1. Introduction
The objective of this paper is to consider an inequality arising from the iterated Laguerre operator on the functions in Laguerre-Pólya class as follows
\begin{equation}
(a_{n+1}a_{n+2} - a_{n}a_{n+3})^2 -(a_{n+1}^2 -a_{n}a_{n+2})(a_{n+2}^2 - a_{n}a_{n+4}) \gt 0.
\end{equation}A real entire function
\begin{eqnarray}
\psi(x)=\sum_{k=0}^{\infty} \gamma_k\frac{x^k}{k!}
\end{eqnarray} is said to be in the Laguerre-Pólya class, denoted by 
$\psi(x) \in \mathcal{LP}$, if it can be represented in the form
\begin{eqnarray*}
\psi(x)=c x^m e^{-\alpha x^{2}+\beta x} \prod_{k=1}^{\infty}\left(1+x/x_{k}\right)e^{- x/x_k},
\end{eqnarray*} where c, β, xk are real numbers, 
$\alpha \geq 0$, m is a non-negative integer and 
$\sum x^{-2}_{k} \lt \infty$.
The 
$\mathcal{LP}$ class has attracted much attention in view of its connection with the Riemann hypothesis. The Riemann Ξ-function is defined by
\begin{equation}
\Xi(z):=\frac{1}{2}\left(-z^2-\frac{1}{4}\right)\pi^{\frac{iz}{2}-\frac{1}{4}}\Gamma
\left(-\frac{iz}{2}+\frac{1}{4}\right)\zeta\left(-iz+\frac{1}{2}\right).
\end{equation}Following [Reference Nuttall20], the Riemann Ξ-function can be written as
\begin{equation*}\Xi(z)=\sum \limits_{n=0}^{\infty}\frac{\gamma(n)}{n!}z^{2n}.\end{equation*} The Riemann hypothesis is equivalent to the statement that 
$\Xi(z)$ belongs to 
$\mathcal{LP}$.
The Laguerre inequality plays an important role in the study of the 
$\mathcal{LP}$ class and the Riemann hypothesis (see [Reference Csordas and Varga7]). A polynomial f(x) is called to satisfy the Laguerre inequality if
\begin{equation}
{f'(x)}^2-f(x)f''(x)\geq 0,
\end{equation} holds. Jensen [Reference Jensen17] found a higher order generalization of the Laguerre inequality, namely, for 
$f\in\mathcal{LP}$ and all real x,
\begin{equation}
L_{n}(f(x))\colon =\frac{1}{2}\sum_{k=0}^{2n}(-1)^{n+k}{{2n}\choose k}f^{(k)}(x)f^{(2n-k)}(x) \geq 0,
\end{equation} where 
$f^{(k)}(x)$ denotes the k-th derivative of f(x). It yields the classical Laguerre inequality when n = 1. Since Csordas and Vishnyakova [Reference Csordas and Vishnyakova8] proved that 
$\phi \in \mathcal{LP}$ is equivalent to 
$L_{k}(\phi(x)) \geq 0$ for all 
$k \in \mathbb{N}$ and 
$x \in \mathbb{R}$, the Laguerre inequality is a characterizing property of functions in the Laguerre-Pólya class. For a discrete sequence, we say it satisfies the Laguerre inequality of order m if
\begin{equation}
L_{m}(a_n)\colon =\frac{1}{2}\sum_{k=0}^{2m}(-1)^{k+m}{{2m}\choose k}a_{n+k}a_{2m-k+n} \geq 0,
\end{equation}see [Reference Wang and Yang30]. For more background on Laguerre inequalities, we refer to [Reference Gasper13], [Reference Patrick25], [Reference Patrick26], [Reference Skovgaard27].
