Analytic Aspects of the Generalized Second-Order Central Trinomial Coefficients

Abstract

The generalized second-order central trinomial coefficients $T_n^{\left(2\right)}(b,c)$ are a special case of the generalized r-order central trinomial coefficients, corresponding to $r=2$. We show that all zeros of the polynomial ${{\mathcal{T}}^{\left(2\right)}}_n\left(x\right)={\sum }_{k=0}^{\lfloor n/2\rfloor }{T}^{\left(2\right)}(n,k)x^k$ are real. And zeros of ${{\mathcal{T}}^{\left(2\right)}}_{n-1}\left(x\right)$ interlace those of ${{\mathcal{T}}^{\left(2\right)}}_{n-2}\left(x\right)$, as well as those of ${{\mathcal{T}}^{\left(2\right)}}_n\left(x\right)$. Using this result, we also discuss the asymptotic normality of $T^{\left(2\right)}(n,k)$.

1. Introduction

The trinomial coefficient ${\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)}_2$ is given by

$${\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)}_2:=\left[x^{n+k}\right](x^2+x+1),$$

where $\left[x^n\right]f\left(x\right)$ denotes the coefficient of $x^n$ in the polynomial $f\left(x\right)$. According to Ref. [1], the trinomial coefficients had been investigated by Euler. Define the central trinomial coefficient as

$$T_n:={\left({\displaystyle \genfrac{}{}{0pt}{}{n}{0}}\right)}_2.$$

i.e., $T\left(n\right)=\left[x^n\right]{(x^2+x+1)}^n$. It is not difficult to prove that

$$T_n=\sum _{k=0}^{\lfloor n/2\rfloor }T(n,k),$$

where $\lfloor x\rfloor :=max\{n\in \mathbb{Z}:n\le x\}$ and

$$T(n,k):={\displaystyle \frac{n!}{k!k!(n-2k)!}}=\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right).$$

Furthermore, $T_n$ satisfies the recurrence relation

$$(n+1)T_{n+1}=(2n+1)T_n+3n{T}_{n-1}.$$

The generating function of $T_n$ is

$$\sum _{n\ge 0}{T}_n{x}^n={\displaystyle \frac{1}{\sqrt{1-2x-3x^2}}}.$$

(1)

In Ref. [2], Liang, Wang, and Wang considered the polynomial

$${\mathcal{T}}_n\left(x\right):=\sum _{k=0}^{\lfloor n/2\rfloor }T(n,k)x^k.$$

Clearly $T\left(n\right)={\mathcal{T}}_n\left(1\right)$. Liang, Wang and Wang [2] proved that all the zeros of ${\mathcal{T}}_n\left(x\right)$ are real, and the zeros of ${\mathcal{T}}_{n-1}\left(x\right)$ are interlaced with the zeros of ${\mathcal{T}}_{n-2}\left(x\right)$ and ${\mathcal{T}}_n\left(x\right)$. Using the Aissen–Schoenberg–Whitney theorem and the Schoenberg–Edrei theorem, they also proved that the numbers $T(n,k)$ are asymptotically normal with the mean ${\mu }_n=n(1/3+o\left(1\right))$ and the variance ${\sigma }_n^2\to \infty$.

Motivated by (1), we define the r-order central trinomial coefficient $T_n^{\left(r\right)}$ by

$$\sum _{n\ge 0}{T}_n^{\left(r\right)}{x}^n={\displaystyle \frac{1}{{(1-2x-3x^2)}^{r-\frac{1}{2}}}}.$$

(2)

Then, $T_n=T_n^{\left(1\right)}$. As we shall see soon, when $r=2$,

$$T_n^{\left(2\right)}=\sum _{k=0}^{\lfloor n/2\rfloor }{T}^{\left(2\right)}(n,k),$$

where

$$T^{\left(2\right)}(n,k):=(2k+1)\left({\displaystyle \genfrac{}{}{0pt}{}{n+2}{2k+2}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right).$$

We define

$${\mathcal{T}}_n^{\left(2\right)}\left(x\right):=\sum _{k=0}^{\lfloor n/2\rfloor }{T}^{\left(2\right)}(n,k)x^k.$$

Thus, $T_n^{\left(2\right)}={\mathcal{T}}_n^{\left(2\right)}\left(1\right)$. In this paper, we investigate the zeros of ${\mathcal{T}}_n^{\left(2\right)}\left(x\right)$.

Theorem 1. 

All zeros of the polynomials ${\mathcal{T}}_n^{\left(2\right)}\left(x\right)$ are real, and the zeros of ${\mathcal{T}}_{n-1}^{\left(2\right)}\left(x\right)$ are interlaced with the zeros of ${\mathcal{T}}_{n-2}^{\left(2\right)}\left(x\right)$ and ${\mathcal{T}}_n^{\left(2\right)}\left(x\right)$.

Furthermore, we define the generalized central trinomial coefficient $T_n(b,c)$ by

$$T_n(b,c):=\left[x^n\right]{(x^2+bx+c)}^n.$$

Clearly, $T_n(1,1)=T_n$. It can be shown that

$$T_n(b,c)=\sum _{k=0}^{\lfloor n/2\rfloor }T(n,k)b^{n-2k}{c}^k,$$

and $T_n(b,c)$ satisfy the recurrence relation

$$(n+1)T_{n+1}(b,c)=(2n+1)b{T}_n(b,c)-n(b^2-4c)T_{n-1}(b,c)$$

with $T_0(b,c)=1$ and $T_1(b,c)=b.$ The generating function of $T_n(b,c)$ is

$$\sum _{n\ge 0}{T}_n(b,c)x^n={\displaystyle \frac{1}{\sqrt{1-2bx+(b^2-4c)x^2}}}.$$

Some properties of $T_n(b,c)$ have been studied by Liang, Wang, and Wang in [2].

We define the generalized r-order central trinomial coefficient $T_n^{\left(r\right)}(b,c)$ by

$$\sum _{n\ge 0}{T}_n^{\left(r\right)}(b,c)x^n={\displaystyle \frac{1}{{(1-2bx+(b^2-4c)x^2)}^{r-\frac{1}{2}}}}.$$

(3)

As we shall see later, when $r=2$,

$$T_n^{\left(2\right)}(b,c)=\sum _{k=0}^{\lfloor n/2\rfloor }{T}^{\left(2\right)}(n,k)b^{n-2k}{c}^k.$$

With the help of ${\mathcal{T}}_n^{\left(2\right)}\left(x\right)$, we can prove that

Theorem 2. 

