D(q)$. Let $(d_{n_i,k_i}(q))_{i\geq 0}$ be such a sequence. Then $(n_i)_{i\geq 0}$ and $(k_i)_{i\geq 0}$ form two arithmetic sequences (see Figure 1). Clearly, the common difference of $(n_i)_{i\geq 0}$ can be assumed to be non-negative. Meanwhile the common difference of $(k_i)_{i\geq 0}$ can also be assumed to be non-negative without loss of generality since the symmetry of $D(q)$ leads to the fact that the sequences $(d_{n_i,k_i}(q))_{i\geq 0}$ and $(d_{n_i,n_i-k_i}(q))_{i\geq 0}$ are the same. Thus, to achieve our aim, it suffices to investigate the strong $q$-log-concavity of the sequence $(d_{n_0+ai,k_0+bi}(q))_{i\geq 0}$ for non-negative integers $a$ and $b$, giving rise to our first main result of this paper.

Theorem 2.1 Let $n_0,$ $k_0,$ $a$ and $b$ be four non-negative integers and $n_0\geq k_0,$ $a+b\neq 0$. Define the sequence

\[ S_i(q)=d_{n_0+ai,k_0+bi}(q), \quad i=0,1,2,\ldots. \]

If $a\leq b,$ then the polynomial sequence $(S_i(q))_{i\geq 0}$ is strongly $q$-log-concave.

Figure 1. The symmetric isosceles triangle $D(q)$ of $q$-coloured Delannoy numbers.

Before a combinatorial proof of this theorem, we need to introduce a few notions. Let $\mathfrak {D}(n,\,k)$ denote the set of all $q$-coloured Delannoy paths from $(s,\,t)$ to $(s+n,\,t+k)$ for fixed $s$ and $t$. Note that $S_i(q)$ count the number of $q$-coloured Delannoy paths from $(0,\,0)$ to $(n_0-k_0+(a-b)i,\,k_0+bi)$. Hence, for convenience we let

\[ \mathfrak{D}_i:=\mathfrak{D}(n_0-k_0+(a-b)i,k_0+bi). \]

Then we have

\[ S_i(q)=\sum_{P\in \mathfrak{D}_i}w(P), \]

where the weight of path $P$, denoted by $w(P)$, is defined as the product of the weights of all its steps. Suppose that $P$ has exactly $k$ diagonal steps (i.e. $(1,\,1)$ steps). Then

\[ w(P)=q^{k}, \]

since the weight of each diagonal step in P is $q$, and the others $1$. Moreover, $w(P,\,Q)=w(P)w(Q)$ is to denote the weight of a pair of $q$-coloured Delannoy paths in the following.

Proof of Theorem 2.1 To show the strong $q$-log-concavity of $(S_i(q))_{i\geq 0}$, it suffices to show that

\[ S_i(q)S_j(q)\geq_q S_{i+1}(q)S_{j-1}(q), \]

for $i\geq j$, i.e.,

(2.1)\begin{equation} \displaystyle\sum_{P\in \mathfrak{D}_i}w(P)\sum_{P\in \mathfrak{D}_j}w(P)\geq_q \sum_{P\in \mathfrak{D}_{i+1}}w(P)\sum_{P\in \mathfrak{D}_{j-1}}w(P). \end{equation}

It is equivalent to

(2.2)\begin{equation} \displaystyle\sum_{(Q_1,Q_2)\in (\mathfrak{D}_i,\mathfrak{D}_j)}w(Q_1,Q_2) \geq_q \sum_{(P_1,P_2)\in (\mathfrak{D}_{i+1},\mathfrak{D}_{j-1})}w(P_1,P_2). \end{equation}

Let $N_k(\mathfrak {D}_i,\,\mathfrak {D}_j)$ denote the number of pairs of paths with exactly $k$ diagonal steps in the set $(\mathfrak {D}_i,\,\mathfrak {D}_j)$. So it needs to prove

(2.3)\begin{equation} \displaystyle\sum_k N_k(\mathfrak{D}_i,\mathfrak{D}_j) q^{k} \geq_q \sum_k N_k(\mathfrak{D}_{i+1},\mathfrak{D}_{j-1}) q^{k}. \end{equation}