Recall that a sequence 
$\{a_n\}_{n\geq1}$ is called to satisfy the Turán inequality if
\begin{equation*}
T(a_n)\colon={a_n}^2-a_{n-1}a_{n+1}\geq0,
\end{equation*} we say that 
$\{a_n\}_{n\geq1}$ satisfies the Turán inequality of order 3 and the double Turán inequality if
\begin{equation*} T_3(a_n)\colon=4({a_n}^2-a_{n-1}a_{n+1})({a_{n+1}}^2-a_na_{n+2})-(a_na_{n+1}-a_{n-1}a_{n+2})^2\geq0
\end{equation*}and
\begin{equation*} T(T(a_n))\colon=({a_n}^2-a_{n-1}a_{n+1})^2-({a_{n-1}}^2-a_{n-2}a_{n})({a_{n+1}}^2-a_na_{n+2})\geq0
\end{equation*}respectively. For more background, see [Reference Chen, Jia and Wang5], [Reference Csordas, Norfolk and Varga6], [Reference Ono, Pujahari and Rolen23], [Reference Pólya and Schur24] and [Reference Skovgaard27].
Let 
$\rho_{a}(x)=\sum_{k\geq0}\frac{a_{k}}{k!}x^k$. Note that even though
\begin{equation*}
L_{1}(\rho^{(n)}_{a}(x))|_{x=0} =\left(\rho_{a}^{(n+1)}(x)^2-\rho_{a}^{(n)}(x)\rho_{a}^{(n+2)}(x)\right)|_{x=0}=a^2_{n+1}-a_{n}a_{n+2}
= T(a_{n+1}),
\end{equation*}their iterations are not equal. This fact has been mentioned by Wagner [Reference Wagner29]. For example,
\begin{align*}
&L_{1}^{(2)}(\rho_{a}^{(n)}(x))|_{x=0}=L_1(L_{1}(\rho_{a}^{(n)}(x)))|_{x=0}\\
=&\Bigg(\left(\left(\rho_{a}^{(n+1)}(x)^2-\rho_{a}^{(n)}(x)\rho_{a}^{(n+2)}(x)\right)'\right)^2\\
&-\left(\rho_{a}^{(n+1)}(x)^2-\rho_{a}^{(n)}(x)\rho_{a}^{(n+2)}(x)\right)
\left(\rho_{a}^{(n+1)}(x)^2-\rho_{a}^{(n)}(x)\rho_{a}^{(n+2)}(x)\right)''\Bigg)\mid_{x=0}\\
=&\Bigg(\left(\rho_{a}^{(n+1)}(x)\rho_{a}^{(n+2)}(x)-\rho_{a}^{(n)}(x)\rho_{a}^{(n+3)}(x)\right)^2\\
&-\left(\rho_{a}^{(n+1)}(x)^2-\rho_{a}^{(n)}(x)\rho_{a}^{(n+2)}(x)\right)
\left(\rho_{a}^{(n+2)}(x)-\rho_{a}^{(n)}(x)\rho_{a}^{(n+4)}(x)\right)\Bigg)|_{x=0}\\
=& (a_{n+1}a_{n+2} - a_{n}a_{n+3})^2 -(a_{n+1}^2 -a_{n}a_{n+2})(a_{n+2}^2 - a_{n}a_{n+4}).
\end{align*} One can see that all 
$L_{1}^{(2)}, T^{(2)}$, and T 3 have the same degree, but are not equal. Note that for a real entire function 
$\rho(x)$ in the 
$\mathcal{LP}$ class, although 
$L_{j}(\rho(x)) \gt 0$, we can not infer that 
$L_{j}(\rho(x))$ belongs to the 
$\mathcal{LP}$ class, thus we can not get 
$L^{(k)}_{j}(\rho(x)) \gt 0$.