The sequence $T^{\left(2\right)}(n,k)$ is asymptotically normal with the mean value

$${\mu }_n={\displaystyle \frac{n-1}{3}}+O\left(1\right)$$

and the variance

$${\sigma }_n^2\to +\infty .$$

Remark 1. 

In Ref. [3], Andrews and Baxter found that some q-analogues of trinomial coefficients play an important role in the hard hexagon model. For example, a q-analogue of the central trinomial coefficient $T\left(n\right)$ ([3], Equation (2.7)) is

$${\left({\displaystyle \genfrac{}{}{0pt}{}{m;B;q}{0}}\right)}_q:=\sum _{k=0}^n{q}^{j(j+B)}{\left[\begin{array}{c}n\\ 2k\end{array}\right]}_q{\left[\begin{array}{c}2k\\ k\end{array}\right]}_q,$$

where ${[ ]}_q$ denotes the q-binomial coefficient. Motivated by the work of Andrews and Baxter, we believe that some q-analogue of the second-order central trinomial coefficient $T_n^{\left(2\right)}$ would have some applications in physics.

The proofs of Theorems 1 and 2 will be given in Section 2 and Section 3, respectively.

2. Zeros of ${\mathcal{T}}_{\mathit{n}}^{\left(\mathbf{2}\right)}\left(\mathit{x}\right)$

Lemma 1. 

The generalized second-order central trinomial coefficient $T_n^{\left(2\right)}(b,c)$ satisfies the recurrence relation

$$n{T}_n^{\left(2\right)}(b,c)=(2n+1)b{T}_{n-1}^{\left(2\right)}(b,c)-(n+1)(b^2-4c)T_{n-2}^{\left(2\right)}(b,c)$$

with $T_0^{\left(2\right)}(b,c)=1$ and $T_1^{\left(2\right)}(b,c)=3b$.

Proof of Lemma 1. 

Based on the generating function of the generalized second-order central trinomial coefficients

$$\sum _{n\ge 0}{T}_n^{\left(2\right)}(b,c)x^n={\displaystyle \frac{1}{{(1-2bx+(b^2-4c)x^2)}^{\frac{3}{2}}}}.$$

differentiating both sides yields

$${(\sum _{n\ge 0}{T}_n^{\left(2\right)}(b,c)x^n)}^{\prime }=\sum _{n\ge 1}n{T}_n^{\left(2\right)}(b,c)x^{n-1}={\displaystyle \frac{3(b-(b^2-4c)x)}{{(1-2bx+(b^2-4c)x^2)}^{\frac{5}{2}}}}.$$

Thus, we obtain that

$$\begin{array}{cc}\hfill & \sum _{n\ge 1}n{T}_n^{\left(2\right)}(b,c)x^{n-1}={\displaystyle \frac{3x(b-(b^2-4c)x)}{{(1-2bx+(b^2-4c)x^2)}^{\frac{5}{2}}}}\hfill \\ \hfill =& 3b\sum _{n\ge 1}{T}_{n-1}^{\left(2\right)}(b,c)x^n-3(b^2-4c)\sum _{n\ge 2}{T}_{n-2}^{\left(2\right)}(b,c)x^n+2b\sum _{n\ge 2}(n-1)T_{n-1}^{\left(2\right)}(b,c)x^n\hfill \\ \hfill & -(b^2-4c)\sum _{n\ge 3}(n-2)T_{n-2}^{\left(2\right)}(b,c)x^n\hfill \\ \hfill =& 3b{T}_0^{\left(2\right)}(b,c)x+\sum _{n\ge 2}[(2n+1)b{T}_{n-1}^{\left(2\right)}(b,c)-(b^2-4c)(n+1)T_{n-2}^{\left(2\right)}(b,c)]x^n\hfill \end{array}$$

□

In light of the generating function of $T_n^{\left(2\right)}(b,c)$, its relation to recurrence can be easily derived. However, the crux lies in determining the explicit expression for the coefficient $T^{\left(2\right)}(n,k)$ of $T_n^{\left(2\right)}(b,c)={\sum }_{k=0}^{\lfloor n/2\rfloor }{T}^{\left(2\right)}(n,k)b^{n-2k}{c}^k$, which requires substantial computation. In the following, we present two approaches for obtaining the expression for coefficient $T^{\left(2\right)}(n,k)$. The first method employs Taylor series expansion, a conventional technique that involves relatively less computation but is more intricate. The second method adopts mathematical induction, which is more intuitive, but computationally more intensive.

Lemma 2. 

The coefficients of the generalized second-order central trinomial equation are

$$T_n^{\left(2\right)}(b,c)=\sum _{k=0}^{⌊{\displaystyle \frac{n}{2}}⌋}{T}^{\left(2\right)}(n,k)b^{n-2k}{c}^k=\sum _{k=0}^{⌊{\displaystyle \frac{n}{2}}⌋}(2k+1)\left({\displaystyle \genfrac{}{}{0pt}{}{n+2}{2k+2}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right)b^{n-2k}{c}^k.$$

Proof of Lemma 2. 

To prove this lemma, we show two different methods that establish this result. On the one hand, the lemma can be proved using Taylor series expansion. However, using inductive assumptions also provides a valid proof.

Method 1: Let

$$f\left(x\right)={\displaystyle \frac{1}{{(1-2bx+(b^2-4c)x^2)}^{\frac{3}{2}}}},$$

and $y=1-2bx+(b^2-4c)x^2$.

Then, $y^{\prime }=-2b+2(b^2-4c)x$, $y^{″}=2(b^2-4c)$.