To this end, we construct an injection from $(\mathfrak {D}_{i+1},\,\mathfrak {D}_{j-1})$ to $(\mathfrak {D}_i,\,\mathfrak {D}_j)$, i.e.,

\[ \phi: \quad (P_1,P_2) \rightarrow (Q_1,Q_2), \]

such that

(2.4)\begin{equation} N_k(\mathfrak{D}_{i+1},\mathfrak{D}_{j-1})\leq N_k(\mathfrak{D}_i,\mathfrak{D}_j). \end{equation}

Each pair of $(P_1,\,P_2)$ in $(\mathfrak {D}_{i+1},\,\mathfrak {D}_{j-1})$, as shown in Figure 2, follows such rules:

1. P₁: $(0,\,0)\rightarrow (n_0-k_0+(a-b)(i+1),\,k_0+b(i+1))$;

2. P₂: $((a-b)(i-j+1),\,b(i-j+1))\rightarrow (n_0-k_0+(a-b)i,\,k_0+bi)$.

Clearly, $P_1$ and $P_2$ must intersect at least one lattice point in the shadow area. Let $A$ denote the first intersection point. Then we define the operation $\phi$ on $(P_1,\,P_2)$ at the point $A$:

“Switch the initial segments of the two paths”,

as shown in Figure 3. With this operation $\phi$, we could obtain a corresponding pair $(Q_1,\,Q_2)\in (\mathfrak {D}_i,\,\mathfrak {D}_j)$, and

1. Q₁: $((a-b)(i-j+1),\,b(i-j+1))\rightarrow (n_0-k_0+(a-b)(i+1),\,k_0+b(i+1))$;

2. Q₂: $(0,\,0)\rightarrow (n_0-k_0+(a-b)i,\,k_0+bi)$.

Figure 2. $(P_1,\,P_2)\in (\mathfrak {D}_{i+1},\,\mathfrak {D}_{j-1})$.

Figure 3. Operation $\phi$ on $(P_1,\,P_2)$ in Figure 2.

For instance, let $n_0=10$, $k_0=3$, $a=0$ and $b=1$, and take $i=2$, $j=1$. Then $(P_1,\,P_2)\in (\mathfrak {D}_{3},\,\mathfrak {D}_{0})$, where $P_1$ goes from $(0,\,0)$ to $(4,\,6)$ and $P_2$ from $(-2,\,2)$ to $(5,\,5)$, as shown in Figure 4. The operation $\phi$ on $(P_1,\,P_2)\in (\mathfrak {D}_{3},\,\mathfrak {D}_{0})$ at the point $A$ will lead to a pair $(Q_1,\,Q_2)\in (\mathfrak {D}_{2},\,\mathfrak {D}_{1})$ as shown in Figure 4.

Figure 4. $(P_1,\,P_2)\in (\mathfrak {D}_{3},\,\mathfrak {D}_{0})\rightarrow (Q_1,\,Q_2)\in (\mathfrak {D}_{2},\,\mathfrak {D}_{1})$.

Note that the location of the first intersection point remains invariant under the operation $\phi$, which means $\phi$ is invertible and so that it is an injection. Meanwhile, it is easy to check that the number of diagonal steps also remains invariant under the injection $\phi$, i.e., the number of diagonal steps in $(P_1,\,P_2)$ is the same as that in $(Q_1,\,Q_2)$. Therefore, (2.4) follows, by which (2.1) can be obtained as desired.

From Theorem 2.1, we have the following corollary immediately.

Note that $D(0)$ and $D(1)$ are Pascal triangle and Delannoy triangle, respectively. The log-concavity of the sequences in these two triangles was mentioned at the beginning of this section. $D(2)$ and $D(3)$ are also Pascal-like triangles and could be found in [Reference Sloane14, A081577, A081578]. By Theorem 2.1, we can get the log-concavity of sequences in these two triangles.