Recently, many researchers focus on a kind of sequence 
$\{\alpha(n)\}$ in the following form
\begin{equation}
\log\left(\frac{\alpha(n+j)}{\alpha(n)}\right)=\sum_{i=1}^NA_i(n)j^i+R(n,j),
\end{equation} which generalizes a large number of sequences including the Maclaurin coefficients of the Riemann Ξ-function (i.e., 
$\gamma(n)$ and 
$\beta(n)$) and various partitions, see [Reference Griffin, Ono, Rolen, Thorner, Tripp and Wagner14, Reference Griffin, Ono, Rolen and Zagier15, Reference O’Sullivan21, Reference O’Sullivan22, Reference Wagner28]. Griffin, Ono, Rolen and Zagier [Reference Griffin, Ono, Rolen and Zagier15] proved that under some conditions, the Jensen polynomials of degree d and shift n 
$J_{\alpha}^{d,n}(x)$ associated with the sequence 
$\{\alpha(n)\}$ are hyperbolic for any d and sufficiently large n.
Theorem 1.1 (Griffin, Ono, Rolen and Zagier)
Let 
$\alpha(n)$, A(n), and 
$\delta(n)$ be sequences of positive real numbers with 
$\delta(n)$ tending to zero and satisfying
\begin{equation*}
\log \left(\frac{\alpha(n+j)}{\alpha(n)}\right)=A(n)j-\delta(n)^2j^2+\sum_{i=3}^{d}g_i(n)j^i+o\left(\delta(n)^d\right) \quad \mbox{as}\quad n\rightarrow \infty,
\end{equation*} for some 
$d\geq1$, all 
$0\leq j\leq d$ and some 
$g_i(n)=o\left(\delta(n)^i\right)$. Then, we have
\begin{equation*}
\lim_{n\rightarrow \infty}\left(\frac{\delta(n)^{-d}}{\alpha(n)}J_{\alpha}^{d,n}\left(\frac{\delta(n)X-1}{\exp
(A(n))}\right)\right)=H_d(X),
\end{equation*} uniformly for X in any compact subset of 
$\mathbb{R}$, where 
$H_d(X)$ are the Hermite polynomials.
Griffin, Ono, Rolen, Thorner, Tripp and Wagner [Reference Griffin, Ono, Rolen, Thorner, Tripp and Wagner14] made this approach effective for the Riemann Ξ-function.
For Laguerre inequality of any order, Wang and Yang [Reference Wang and Yang32] proved the following result. For the Maclaurin coefficients 
$\gamma(n)$ of the Riemann Ξ-function and the partition function p(n), Wang and Yang proved that 
$L_r(\gamma(n)) \gt 0$ for 
$n\geq cr^3$ for certain positive c and 
$L_r(p(n)) \gt 0$ for 
$n\geq 6r^4$, respectively.
Theorem 1.2 (Wang and Yang)
Let 
$\{\alpha(n)\},\{\delta(n)\},\{A_i(n)\}$ be sequences with 
$\alpha(n)$ positive, 
$\delta(n)\to0^+$, 
$A_2(n) \lt 0$, 
$A_2(n)=\Theta(\delta(n)^{t})$ and 
$A_{2i}(n)=o(\delta(n)^{it})\ (i\geq2)$ for some positive t, and for 
$-r \lt j \lt r$,
\begin{equation}
\log\left(\frac{\alpha(n+j)}{\alpha(n)}\right)=\sum_{i=1}^{2r}A_i(n)j^i+o\left(\delta(n)^{tr}\right),
\end{equation}then for sufficiently large n,
\begin{equation}
L_r(\alpha(n))=\frac{1}{2}\sum_{k=0}^{2r}(-1)^{k+r}{2r\choose k}\alpha(n+k)\alpha(n+2r-k) \gt 0.
\end{equation}Let
\begin{equation}
\mathcal{L}^{(m)}_r(a_n)\colon =L_{r}^{(m)}(\rho_{a}^{(n)}(x))|_{x=0}.
\end{equation}In this paper, we will concern ourselves with the following inequality
\begin{equation}
\mathcal{L}^{(2)}_1(a_n)\colon =(a_{n+1}a_{n+2} - a_{n}a_{n+3})^2 -(a_{n+1}^2 -a_{n}a_{n+2})(a_{n+2}^2 - a_{n}a_{n+4}) \gt 0.