So,

$$\begin{array}{cc}\hfill f^{\left(n\right)}\left(x\right)& =\sum _{i\ge 0}(2i-1)!!\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2i}}\right)\left[{\displaystyle \frac{d^{(n-i)}}{d{y}^{(n-i)}}}f\left(x\right)\right]{\left(y^{\prime }\right)}^{(n-2i)}{\left(y^{″}\right)}^i\hfill \\ \hfill & =\sum _{i\ge 0}(2i-1)!!\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2i}}\right)\left(-{\displaystyle \frac{3}{2}}\right)\left(-{\displaystyle \frac{5}{2}}\right)\cdots \left(-{\displaystyle \frac{2n-2i+1}{2}}\right)y^{-{\displaystyle \frac{2n-2i+3}{2}}}{\left(y^{\prime }\right)}^{(n-2i)}{\left(y^{″}\right)}^i.\hfill \end{array}$$

The coefficient of $x^n$ in the Taylor expansion [4] of $f\left(x\right)$ is

$$\begin{array}{cc}\hfill T_n^3(b,c)& ={\displaystyle \frac{f^{\left(n\right)}\left(0\right)}{n!}}=\sum _{i\ge 0}{\displaystyle \frac{(2i-1)!!}{n!}}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2i}}\right){(-1)}^{(n-i)}{\displaystyle \frac{(2n-2i+1)!!}{2^{n-i}}}{(-2b)}^{n-2i}{\left[2(b^2-4c)\right]}^i\hfill \\ \hfill & =\sum _{i\ge 0}\sum _{k=0}^i{\displaystyle \frac{(2i-1)!!(2n-2i+1)!!}{\left(2i\right)!(n-2i)!}}\left({\displaystyle \genfrac{}{}{0pt}{}{i}{k}}\right){(-1)}^{i-k}{4}^k{b}^{n-2k}{c}^k.\hfill \end{array}$$

Thus,

$$\begin{array}{cc}\hfill & T^{\left(2\right)}(n,k)=\sum _{i\ge 0}{\displaystyle \frac{(2i-1)!!(2n-2i+1)!!}{\left(2i\right)!(n-2i)!}}\left({\displaystyle \genfrac{}{}{0pt}{}{i}{k}}\right){(-1)}^{i-k}{4}^k\hfill \\ \hfill & =\sum _{i\ge 0}{\displaystyle \frac{2n-2k+1}{2^{n-2k}}}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{i}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2n-2i}{n}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{i}{k}}\right){(-1)}^{i-k}+\sum _{i\ge 0}{\displaystyle \frac{i-k}{2^{n-2k-1}}}\left({\displaystyle \genfrac{}{}{0pt}{}{2n-2i}{n}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{n-k}{n-i}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){(-1)}^{i-k+1}.\hfill \end{array}$$

Using formula [5]

$$\sum _{k\ge 0}{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2n-2k}{n}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{k}{m}}\right)={(-1)}^m{2}^{n-2m}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2m}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2m}{m}}\right),$$

we can obtain

$$\begin{array}{cc}\hfill & T^{\left(2\right)}(n,k)=(2n-2k+1)\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right)+{\displaystyle \frac{n-k}{2^{n-2k-1}}}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\sum _{i\ge 0}{(-1)}^{i-k+1}\left({\displaystyle \genfrac{}{}{0pt}{}{2n-2i}{n}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{n-k-1}{n-i}}\right)\hfill \\ \hfill & =(2n-2k+1)\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right)+{\displaystyle \frac{k+1}{2}}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2k+2}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2k+2}{k+1}}\right)=(2k+1)\left({\displaystyle \genfrac{}{}{0pt}{}{n+2}{2k+2}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right).\hfill \end{array}$$

Method 2: For $T_n^{\left(2\right)}(b,c)={\sum }_{k=0}^{⌊{\displaystyle \frac{n}{2}}⌋}{T}^{\left(2\right)}(n,k)b^{n-2k}{c}^k$, applying Lemma 1 yields the following:

$$\begin{array}{cc}\hfill & n\sum _{k=0}^{⌊{\displaystyle \frac{n}{2}}⌋}{T}^{\left(2\right)}(n,k)b^{n-2k}{c}^k\hfill \\ \hfill =& (2n+1)b\sum _{k=0}^{⌊{\displaystyle \frac{n-1}{2}}⌋}{T}^{\left(2\right)}(n-1,k)b^{n-1-2k}{c}^k-(n+1)(b^2-4c)\sum _{k=0}^{⌊{\displaystyle \frac{n-2}{2}}⌋}{T}^{\left(2\right)}(n-2,k)b^{n-2-2k}{c}^k.\hfill \end{array}$$

To prove Lemma 2, we will divide n into odd and even cases for the sake of our discussion.

Assuming that $n=2i$ is even, we have

$$\begin{array}{cc}\hfill n\sum _{k=0}^i{T}^{\left(2\right)}(n,k)b^{n-2k}{c}^k=& (2n+1)\sum _{k=0}^{i-1}{T}^{\left(2\right)}(n-1,k)b^{n-2k}{c}^k-(n+1)\sum _{k=0}^{i-1}{T}^{\left(2\right)}(n-2,k)b^{n-2k}{c}^k\hfill \\ \hfill & +4(n+1)\sum _{k=0}^i{T}^{\left(2\right)}(n-2,k-1)b^{n-2k}{c}^k\hfill \end{array}$$

where

$$\begin{array}{cc}\hfill n{T}^{\left(2\right)}(n,0)& =(2n+1)T^{\left(2\right)}(n-1,0)-(n+1)T^{\left(2\right)}(n-2,0),k=0;\hfill \end{array}$$

(4)

$$n{T}^{\left(2\right)}(n,i)=4(n+1)T^{\left(2\right)}(n-2,i-1),k=i;$$

(5)

$$n{T}^{\left(2\right)}(n,k)=(2n+1)T^{\left(2\right)}(n-1,k)-(n+1)T^{\left(2\right)}(n-2,k)+4(n+1)T^{\left(2\right)}(n-2,k-1),$$

$$\begin{array}{cc}\hfill & 1\le k\le ⌊{\displaystyle \frac{n-2}{2}}⌋.\hfill \end{array}$$

(6)

Similarly, if $n=2i+1$ is odd, then we have

$$\begin{array}{cc}\hfill & n\sum _{k=0}^i{T}^{\left(2\right)}(n,k)b^{n-2k}{c}^k\hfill \\ \hfill =& (2n+1)\sum _{k=0}^i{T}^{\left(2\right)}(n-1,k)b^{n-2k}{c}^k-(n+1)\sum _{k=0}^{i-1}{T}^{\left(2\right)}(n-2,k)b^{n-2k}{c}^k\hfill \\ \hfill & +4(n+1)\sum _{k=1}^i{T}^{\left(2\right)}(n-2,k-1)b^{n-2k}{c}^k\hfill \end{array}$$

where

$$\begin{array}{cc}\hfill n{T}^{\left(2\right)}(n,i)& =(2n+1)T^{\left(2\right)}(n-1,i)+4(n+1)T^{\left(2\right)}(n-2,i-1),k=i;\hfill \end{array}$$

(7)

For odd $n=2i+1$, the equations of (4) and (6) are preserved. Notably, the behavior at $k=0$ and $1\le k\le ⌊{\displaystyle \frac{n-2}{2}}⌋$ is consistent across both cases of parity of n.