3. $q$-total positivity of $D(q)$

Let $f(q)$ and $g(q)$ be two real polynomials in $q$. Let $M(q)=[m_{n,k}]_{n,k\geq 0}$ be the matrix whose entries are all real polynomials in $q$. We say that $M(q)$ is $q$-totally positive ( $q$-TP for short) if all minors are $q$-non-negative.

Note that, since (1.4), the square matrix $[D_{n,k}(q)]_{n,k\geq 0}=PDP^{T}$, where $P$ is the Pascal triangle and $D={\rm diag}(1,\,1+q,\,(1+q)^{2},\,(1+q)^{3},\,\ldots )$. Hence the $q$-total positivity of $[D_{n,k}(q)]_{n,k\geq 0}$ follows immediately from the Cauchy–Binet formula and the total positivity of the Pascal triangle (i.e., all its minors are non-negative).

It is known that the triangle $D(q)$ is a Riordan array $(\frac {1}{1-x},\,\frac {x+qx^{2}}{1-x})$ (see [Reference Mu and Zheng12] for details). A (proper) Riordan array, denoted by $(d(x),\, h(x))$, is an infinite lower triangular matrix whose generating function of the $k$th column is $d(x)h^{k}(x)$ for $k=0,\,1,\,2,\,\ldots$, where $d(0)=1$, $h(0)=0$ and $h'(0)\neq 0$. In this section, we consider the $q$-total positivity of $D(q)$. We first prove a lemma which is a $q$-analogy of Theorem 3 in [Reference Mao, Mu and Wang11].

Lemma 3.1 Let $M(q)=(d(x),\,h(x))$ be a Riordan array, where $d(x)=\sum \nolimits _{n\ge 0}d_n(q)x^{n}$ and $h(x)=\sum \nolimits _{n\ge 0}h_n(q)x^{n}$. If the matrix

\[ \left[\begin{array}{ccccc} d_0(q) & h_0(q) & & & \\ d_1(q) & h_1(q) & h_0(q) & & \\ d_2(q) & h_2(q) & h_1(q) & h_0(q) & \\ \vdots & & & & \ddots \end{array}\right] \]

is $q$-TP, then so is the Riordan array $M(q)$.

Proof. Let $T(q)=(h(x),\,x)=[h_{i-j}(q)]_{i,j\ge 0}$ and $v(q)=(d_0(q),\,d_1(q),\,\ldots )^{T}$. Then

\[ M(q)=(d(x),d(x)h(x),d(x)h^{2}(x),\ldots)=(v(q),T(q)v(q),T(q)^{2}v(q),\ldots). \]

Let $M_k(q)$ denote the submatrix $(v(q),\,T(q)v(q),\,\ldots,\,T(q)^{k-1}v(q))$ consisting of the first $k$ columns of $M(q)$. Then

\begin{align*} M_{k+1}(q)& =(v(q),T(q)v(q),\ldots,T(q)^{k}v(q))=(v(q),T(q)M_k(q))\\ & =(v(q),T(q)) \left[\begin{array}{cc} 1 & 0 \\ 0 & M_k(q) \\ \end{array}\right]. \end{align*}

If $M_k(q)$ is $q$-TP, then so is $\left [\begin {smallmatrix} 1 & 0 \\ 0 & M_k(q) \\ \end {smallmatrix}\right ]$. The condition states that $(v(q),\,T(q))$ is $q$-TP. It follows that the product $M_{k+1}(q)$ is also $q$-TP from the classic Cauchy–Binet formula. Thus, the statement follows.