\end{equation} In 
$\S$ 2, we will show the sequences 
$\{\alpha(n)\}$ of the form (1.8) asymptotically satisfy (1.11). We also show the companion version of this inequality. As a consequence, we show that 
$\gamma(n)$ satisfy (1.11) and the companion inequality for sufficiently large n. In 
$\S$ 3–5, we make this approach effective for the partition function, the overpartition and the smallest part function and give the thresholds of the positivity of (1.11) and its companion inequality for them, respectively. At last, we will further consider the inequality 
$\mathcal{L}^{(2)}_2(\alpha(n)) \gt 0$ and propose an open problem.
2. Iterated laguerre inequalities
In this section, we will show that 
$\alpha(n)$ of the form given in Theorem 1.1 asymptotically satisfies the inequality (1.11). In fact, we get the following detailed result.
Theorem 2.1. Let 
$\{\alpha(n)\},\{\delta(n)\},\{A_i(n)\}$ be sequences such that as 
$n\to\infty$, 
$\alpha(n)$ stays positive, 
$\delta(n)\to0^+$, 
$A_2(n)=-\delta(n)^{2}$ and 
$A_{i}(n)=o\left(\delta(n)^{i}\right) (i\geq 3)$, and for 
$0\leq j\leq 8$,
\begin{equation}
\log\left(\frac{\alpha(n+j)}{\alpha(n)}\right)=\sum_{i=1}^{8}A_i(n)j^i+o\left(\delta(n)^{8}\right),
\end{equation}then for sufficiently large n and any real number ɛ > 0,
\begin{equation}
\begin{aligned}
&(\alpha(n)\alpha(n+1)-\alpha(n-1)\alpha(n+2))^2\\
&-\left(\alpha(n)^2-\alpha(n-1)\alpha(n+1)\right)\left(\alpha(n+1)^2-\alpha(n-1)\alpha(n+3)\right) \gt 0
\end{aligned}
\end{equation}and
\begin{equation}
\begin{aligned}
&\left(1+(1-\varepsilon)\delta(n)^2\right)\left(\alpha(n)^2-\alpha(n-1)\alpha(n+1)\right)\left(\alpha(n+1)^2-\alpha(n-1)\alpha(n+3)\right)\\
& \gt (\alpha(n)\alpha(n+1)-\alpha(n-1)\alpha(n+2))^2.
\end{aligned}
\end{equation}Proof. Since the left-side of (2.2) is 
$\mathcal{L}_{1}^{(2)}({\alpha}(n-1))$, (2.2) is equivalent to
\begin{equation}
\mathcal{L}_{1}^{(2)}({\alpha}(n-1)) \gt 0.
\end{equation} Without loss of generality, we suppose 
$A_1(n)=0$ (otherwise we multiply 
$\alpha(n)$ by 
$\exp(-A_1(n))$.) We proceed to estimate each term on the left-hand side of the inequality above. The term 
$(\alpha(n)\alpha(n+1))^2$ can be represented as
\begin{equation}
\alpha(n)^4\exp\left(\log\left(\frac{\alpha(n+1)^2}{\alpha(n)^2}\right)\right)=\alpha(n)^4\exp\left(2\sum_{i=2}^8A_i(n)+o\left(\delta(n)^{8}\right)\right).
\end{equation}It implies that
\begin{equation}
\frac{(\alpha(n)\alpha(n+1))^2}{\alpha(n)^4}=\exp\left(2\sum_{i=2}^8A_i(n)\right)+o\left(\delta(n)^{8}\right).