• (i) It follows from the Equation (4) that

$$n\left[T^{\left(2\right)}(n,0)-T^{\left(2\right)}(n-1,0)\right]=(n+1)\left[T^{\left(2\right)}(n-1,0)-T^{\left(2\right)}(n-2,0)\right].$$

Clearly,

$${\displaystyle \frac{T^{\left(2\right)}(n,0)-T^{\left(2\right)}(n-1,0)}{T^{\left(2\right)}(1,0)-T^{\left(2\right)}(0,0)}}={\displaystyle \frac{n+1}{2}},$$

since $T^{\left(2\right)}(n,0)-T^{\left(2\right)}(n-1,0)=n+1.$

Therefore,

$$T^{\left(2\right)}(n,0)=\left({\displaystyle \genfrac{}{}{0pt}{}{n+2}{2}}\right).$$

• (ii) Substituting $n=2i$ into Equation (5) gives the following:

$$T^{\left(2\right)}(2i,i)=(2i+1)\left({\displaystyle \genfrac{}{}{0pt}{}{2i}{i}}\right).$$

Since

$${\displaystyle \frac{T^{\left(2\right)}(2i,i)}{T^{\left(2\right)}(2i-2,i-1)}}={\displaystyle \frac{2(2i+1)}{i}}.$$

It follows that ${\displaystyle \frac{T^{\left(2\right)}(2i,i)}{T^{\left(2\right)}(0,0)}}={\displaystyle \frac{2^i(2i+1)!!}{i!}}$.

• (iii) Substituting $n=2i+1$ into Equation (7) gives the following:

$$T^{\left(2\right)}(2i+1,i)=(2i+1)\left({\displaystyle \genfrac{}{}{0pt}{}{2i+3}{2i+2}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2i}{i}}\right),$$

since

$$\begin{array}{cc}\hfill (2i+1)T^{\left(2\right)}(2i+1,i)& =(4i+3)T^{\left(2\right)}(2i,i)+4(2i+2)T^{\left(2\right)}(2i-1,i-1)\hfill \\ \hfill & =(4i+3)(2i+1)\left({\displaystyle \genfrac{}{}{0pt}{}{2i}{i}}\right)+4(2i+2)T^{\left(2\right)}(2i-1,i-1),\hfill \end{array}$$

i.e.,

$$\begin{array}{cc}\hfill & (2i+1)\left[T^{\left(2\right)}(2i+1,i)-(2i+3)(2i+1)\left({\displaystyle \genfrac{}{}{0pt}{}{2i}{i}}\right)\right]\hfill \\ \hfill & =4(2i+2)\left[T^{\left(2\right)}(2i-1,i-1)-(2i-1)(2i+1)\left({\displaystyle \genfrac{}{}{0pt}{}{2i-2}{i-1}}\right)\right].\hfill \end{array}$$

For the special case when $i=0$, we have $T^{\left(2\right)}(1,0)-3=0$.

Clearly,

$$T^{\left(2\right)}(2i+1,i)-(2i+3)(2i+1)\left({\displaystyle \genfrac{}{}{0pt}{}{2i}{i}}\right)\equiv 0.$$

• (iv) Based on the case when $1\le k\le ⌊{\displaystyle \frac{n-2}{2}}⌋$, simplifying Equation (6) yields the following:

$${\displaystyle \frac{T^{\left(2\right)}(n,k)-T^{\left(2\right)}(n-1,k)}{n+1}}-{\displaystyle \frac{T^{\left(2\right)}(n-1,k)-T^{\left(2\right)}(n-2,k)}{n}}={\displaystyle \frac{4}{n}}{T}^{\left(2\right)}(n-2,k-1),$$

therefore,

$${\displaystyle \frac{T^{\left(2\right)}(n,k)-T^{\left(2\right)}(n-1,k)}{n+1}}-{\displaystyle \frac{T^{\left(2\right)}(2k+1,k)-T^{\left(2\right)}(2k,k)}{2k+2}}=4\sum _{i=2}^{n-2k}{\displaystyle \frac{T^{\left(2\right)}(n-i,k-1)}{n+2-i}}.$$

So

$${\displaystyle \frac{T^{\left(2\right)}(n,k)-T^{\left(2\right)}(n-1,k)}{n+1}}=(2k+1)\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right)+4\sum _{i=2}^{n-2k}{\displaystyle \frac{T^{\left(2\right)}(n-i,k-1)}{n+2-i}}.$$

Using mathematical induction on k, we shall prove that

$$T^{\left(2\right)}(n,k)=(2k+1)\left({\displaystyle \genfrac{}{}{0pt}{}{n+2}{2k+2}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2i}{i}}\right).$$

(1) When $k=1$, we have $T^{\left(2\right)}(n,1)=3\left({\displaystyle \genfrac{}{}{0pt}{}{n+2}{4}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2}{1}}\right).$

Due to

$$\begin{array}{c}\hfill {\displaystyle \frac{T^{\left(2\right)}(n,1)-T^{\left(2\right)}(n-1,1)}{n+1}}=3·\left({\displaystyle \genfrac{}{}{0pt}{}{2}{1}}\right)+4\sum _{i=2}^{n-2}{\displaystyle \frac{T^{\left(2\right)}(n-i,0)}{n+2-i}}=n(n-1),\end{array}$$

i.e., $T^{\left(2\right)}(n,1)-T^{\left(2\right)}(n-1,1)=6\left({\displaystyle \genfrac{}{}{0pt}{}{n+1}{3}}\right)$.