Proof. Note that $D(q)=(d(x),\,h(x))=(\frac {1}{1-x},\,\frac {x+qx^{2}}{1-x})$. Let $T(q)=(h(x),\,x)$ and $v(q)=(d_0(q),\,d_1(q),\,d_2(q),\,\ldots )^{T}$. By Lemma 3.1, it suffices to show that $(v(q),\,T(q))$ is $q$-TP. We have

\begin{align*} (v(q),T(q))& =\left[\begin{array}{ccccc} 1 & & & & \\ 1 & 1 & & & \\ 1 & 1+q & 1 & & \\ 1 & 1+q & 1+q & 1 & \\ \vdots & & & & \ddots\end{array}\right]\\ & =\left[\begin{array}{ccccc} 1 & & & & \\ 1 & 1 & & & \\ 1 & 1 & 1 & & \\ 1 & 1 & 1 & 1 & \\ \vdots & & & & \ddots \end{array}\right] \left[\begin{array}{ccccc} 1 & & & & \\ & 1 & & & \\ & q & 1 & & \\ & & q & 1 & \\ & & & \ddots & \ddots \end{array}\right]. \end{align*}

One can check that both matrices on the right-hand side are $q$-TP. Therefore, $(v(q),\,T(q))$ is $q$-TP by the classic Cauchy–Binet formula, as required.

4. Zeros of row sums

Let $R_n(q)=\sum \nolimits _i r_{n,i}q^{i}$ be the sum of the $n$th row of $D(q)$, i.e.,

\[ R_n(q)=\displaystyle\sum_{k=0}^{n}d_{n,k}(q). \]

The first few entries of $(R_n(q))_{n\geq 0}$ are $(1,\,2,\,4+q,\,8+4q,\,\ldots )$. The coefficient matrix of $R_n(q)$ is defined by the matrix

\[ [r_{n,i}]_{n,i\geq 0}= \left[\begin{array}{ccc} 1 & \\ 2 & \\ 4 & 1 & \\ 8 & 4 & \\ \vdots & & \ddots \\ \end{array}\right] . \]

Note that the polynomial $D_{n,k}(q)$ satisfies the recurrence (1.3), hence

(4.1)\begin{equation} d_{n,k}(q)=d_{n-1,k-1}(q)+d_{n-1,k}(q)+q d_{n-2,k-1}(q). \end{equation}

Thus, the row sum $R_n(q)$ satisfies the simple recurrence

\[ R_n(q)=2R_{n-1}(q)+qR_{n-2}(q) \]

with $R_1(q)=1$, $R_2(q)=2$.

Let $(f_n(z))_{n\ge 0}$ be a sequence of complex polynomials. We say that the complex number $z$ is a limit of zeros of the sequence $(f_n(z))_{n\ge 0}$ if there is such a sequence $(z_n)_{n\ge 0}$ that $f_n(z_n)=0$ and $z_n\rightarrow z$ as $n\rightarrow +\infty$. Suppose now that $(f_n(z))_{n\ge 0}$ is a sequence of polynomials satisfying the recursion

\[ f_{n+k}(z)={-}\sum_{j=1}^{k}c_j(z)f_{n+k-j}(z) \]

where $c_j(z)$ are polynomials in $z$. Let $\lambda _j(z)$ be all roots of the associated characteristic equation $\lambda ^{k}+\sum \nolimits _{j=1}^{k}c_j(z)\lambda ^{k-j}=0$. It is well known that if $\lambda _j(z)$ are distinct, then

(4.2)\begin{equation} f_n(z)=\displaystyle\sum_{j=1}^{k}\alpha_j(z)\lambda_j^{n}(z), \end{equation}

where $\alpha _j(z)$ is determined from the initial conditions.

Proof. We first need to prove that

(4.3)\begin{equation} R_n(q)=4\prod_{k=1}^{\lfloor n/2\rfloor}\left(1+q\cos^{2}\displaystyle\frac{k\pi}{n+1}\right), \end{equation}

for which we only demonstrate the case that $n$ is even in the following since it is quite similar for odd $n$.