\end{equation}The other terms can be represented with the same method. Thus, we have
\begin{equation}
\begin{aligned}
&\frac{\mathcal{L}_{1}^{(2)}(({\alpha}(n-1))}{\alpha(n)^4}\\
=&\exp\left(\log\left(\frac{\alpha(n-1)\alpha(n+3)}{\alpha(n)^2}\right)\right)+\exp\left(\log\left(\frac{\alpha(n-1)\alpha(n+1)^3}{\alpha(n)^4}\right)\right)\\
&+\exp\left(\log\left(\frac{\alpha(n-1)^2\alpha(n+2)^2}{\alpha(n)^4}\right)\right)\\
&-2\exp\left(\log\left(\frac{\alpha(n-1)\alpha(n+1)\alpha(n+2)}{\alpha(n)^3}\right)\right)
-\exp\\ &\times\left(\log\left(\frac{\alpha(n-1)^2\alpha(n+1)\alpha(n+3)}{\alpha(n)^4}\right)\right)\\
=&\exp\left(\sum_{i=1}^8A_i(n)\left((-1)^i+3^i\right)\right)+\exp\left(\sum_{i=1}^8A_i(n)\left((-1)^i+3\right)\right)
+\exp\\ &\times\left(2\sum_{i=2}^8A_i(n)\left((-1)^i+2^i\right)\right)\\
&-2\exp\left(\sum_{i=1}^8A_i(n)\left((-1)^i+1+2^i\right)\right)-\exp\left(\sum_{i=1}^8A_i(n)\left((-1)^i+(-1)^i+1+3^i\right)\right)\\ &+o\left(\delta(n)^{8}\right).
\end{aligned}
\end{equation} Let 
$\varphi(x)=1+x+\frac{x^2}{2}+\frac{x^3}{6}=\exp(x)+o(x^3)$. In view of
\begin{equation}
A_i(n)=\left\{
\begin{aligned}
&-\delta(n)^{2}, \ &i=2,\\
&o(\delta(n)^{i}),\ &i\geq 3.
\end{aligned}
\right.
\end{equation} and 
$A_2(n)$ contributes to the main term, we get that
\begin{equation}
\begin{aligned}
&\frac{\mathcal{L}_{1}^{(2)}({\alpha}(n-1))}{\alpha(n)^4}\\
=&\varphi\left(\sum_{i=1}^8A_i(n)\left((-1)^i+3^i\right)\right)+\varphi\left(\sum_{i=1}^8A_i(n)\left((-1)^i+3\right)\right)
+\varphi\\ &\times\left(2\sum_{i=2}^8A_i(n)\left((-1)^i+2^i\right)\right)\\
&-2\varphi\left(\sum_{i=1}^8A_i(n)\left((-1)^i+1+2^i\right)\right)-\varphi\left(\sum_{i=1}^8A_i(n)\left((-1)^i+(-1)^i+1+3^i\right)\right)\\ &+o\left(\delta(n)^{8}\right)\\
=&-16A_2(n)^3+36A_3(n)^2-144A_2(n)^2A_3(n)+o\left(\delta(n)^{7}\right)\\
=&16\delta(n)^6+36A_3(n)^2-144\delta(n)^4A_3(n)+o\left(\delta(n)^{7}\right),
\end{aligned}
\end{equation}which is positive by (2.1). Thus, (2.4) can be obtained immediately.
On the other hand, (2.3) can be rewritten as
\begin{align}
&(1-\varepsilon)\delta(n)^2\left(\alpha(n)^2-\alpha(n-1)\alpha(n+1)\right)\left(\alpha(n+1)^2-\alpha(n-1)\alpha(n+3)\right)\nonumber\\ &\quad \gt \mathcal{L}_{1}^{(2)}\left(\alpha(n-1)\right).