So

$$T^{\left(2\right)}(n,1)-T^{\left(2\right)}(2,1)=6\sum _{i=0}^{n-3}\left({\displaystyle \genfrac{}{}{0pt}{}{n+1-i}{3}}\right).$$

Using the formula

$$\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{n-1}{k}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{n-2}{k}}\right)+\cdots +\left({\displaystyle \genfrac{}{}{0pt}{}{k}{k}}\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{n+1}{k+1}}\right),$$

we obtain the following:

$$T^{\left(2\right)}(n,1)=6\sum _{i=0}^{n-2}\left({\displaystyle \genfrac{}{}{0pt}{}{n+1-i}{3}}\right)=3\left({\displaystyle \genfrac{}{}{0pt}{}{n+2}{4}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2}{1}}\right).$$

(2) When $k=2$, we have $T^{\left(2\right)}(n,2)=5\left({\displaystyle \genfrac{}{}{0pt}{}{n+2}{6}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{4}{2}}\right).$ Because of

$$\begin{array}{c}\hfill {\displaystyle \frac{T^{\left(2\right)}(n,2)-T^{\left(2\right)}(n-1,2)}{n+1}}=5·\left({\displaystyle \genfrac{}{}{0pt}{}{4}{2}}\right)+4\sum _{i=2}^{n-4}{\displaystyle \frac{T^{\left(2\right)}(n-i,1)}{n+2-i}}=6\left({\displaystyle \genfrac{}{}{0pt}{}{n}{4}}\right),\end{array}$$

it follows that $T^{\left(2\right)}(n,2)-T^{\left(2\right)}(n-1,2)=30\left({\displaystyle \genfrac{}{}{0pt}{}{n+1}{5}}\right).$

So

$$T^{\left(2\right)}(n,2)-T^{\left(2\right)}(4,2)=30\sum _{i=0}^{n-5}\left({\displaystyle \genfrac{}{}{0pt}{}{n+1-i}{5}}\right).$$

Hence,

$$T^{\left(2\right)}(n,2)=30\sum _{i=0}^{n-4}\left({\displaystyle \genfrac{}{}{0pt}{}{n+1-i}{5}}\right)=5\left({\displaystyle \genfrac{}{}{0pt}{}{n+2}{6}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{4}{2}}\right).$$

(3) We conjecture that the quantity $T^{\left(2\right)}(n,k)$ admits the following explicit expression:

$$T^{\left(2\right)}(n,k)=(2k+1)\left({\displaystyle \genfrac{}{}{0pt}{}{n+2}{2k+2}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right).$$

Under the assumption that this formula holds for all cases where $k<n$, we now proceed to rigorously prove the following:

$$T^{\left(2\right)}(n,k)=(2k+1)\left({\displaystyle \genfrac{}{}{0pt}{}{n+2}{2k+2}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right).$$

Then,

$$\begin{array}{cc}\hfill & {\displaystyle \frac{T^{\left(2\right)}(n,k)-T^{\left(2\right)}(n-1,k)}{n+1}}=(2k+1)\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right)+4\sum _{i=2}^{n-2k}{\displaystyle \frac{T^3(n-i,k-1)}{n+2-i}}\hfill \\ \hfill =& \sum _{i=2}^{n-2(k-1)}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{n+1-i}{2k-1}}\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2k}}\right).\hfill \end{array}$$

Therefore,

$$T^{\left(2\right)}(n,k)-T^{\left(2\right)}(n-1,k)=(2k+1)\left({\displaystyle \genfrac{}{}{0pt}{}{n+1}{2k+1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right),$$

since

$$T^{\left(2\right)}(n,k)-T^{\left(2\right)}(2k,k)=\sum _{i=2k+2}^{n+1}(2k+1)\left({\displaystyle \genfrac{}{}{0pt}{}{i}{2k+1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right).$$

Thus,

$$T^{\left(2\right)}(n,k)=\sum _{i=2k+1}^{n+1}(2k+1)\left({\displaystyle \genfrac{}{}{0pt}{}{i}{2k+1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right)=(2k+1)\left({\displaystyle \genfrac{}{}{0pt}{}{n+1}{2k+1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right).$$

□

Let $h\left(x\right)$ be a real-rooted polynomial, that is, a real polynomial with only real roots. Denote by $r_i\left(h\right)$ the roots of f sorted in non-increasing order as follows: $r_1\left(h\right)\ge r_2\left(h\right)\ge r_3\left(h\right)\ge \cdots$. Let h,g be two real-rooted polynomials and $degg\le degh\le degg+1$. We say that g interlaces h, denoted by $g⪯h$, if

$$r_1\left(h\right)\ge r_1\left(g\right)\ge r_2\left(h\right)\ge r_2\left(g\right)\ge \cdots .$$

If all inequalities in

$$r_1\left(h\right)\ge r_1\left(g\right)\ge r_2\left(h\right)\ge r_2\left(g\right)\ge \cdots$$

are strict, then we say that g strictly binds to h and denote it by $g\prec h$. For notational convenience, we say that a real number a is real rooted and $a\prec bx+c$ for a, $b+c>0$, and $b,c\ge 0$[2]. If $g_1\prec h$ and $h\prec g_2$, then h is said to be interlaced between $g_1$ and $g_2$.

Proof of Theorem 1. 

By mathematical induction in n, when $n=0$, ${\mathcal{T}}_0^{\left(2\right)}\left(x\right)=T_0^{\left(2\right)}(1,x)=1$; when $n=1$, we have ${\mathcal{T}}_1^{\left(2\right)}\left(x\right)=T_1^{\left(2\right)}(1,x)=3$; when $n=2$, we have ${\mathcal{T}}_2^{\left(2\right)}\left(x\right)=T_2^{\left(2\right)}(1,x)=6+6x$; and when $n=3$, we have ${\mathcal{T}}_3^{\left(2\right)}\left(x\right)=T_3^{\left(2\right)}(1,x)=10+30x$. Obviously, for $n\le 3$, the conclusions of Theorem 1 are true. We assume that both ${\mathcal{T}}_{n-2}^{\left(2\right)}\left(x\right)$ and ${\mathcal{T}}_{n-1}^{\left(2\right)}\left(x\right)$ have real roots, and ${\mathcal{T}}_{n-2}^{\left(2\right)}\left(x\right)\prec {\mathcal{T}}_{n-1}^{\left(2\right)}\left(x\right)$, to prove that ${\mathcal{T}}_n^{\left(2\right)}\left(x\right)$ has real roots and ${\mathcal{T}}_{n-1}^{\left(2\right)}\left(x\right)\prec {\mathcal{T}}_n^{\left(2\right)}\left(x\right)$.