Note that

\[ R_n(q)=2R_{n-1}(q)+qR_{n-2}(q) \]

with $R_1=1$, $R_2=2$. Hence the Binet form of the row sums is

(4.4)\begin{equation} R_n(q)=\displaystyle\frac{\lambda_1^{n+1}-\lambda_2^{n+1}}{\lambda_1-\lambda_2}, \end{equation}

where

(4.5)\begin{equation} \lambda_{1,2}=1\pm \sqrt{1+q} \end{equation}

are the roots of the characteristic equation $\lambda ^{2}-2\lambda -q=0$. Let $\omega _k=e^{\frac {2k\pi i}{n+1}}$. Then $\lambda ^{n+1}-1=\prod _{k=1}^{n+1}(\lambda -\omega _k)$. Note that

\[ (\lambda-\omega_k)(\lambda-\omega_{n+1-k})=\lambda^{2}-2\lambda \cos\frac{k\pi}{n+1}+1=(\lambda+1)^{2}-4\lambda\cos^{2}\frac{k\pi}{n+1}. \]

Since $n$ is even, we have

\[ \lambda^{n+1}-1=(\lambda-1)\prod_{k=1}^{n/2}\left((\lambda+1)^{2}-4\lambda\cos^{2}\frac{k\pi}{n+1}\right), \]

and hence

\[ \lambda_1^{n+1}-\lambda_2^{n+1}=(\lambda_1-\lambda_2)\prod_{k=1}^{n/2}\left((\lambda_1+\lambda_2)^{2}-4\lambda_1\lambda_2\cos^{2}\frac{k\pi}{n+1}\right). \]

Since $\lambda _1+\lambda _2=2$ and $\lambda _1\lambda _2=-q$, we have

\[ R_n(q)=\frac{\lambda_1^{n+1}-\lambda_2^{n+1}}{\lambda_1-\lambda_2}=\prod_{k=1}^{n/2}\left(4+4q\cos^{2}\frac{k\pi}{n+1}\right). \]

Denote $z_{n,k}=-1/\cos ^{2}\frac {k\pi }{n+1},\,\ k=1,\,2,\,\cdots,\,n/2.$ Then the polynomial $R_n(q)$ has distinct real zeros $z_{n,1}>z_{n,2}>\cdots >z_{n,n/2}$. Since

\[ \lim_{n\rightarrow \infty}z_{n,1}={-}\infty\ \text{and} \lim_{n\rightarrow \infty}z_{n,n/2}={-}1, \]

all zeros of $R_n(q)$ are in $(-\infty,\, -1)$.

We proceed to prove that each $q\in (-\infty,\, -1]$ is a limit of zeros of the sequence $(R_n(q))_{n\ge 0}$. The non-degeneracy conditions of Lemma 4.1 are clearly satisfied by (4.4). So the limits of zeros of $(R_n(q))_{n\ge 0}$ are those $q$ for which $|\lambda _1(q)|=|\lambda _2(q)|$, i.e.,

\[ |1+\sqrt{q+1}|=|1-\sqrt{q+1}| \]

by (4.5). In other words, $\sqrt {q+1}$ must be a pure imaginary. It follows that $q+1\le 0$, i.e., $q\le -1$. Then the proof is completed.

Let $A_n(q)$ be the sum of the $n$th antidiagonal row of $D(q)$, i.e.,

\[ A_n(q)=\displaystyle\sum_{k=0}^{\lfloor n/2 \rfloor}d_{n-k,k}(q). \]

The first few entries of $(A_n(q))_n$ are $(1,\,1,\,2,\,3+q,\,5+2q,\,\ldots )$. By (4.1), it is easy to check that $A_n(q)$ satisfies

(4.6)\begin{equation} A_n(q)=A_{n-1}(q)+A_{n-2}(q)+qA_{n-3}(q), \end{equation}

with $A_1=1$, $A_2=1$ and $A_3=2$.

To prove this, we need the following lemma which can be found in [Reference Tran and Zumba20, Theorem 3].

Lemma 4.4 Consider the sequence of polynomials $\{P_n(q)\}_{n=0}^{\infty }$ generated by

(4.7)\begin{equation} \displaystyle\sum_{n=0}^{\infty}P_n(q)x^{n}=\displaystyle\frac{1}{1+x+ax^{2}+qx^{3}}, \end{equation}

where $a\in \mathbb {R}$. If $-1\leq a\leq 1/3$, then all the zeros of $P_n(q)$ are in the real interval

\[ I_a=\left(-\infty,\frac{-2+9a-2\sqrt{(1-3a)^{3}}}{27}\right]. \]

and are also dense in $I_a$.