\end{align}With a similar argument, we have
\begin{equation*}
\begin{aligned}
&\frac{\left(\alpha(n)^2-\alpha(n-1)\alpha(n+1)\right)\left(\alpha(n+1)^2-\alpha(n-1)\alpha(n+3)\right)}{\alpha(n)^4}\\
=&\exp\left(\log\left(\frac{\alpha(n+1)^2}{\alpha(n)^2}\right)\right)
-\exp\left(\log\left(\frac{\alpha(n-1)\alpha(n+3)}{\alpha(n)^2}\right)\right)\\
&-\exp\left(\log\left(\frac{\alpha(n-1)\alpha(n+1)^3}{\alpha(n)^4}\right)\right)
+\exp\left(\log\left(\frac{\alpha(n-1)^2\alpha(n+1)\alpha(n+3)}{\alpha(n)^4}\right)\right)\\
=&\exp\left(2\sum_{i=1}^8A_i(n)\right)-\exp\left(\sum_{i=1}^8A_i(n)\left((-1)^i+3^i\right)\right)-\exp\left(\sum_{i=1}^8A_i(n)\left((-1)^i+3\right)\right)\\
&+\exp\left(\sum_{i=1}^8A_i(n)\left((-1)^i+(-1)^i+1+3^i\right)\right)+o\left(\delta(n)^{8}\right)\\
=&16A_2(n)^2+48A_2(n)A_3(n)+o\left(\delta(n)^{5}\right)\\
=&16\delta(n)^4-48\delta(n)^2A_3(n)+o\left(\delta(n)^{5}\right).
\end{aligned}
\end{equation*}Combining the equality above and (2.7), we get (2.3). This completes the proof.
The Maclaurin coefficients 
$\gamma(n)$ of the Riemann Ξ-function have been proven to satisfy all the conditions in the theorem (for details, refer to [Reference Griffin, Ono, Rolen and Zagier15, Reference Wang and Yang33]). For 
$\gamma(n)$, we may refer to [Reference Griffin, Ono, Rolen, Thorner, Tripp and Wagner14] to see that
\begin{equation*}
A_2(n)=-\frac{1}{2}\left(1-\frac{\gamma(n-2)\gamma(n)}{\gamma(n-1)^2}\right)+O\left(n^{-2}\right),
\end{equation*}which gives that
\begin{equation*}
A_2(n)=-\frac{1}{2n}+\frac{2}{n\log n}+O\left(n^{-2}\right).
\end{equation*}Hence, we can deduce the following theorem from Theorem 2.1.
Theorem 2.2. For sufficiently large n and any real number ɛ > 0, we have
\begin{equation*}
\begin{aligned}
&1 \lt \frac{(\gamma(n)\gamma(n+1)-\gamma(n-1)\gamma(n+2))^2}{\left(\gamma(n)^2-\gamma(n-1)\gamma(n+1)\right)\left(\gamma(n+1)^2-\gamma(n-1)\gamma(n+3)\right)} \lt 1+\frac{1-\varepsilon}{2n}.
\end{aligned}
\end{equation*}3. Partition function
In this section, we will give thresholds for the positivity of the iterated Laguerre inequality and its companion inequality of order 2 for the partition function. An integer partition of a positive integer n is a non-increasing sequence 
$(\lambda_1,\lambda_2,\ldots, \lambda_r)$ of positive integers such that 
$ \lambda_1+\lambda_2+\cdots+\lambda_r=n$. We denote the number of integer partitions of n by p(n) and set
\begin{equation}
\mu(n)=\frac{\pi}{6}\sqrt{24n-1}.
\end{equation}Based on the Hardy-Ramanujan-Rademacher formula [Reference Hardy and Ramanujan16] and Lehmer’s error bound [Reference Lehmer18, Reference Lehmer19], Wang and Yang [Reference Wang and Yang31, Lemma 2.1] gave the following estimate of p(n) which is a key ingredient in our proofs.
Lemma 3.1. For any given integer t, there exists N(t) such that for all 
$n\geq N(t)$,
\begin{equation}
\frac{\sqrt{12}\pi^2e^{\mu(n)}}{36\mu(n)^2}\left(1-\frac{1}{\mu(n)}-\frac{1}{\mu(n)^{t}}\right) \lt p(n) \lt \frac{\sqrt{12}\pi^2e^{\mu(n)}}{36\mu(n)^2}\left(1-\frac{1}{\mu(n)}+\frac{1}{\mu(n)^{t}}\right).