Denote by $sgn\left(k\right)$ the sign of a real number t, i.e.,

$$\begin{array}{c}\hfill \left\{\begin{array}{cc}& +1,k>0;\\ & 0,k=0;\\ & -1,k<0.\end{array}\right.\end{array}$$

Let $r_1^{(n-1)}>r_2^{(n-1)}>r_3^{(n-1)}>\cdots$ be the real roots of ${\mathcal{T}}_{n-1}^{\left(2\right)}\left(x\right)$, and let $x_1^{(n-2)},x_2^{(n-2)},\cdots ,x_{⌊\frac{n-2}{2}⌋}^{(n-2)}$ be the real roots of ${\mathcal{T}}_{n-2}^{\left(2\right)}\left(x\right)$. Then, $x_0^{(n-2)}={\displaystyle \frac{1}{4}},x_1^{(n-2)},x_2^{(n-2)},$$\cdots ,x_{⌊\frac{n-2}{2}⌋}^{(n-2)}$ are the real roots of $(1-4x){\mathcal{T}}_{n-2}^{\left(2\right)}\left(x\right)$. When $x=x_i^{(n-2)}$, for $i=0,1,2,\cdots ,⌊{\displaystyle \frac{n-2}{2}}⌋$, then we have

$$n{\mathcal{T}}_n^{\left(2\right)}\left(x_i^{(n-2)}\right)=(2n+1){\mathcal{T}}_{n-1}^{\left(2\right)}\left(x_i^{(n-2)}\right).$$

Therefore, $sgn{\mathcal{T}}_n^{\left(2\right)}\left(x_i^{(n-2)}\right)={(-1)}^i,i=0,1,2,\cdots ,⌊{\displaystyle \frac{n-2}{2}}⌋$.

According to the zero-point existence theorem, it can be inferred that ${\mathcal{T}}_n^{\left(2\right)}\left(x\right)$ has at least $⌊{\displaystyle \frac{n}{2}}⌋-1$ zeros. But ${\mathcal{T}}_n^{\left(2\right)}\left(x\right)$ has no more $⌊{\displaystyle \frac{n}{2}}⌋$ zeros. And virtual roots appear in pairs, so the roots of ${\mathcal{T}}_n^{\left(2\right)}\left(x\right)$ are all real.

As ${\mathcal{T}}_{n-2}^{\left(2\right)}\left(x\right)\prec {\mathcal{T}}_{n-1}^{\left(2\right)}\left(x\right)$, we have $sgn{\mathcal{T}}_{n-2}^{\left(2\right)}\left(r_i^{(n-1)}\right)={(-1)}^{i-1},i=0,1,2,\cdots ,⌊{\displaystyle \frac{n-1}{2}}⌋$. And $n{\mathcal{T}}_n^{\left(2\right)}\left(x\right)=(2n+1){\mathcal{T}}_{n-1}^{\left(2\right)}\left(x\right)-(1-4x)(n+1){\mathcal{T}}_{n-2}^{\left(2\right)}\left(x\right)$. Let $x=r_i^{(n-1)}$, then

$$n{\mathcal{T}}_n^{\left(2\right)}\left(r_i^{(n-1)}\right)=-(1-4r_i^{(n-1)})(n+1){\mathcal{T}}_{n-2}^{\left(2\right)}\left(r_i^{(n-1)}\right).$$

Therefore, $sgn{\mathcal{T}}_n^{\left(2\right)}\left(r_i^{(n-1)}\right)={(-1)}^i,i=1,2,\cdots ,⌊{\displaystyle \frac{n-1}{2}}⌋$. Thus, ${\mathcal{T}}_{n-1}^{\left(2\right)}\left(x\right)\prec {\mathcal{T}}_n^{\left(2\right)}\left(x\right).$

Next, we prove that ${\mathcal{T}}_{n-2}^{\left(2\right)}\left(x\right)\prec {\mathcal{T}}_n^{\left(2\right)}\left(x\right)$. When $n\ge 3$, according to ${\mathcal{T}}_{n-2}^{\left(2\right)}\left(x\right)\prec {\mathcal{T}}_{n-1}^{\left(2\right)}\left(x\right)$, we have $sgn{\mathcal{T}}_n^{\left(2\right)}\left(x_i^{(n-2)}\right)={(-1)}^i,i=1,2,\cdots ,⌊{\displaystyle \frac{n-2}{2}}⌋$. Let $x=x_i^{(n-2)}$, then $n{\mathcal{T}}_n^{\left(2\right)}\left(x_i^{(n-2)}\right)=(2n+1){\mathcal{T}}_{n-1}^{\left(2\right)}\left(x_i^{(n-2)}\right)$. So $sgn{\mathcal{T}}_n^{\left(2\right)}\left(x_i^{(n-2)}\right)={(-1)}^i,i=1,2,\cdots ,⌊{\displaystyle \frac{n-2}{2}}⌋$. Therefore, ${\mathcal{T}}_{n-2}^{\left(2\right)}\left(x\right)\prec {\mathcal{T}}_n^{\left(2\right)}\left(x\right)$. □

3. Asymptotic Normality of ${\mathit{T}}^{\left(\mathbf{2}\right)}(\mathit{n},\mathit{k})$

The real roots of polynomials with nonnegative coefficients are closely related to totally positive matrices. Following Karlin, we say that a matrix A (finite or infinite) is a total positive of order r (denoted by $T{P}_r$) if all its minors of orders $\le r$ are nonnegative. It is called totally positive ($TP$) if all minors of all orders are non-negative. Let ${\left(a_k\right)}_{k\ge 0}$ be a sequence of nonnegative numbers (finite sequences are implicitly extended to infinite by zero-padding). We say that the sequence is a Pólya frequency ($PF$ for short) sequence if the corresponding infinite Toeplitz matrix

$${\left[a_{n-k}\right]}_{n,k\ge 0}=\left(\begin{array}{ccccc}{a}_0& 0& 0& 0& \cdots \\ a_1& a_0& 0& 0& \cdots \\ a_2& a_1& a_0& 0& \cdots \\ a_3& a_2& a_1& a_0& \cdots \\ ⋮& ⋮& ⋮& ⋮& \ddots \end{array}\right)$$

is $TP$. The following is a fundamental characterization for $PF$ sequences [2].

Lemma 3 

(Schoeber–Edrei Theorem [2]). A sequence ${\left(a_k\right)}_{k\ge 0}$ is $PF$ if and only if its generating function is

$$\sum _{k\ge 0}{a}_k{x}^k=a{x}^m{e}^{\gamma x}{\displaystyle \frac{{\prod }_{j\ge 0}(1+{\alpha }_{j}x)}{{\prod }_{j\ge 0}(1-{\beta }_{j}x)}}.$$

Here, $a>0,m\in N,{\alpha }_j,{\beta }_j,\gamma \ge 0,{\sum }_{j\ge 0}({\alpha }_j+{\beta }_j)<+\infty .$

Theorem 3. 