Proof of Theorem 4.3 By (4.6), the generating function of $A_n(q)$ follows that

\[ \sum_{n=0}^{\infty}A_n(q)x^{n}=\frac{1}{1-x-x^{2}-qx^{3}}, \]

which can also be derived from (4.7) with substitutions $x\rightarrow -x$ and $a\rightarrow -1$. So $A_n(q)$ meets the condition of Lemma 4.4, and therefore, the zeros of $A_n(q)$ are in $(-\infty,\, -1]$ and are dense there.

5. Remarks

In this section, we give some remarks on the asymptotic normality of coefficients of row sums. Let $a_{n,k}$ be a double-indexed sequence of non-negative numbers and let

\[ p_{n,k}=\displaystyle\frac{a_{n,k}}{\displaystyle\sum_{j=0}^{n}a_{n,j}} \]

denote the normalized probabilities. Following Bender [Reference Bender3], we say that the sequence $a_{n,k}$ is asymptotically normal by a central limit theorem if

(5.1)\begin{equation} \lim_{n\rightarrow\infty}\sup_{x\in\mathbb{R}}\left|\displaystyle\sum_{k\le\mu_n+x\sigma_n}p_{n,k}-\displaystyle\frac{1}{\sqrt{2\pi}}\displaystyle\int_{-\infty}^{x}e^{{-}t^{2}/2}dt\right|=0, \end{equation}

where $\mu _n$ and $\sigma ^{2}_n$ are the mean and variance of $a_{n,k}$, respectively. We say that $a_{n,k}$ is asymptotically normal by a local limit theorem on $\mathbb {R}$ if

(5.2)\begin{equation} \lim_{n\rightarrow\infty}\sup_{x\in\mathbb{R}}\left|\sigma_np_{n,\lfloor\mu_n+x\sigma_n\rfloor}-\displaystyle\frac{1}{\sqrt{2\pi}}e^{{-}x^{2}/2}\right|=0. \end{equation}

In this case,

\[ a_{n,k}\sim \displaystyle\frac{e^{{-}x^{2}/2}\displaystyle\sum_{j=0}^{n}a_{n,j}}{\sigma_n\sqrt{2\pi}} \textrm{ as } n\rightarrow \infty, \]

where $k=\mu _n+x\sigma _n$ and $x=O(1)$. Clearly, the validity of (5.2) implies that of (5.1).

Many well-known combinatorial sequences enjoy central and local limit theorems, such as the binomial coefficients $\binom {n}{k}$, the signless Stirling numbers $c(n,\,k)$ of the first kind, the Stirling numbers $S(n,\,k)$ of the second kind, the Eulerian numbers $A(n,\,k)$ [Reference Canfield6], and the Delannoy numbers $d(n,\,k)$ [Reference Wang, Zheng and Chen21]. Besides, the asymptotic normality of Laplacian coefficients of graphs was discovered in [Reference Wang, Zhang and Zhu22]. A standard approach to demonstrating asymptotic normality is the following criterion (see [Reference Bender3, Theorem 2] for instance and [Reference Canfield6, Example 3.4.2] for historical remarks).

Lemma 5.1 Suppose that $S_n(q)=\sum \nolimits _{k=0}^{n}a_{n,k}q^{k}$ have only real zeros and $S_n(q)=\prod _{i=1}^{n}(q+r_i),$ where all $a_{n,k}$ and $r_i$ are non-negative. Let

\[ \mu_n=\sum_{i=1}^{n}\frac{1}{1+r_i} \]

and

\[ \sigma^{2}_n=\sum_{i=1}^{n}\frac{r_i}{(1+r_i)^{2}}. \]

Then if $\sigma _n^{2}\rightarrow +\infty,$ the numbers $a_{n,k}$ are asymptotically normal (by central and local limit theorems) with the mean $\mu _n$ and variance $\sigma _n^{2}$.

For the asymptotic normality of $r_{n,i}$ (the coefficients of row sums $R_n(q)$), we have the following result.

Its proof can be similarly produced by referring to [Reference Wang, Zheng and Chen21, Theorem 3.2].