\end{equation}Now we proceed to prove these two inequalities for p(n).
Theorem 3.2. For 
$n\geq 220$, p(n) satisfies the iterated Laguerre inequality of order 2, i.e.,
\begin{eqnarray}
\mathcal{L}_1^{(2)}\left(p\left(n\right)\right) \gt 0.
\end{eqnarray}Proof. Setting t = 10 in Lemma 3.1 gives that for 
$n\geq 1520$,
\begin{eqnarray}
e^{\mu} \frac{\sqrt{12}\beta(\mu) \pi^{2} }{36 \mu^{12}} \lt p(n) \lt e^{\mu} \frac{\sqrt{12}\alpha(\mu) \pi^{2} }{36 \mu^{12}},
\end{eqnarray} where µ is the abbreviation of 
$\mu(n)$ and
\begin{equation}
\alpha(t)=t^{10}-t^{9}+1,~~ \beta(t)=t^{10}-t^{9}-1.
\end{equation}Let
\begin{eqnarray}
f(n)\colon =e^{\mu} \frac{\sqrt{12}\beta(\mu) \pi^{2} }{36 \mu^{12}}
\quad \mbox{and} \quad
g(n)\colon =e^{\mu} \frac{\sqrt{12}\alpha(\mu) \pi^{2} }{36 \mu^{12}}.
\end{eqnarray}To prove (3.3), it suffices to show that
\begin{equation}
\begin{aligned} f(n)f(n+2)^3-2g(n)g(n+1)g(n+2)g(n+3)+f(n)f(n+1)^2f(n+4)\\-g(n)^2g(n+2)g(n+4)+f(n)^2f(n+3)^2 \gt 0.
\end{aligned}
\end{equation}For convenience, we denote
\begin{equation}
y=\mu(n),z=\mu(n+1),x=\mu(n+2), w=\mu(n+3),i=\mu(n+4)
\end{equation}and reformulate (3.7) as
\begin{equation}
\frac{h_1+e^{2z+i-3x}h_2+e^{2w+y-3x}h_3-2e^{z+w-2x}h_4-e^{y+i-2x}h_5}{y^{24}z^{24}x^{36}w^{24}i^{12}e^{-(3x+y)}} \gt 0,
\end{equation}where
\begin{align}
&h_{1}=\beta(y)\beta(x)^3y^{12}z^{24}w^{24}i^{12},\ h_{2}=\beta(y)\beta(i)\beta(z)^2y^{12}x^{36}w^{24},
\end{align}
\begin{align*}
&h_{3}=\beta(y)^2\beta(w)^2z^{24}x^{36}i^{12},\ h_{4}=\alpha(y)\alpha(z)\alpha(x)\alpha(w)y^{12}z^{12}x^{24}w^{12}i^{12},\\
&h_{5}=\alpha(x)\alpha(i)\alpha(y)^2z^{24}x^{24}w^{24}.
\end{align*} Since 
$y^{24}z^{24}x^{36}w^{24}i^{12}e^{-(3x+y)}$ is positive for all 
$n\geq 1$, it is sufficient to prove
\begin{equation}
h_1+e^{2z+i-3x}h_2+e^{2w+y-3x}h_3-2e^{z+w-2x}h_4-e^{y+i-2x}h_5 \gt 0.
\end{equation} To achieve this aim, we need to estimate 
$h_{1}, h_{2}, h_{3},h_{4},h_5,e^{2z+i-3x},e^{2w+y-3x}$, 
$e^{z+w-2x}$ and 
$e^{y+i-2x}$. We notice that 
$y,z,w$ and i can be expressed as functions of x, i.e., for 
$n\geq 2$,
\begin{equation}
y=\sqrt{x^2-\frac{4 \pi ^2}{3}},z=\sqrt{x^2-\frac{2 \pi ^2}{3}},
w=\sqrt{x^2+\frac{2 \pi ^2}{3}}, i=\sqrt{x^2+\frac{4 \pi ^2}{3}}.