The matrix $T={\left[T^{\left(2\right)}(n,k)\right]}_{n,k\ge 0}$ is $TP$.

Proof of Theorem 3. 

According to Lemma (3), ${\left({\displaystyle \frac{1}{k!}}\right)}_{k\ge 0}$ is $PF$, since

$$\sum _{k\ge 0}{\displaystyle \frac{1}{k!}}{x}^k=e^x.$$

So the Toeplitz matrix ${\left({\displaystyle \frac{1}{(n-k)!}}\right)}_{n,k\ge 0}$ is $TP$. Therefore, the submatrix ${\left({\displaystyle \frac{1}{(n-2k)!}}\right)}_{n,k\ge 0}$ is $TP$, which consists of even columns of the Toeplitz matrix.

As ${\left({\displaystyle \frac{1}{(n-2k)!}}\right)}_{n,k\ge 0}$ is $TP$, the sequences ${\left({\displaystyle \frac{1}{k!(k+1)!}}\right)}_{k\ge 0},{\left({\displaystyle \frac{(n+2)!}{2}}\right)}_{n\ge 0}$ are $PF$. So, the matrix

$$T={\left[T^{\left(2\right)}(n,k)\right]}_{n,k\ge 0}={\displaystyle \frac{(n+2)!}{2(k+1)!k!(n-2k)!}}$$

is $TP$. □

Lemma 4. 

Suppose that ${\ell }_n\left(x\right)={\sum }_{k=0}^{n}a(n,k)x^k$ has only real roots, where $a(n,k)>0$. Let ${\mu }_n={\displaystyle \frac{{\ell }_n^{\prime }\left(1\right)}{{\ell }_n\left(1\right)}}$ and ${\sigma }_n^2={\displaystyle \frac{{\ell }_n^{″}\left(1\right)}{{\ell }_n\left(1\right)}}+{\mu }_n-{\mu }_n^2$. If ${\sigma }_n^2\to +\infty$, then the numbers $a(n,k)$ are asymptotically normal with mean ${\mu }_n$ and variance ${\sigma }_n^2$.

Proof of Theorem 2. 

By Lemma (4), we need to show that ${\sigma }_n^2\to +\infty$.

$$\begin{array}{cc}\hfill & {\left(T_n^{\left(2\right)}\right)}^{\prime }=\sum _{k}k(2k+1)\left({\displaystyle \genfrac{}{}{0pt}{}{n+2}{2k+2}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right)={\displaystyle \frac{n}{2}}{T}_n^{\left(2\right)}-{\displaystyle \frac{n+2}{2}}{T}_{n-1}^{\left(2\right)}.\hfill \end{array}$$

Let

$$a_k=(2k+1)\left({\displaystyle \genfrac{}{}{0pt}{}{n+2}{2k+2}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right),$$

then $T_n^{\left(2\right)}={\sum }_k{a}_k,T_{n-1}^{\left(2\right)}={\sum }_k{\displaystyle \frac{n-2k}{n+2}}{a}_k,T_{n+1}^{\left(2\right)}={\sum }_k{\displaystyle \frac{n+3}{n+1-2k}}{a}_k.$

Let $m\left(t\right)=T_{n+1}^{\left(2\right)}{t}^2+2T_n^{\left(2\right)}t+T_{n-1}^{\left(2\right)}={\sum }_k({\displaystyle \frac{n+3}{n+1-2k}}{t}^2+2t+{\displaystyle \frac{n-2k}{n+2}})a_k$.

The equation $m\left(t\right)=0$ is a quadratic equation, where

$$\Delta =4-{\displaystyle \frac{4(n-2k)(n+3)}{(n+2)(n+1-2k)}}={\displaystyle \frac{4(n+3)}{n+1-2k}}·{\displaystyle \frac{(2+2k)}{(n+3)(n+2)}}>0.$$

So ${\displaystyle \frac{n+3}{n+1-2k}}{t}^2+2t+{\displaystyle \frac{n-2k}{n+2}}=0$ has two different roots.

There exists a root $t_0$, such that

$${\displaystyle \frac{n+3}{n+1-2k}}{t}_0^2+2t_0+{\displaystyle \frac{n-2k}{n+2}}<0.$$

Then, $m\left(t_0\right)<0$. But

$$m\left(1\right)=T_{n+1}^{\left(2\right)}+2T_n^{\left(2\right)}+T_{n-1}^{\left(2\right)}>0.$$

So $m\left(t\right)=0$ has a real root, that is ${\left(T_n^{\left(2\right)}\right)}^2\ge T_{n+1}^{\left(2\right)}{T}_{n-1}^{\left(2\right)}$.

Hence, sequence $\left\{{\displaystyle \frac{T_{n-1}^{\left(2\right)}}{T_n^{\left(2\right)}}}\right\}$ increases monotonically. But $T_{n-1}^{\left(2\right)}={\sum }_k{\displaystyle \frac{n-2k}{n+2}}{a}_k<{\sum }_k{a}_k=T_n^{\left(2\right)}$, then ${\displaystyle \frac{T_{n-1}^{\left(2\right)}}{T_n^{\left(2\right)}}}<1$. Thus, sequence $\left\{{\displaystyle \frac{T_{n-1}^{\left(2\right)}}{T_n^{\left(2\right)}}}\right\}$ is convergent. As $n{T}_n^{\left(2\right)}=(2n+1)T_{n-1}^{\left(2\right)}+3(n+1)T_{n-2}^{\left(2\right)}$, we can obtain that

$${\displaystyle \frac{T_{n-1}^{\left(2\right)}}{T_n^{\left(2\right)}}}\to {\displaystyle \frac{1}{3}}.$$

Therefore,

$${\mu }_n={\displaystyle \frac{{\left[{\mathcal{T}}_n^{\left(2\right)}\left(1\right)\right]}^{\prime }}{{\mathcal{T}}_n^{\left(2\right)}\left(1\right)}}={\displaystyle \frac{n{T}_n^{\left(2\right)}-(n+3)T_{n-1}^{\left(2\right)}}{2T_n^{\left(2\right)}}}={\displaystyle \frac{n-1}{3}}+\mathrm{O}\left(1\right).$$

On the other hand,

$$\begin{array}{cc}\hfill & {\left[{\mathcal{T}}_n^{\left(2\right)}\left(1\right)\right]}^{″}+{\left[{\mathcal{T}}_n^{\left(2\right)}\left(1\right)\right]}^{\prime }=\sum _k{k}^2(2k+1)\left({\displaystyle \genfrac{}{}{0pt}{}{n+2}{2k+2}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right)\hfill \\ \hfill & ={\displaystyle \frac{n^2}{4}}{T}_n^{\left(2\right)}-{\displaystyle \frac{(2n-1)(n+2)}{4}}{T}_{n-1}^{\left(2\right)}+{\displaystyle \frac{(n+2)(n+1)}{4}}{T}_{n-2}^{\left(2\right)}.\hfill \end{array}$$

Hence,

$$\begin{array}{cc}\hfill & {\sigma }_n^2={\displaystyle \frac{{\left[{\mathcal{T}}_n^{\left(2\right)}\left(1\right)\right]}^{″}+{\left[{\mathcal{T}}_n^{\left(2\right)}\left(1\right)\right]}^{\prime }}{{\mathcal{T}}_n^{\left(2\right)}\left(1\right)}}-{\mu }_n^2\hfill \\ \hfill & ={\displaystyle \frac{n^2}{4}}-{\displaystyle \frac{(2n-1)(n+2)}{4}}{\displaystyle \frac{T_{n-1}^{\left(2\right)}}{T_n^{\left(2\right)}}}+{\displaystyle \frac{(n+2)(n+1)}{4}}{\displaystyle \frac{T_{n-2}^{\left(2\right)}}{T_n^{\left(2\right)}}}-{\left[{\displaystyle \frac{n{T}_n^{\left(2\right)}-(n+2)T_{n-1}^{\left(2\right)}}{2T_n^{\left(2\right)}}}\right]}^2\hfill \\ \hfill & ={\displaystyle \frac{n+2}{4}}{\displaystyle \frac{T_{n-1}^{\left(2\right)}}{T_n^{\left(2\right)}}}+{\displaystyle \frac{(n+2)(n+1)}{4}}{\displaystyle \frac{T_{n-2}^{\left(2\right)}}{T_n^{\left(2\right)}}}+{\displaystyle \frac{{(n+2)}^2}{4}}{\left({\displaystyle \frac{T_{n-2}^{\left(2\right)}}{T_n^{\left(2\right)}}}\right)}^2,\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill {\sigma }_n^2& ={\displaystyle \frac{n+2}{4}}·{\displaystyle \frac{T_{n-1}^{\left(2\right)}{T}_n^{\left(2\right)}+(2n+3)T_{n-2}^{\left(2\right)}{T}_n^{\left(2\right)}+(n+2){\left[T_{n-1}^{\left(2\right)}\right]}^2-(n+2)T_{n-2}^{\left(2\right)}{T}_n^{\left(2\right)}}{{\left[T_n^{\left(2\right)}\right]}^2}}\hfill \\ \hfill & \ge {\displaystyle \frac{n+2}{4}}·{\displaystyle \frac{T_{n-1}^{\left(2\right)}+(2n+3)T_{n-2}^{\left(2\right)}}{T_n^{\left(2\right)}}}.\hfill \end{array}$$

Since

$${\left(T_n^{\left(2\right)}\right)}^2\ge T_{n+1}^{\left(2\right)}{T}_{n-1}^{\left(2\right)}.$$

we can obtain

$$\begin{array}{c}\hfill {\displaystyle \frac{n+2}{4}}·{\displaystyle \frac{T_{n-1}^{\left(2\right)}+(2n+3)T_{n-2}^{\left(2\right)}}{T_n^{\left(2\right)}}}={\displaystyle \frac{n+2}{4}}·\left[{\displaystyle \frac{2(n+3)}{9}}+\mathrm{O}\left(1\right)\right]={\displaystyle \frac{(n+2)(n+3)}{18}}+\mathrm{O}\left(n\right).\end{array}$$

Thus, ${\sigma }_n^2\to +\infty$, as required. □

4. Conclusions

This paper investigates the analytic properties of generalized second-order central trinomial coefficients $T_n^{\left(2\right)}(b,c)$. Our main contributions are as follows.

1.

Real zeros and interlacing properties (Theorem 1): For the polynomial

$${\mathcal{T}}_n^{\left(2\right)}\left(x\right)=\sum _{k=0}^{\lfloor n/2\rfloor }{T}^{\left(2\right)}(n,k)x^k,$$

we prove the following:

• All zeros are real;
• The zeros of ${\mathcal{T}}_{n-1}^{\left(2\right)}\left(x\right)$ strictly interlace with those of both ${\mathcal{T}}_{n-2}^{\left(2\right)}\left(x\right)$ and ${\mathcal{T}}_n^{\left(2\right)}\left(x\right)$.

2.

Asymptotic normality (Theorem 2): The coefficients $T^{\left(2\right)}(n,k)$ are asymptotically normal with

$${\mu }_n={\displaystyle \frac{n-1}{3}}+O\left(1\right),{\sigma }_n^2\to +\infty .$$

The paper develops these results through the following:

• Section 2: Zeros of ${\mathcal{T}}_n^{\left(2\right)}\left(x\right)$
  The explicit expression for the generalized second-order central trinomial coefficients $T_n^{\left(2\right)}(b,c)$ is given by two different methods.

  –

  Generating function approach: Taylor series expansion of

  $${\displaystyle \frac{1}{{(1-2bx+(b^2-4c)x^2)}^{3/2}}}$$

  –

  Mathematical induction on the recurrence relation: Perform induction on k for $T^{\left(2\right)}(n,k)$.

  Based on this, we investigate the zeros of the generalized second-order central trinomial polynomial using the Intermediate Value Theorem and to prove Theorem 1.
• Section 3: Asymptotic normality of $T^{\left(2\right)}(n,k)$
  Starting from the real zeros of the generalized second-order centered trinomial polynomial, we investigate whether the variance tends to $+\infty$, thereby proving that the generalized second-order centered trinomial numbers $T^{\left(2\right)}(b,c)$ are asymptotically normal, and we determine their mean ${\mu }_n$ and variance ${\sigma }_n^2$.

Contributions and Extensions

Our work generalizes Liang, Wang, and Wang’s (2023) [2] results on classical trinomial coefficients $T_n$ to the second-order case. Future directions include the following:

• Combinatorics: Extension to higher-order coefficients $T_n^{\left(r\right)}(b,c)$ for $r\ge 3$.
• Physics: The potential q-analogue of $T_n^{\left(2\right)}$ may provide insights into hard-core lattice models and quantum spin chains with higher-order interactions.