\end{equation} By the expansions of 
$y,z,w,i$, we have that for 
$x \geq 18$,
\begin{equation}
y_1 \lt y \lt y_2,\ z_1 \lt z \lt z_2,\ w_1 \lt w \lt w_2,\ i_1 \lt i \lt i_2,
\end{equation}where
\begin{align}
&y_1=x-\frac{2 \pi ^2}{3 x}-\frac{2 \pi ^4}{9 x^3}-\frac{4 \pi ^6}{27 x^5}-\frac{10 \pi ^8}{81 x^7}- \frac{28\pi^{10}}{243x^9}-\frac{29\pi^{12}}{243x^{11}},
\end{align}
\begin{align*}
&y_2=x-\frac{2 \pi ^2}{3 x}-\frac{2 \pi ^4}{9 x^3}-\frac{4 \pi ^6}{27 x^5}-\frac{10 \pi ^8}{81 x^7}- \frac{28\pi^{10}}{243x^9}-\frac{28\pi^{12}}{243x^{11}},
\end{align*}
\begin{align*}
&z_1=x-\frac{\pi ^2}{3 x}-\frac{\pi ^4}{18 x^3}-\frac{\pi ^6}{54 x^5}-\frac{5 \pi ^8}{648 x^7}-\frac{7 \pi ^{10}}{1944 x^9}-\frac{8\pi^{12}}{3888x^{11}},
\end{align*}
\begin{align*}
&z_2=x-\frac{\pi ^2}{3 x}-\frac{\pi ^4}{18 x^3}-\frac{\pi ^6}{54 x^5}-\frac{5 \pi ^8}{648 x^7}-\frac{7 \pi ^{10}}{1944 x^9}-\frac{7\pi^{12}}{3888x^{11}},
\end{align*}
\begin{equation*}
\begin{aligned}
&w_1=x+\frac{\pi ^2}{3 x}-\frac{\pi ^4}{18 x^3}+\frac{\pi ^6}{54 x^5}-\frac{5 \pi ^8}{648 x^7}+\frac{7 \pi ^{10}}{1944 x^9}-\frac{7\pi^{12}}{3888x^{11}},\\
&w_2=x+\frac{\pi ^2}{3 x}-\frac{\pi ^4}{18 x^3}+\frac{\pi ^6}{54 x^5}-\frac{5 \pi ^8}{648 x^7}+\frac{7 \pi ^{10}}{1944 x^9},\\
&i_1=x+\frac{2 \pi ^2}{3 x}-\frac{2 \pi ^4}{9 x^3}+\frac{4 \pi ^6}{27 x^5}-\frac{10 \pi ^8}{81 x^7}+\frac{28 \pi ^{10}}{243 x^9}-\frac{28 \pi ^{12}}{243 x^{11}},\\
&i_2=x+\frac{2 \pi ^2}{3 x}-\frac{2 \pi ^4}{9 x^3}+\frac{4 \pi ^6}{27 x^5}-\frac{10 \pi ^8}{81 x^7}+\frac{28 \pi ^{10}}{243 x^9}.
\end{aligned}
\end{equation*}To make the following polynomials more concise, we define
\begin{equation}
\alpha_2(t)=t^{10}-t_1t^8+1,~~\beta_1(t)=t^{10}-t_2t^8-1.
\end{equation} Because 
$\alpha_2(t) \gt \alpha(t),\beta_1(t) \lt \beta(t)$ for 
$t_1 \lt t \lt t_2$ and 
$y,z,w,i$ are positive for all 
$n\geq 1$, we obtain
\begin{equation}
\begin{aligned}
&h_{1} \gt y^{12}z^{24}w^{24}i^{12}\beta_1(y)\beta(x)^3,\ h_{2} \gt y^{12}x^{36}w^{24}\beta_1(i)\beta_1(y)\beta_1(z)^2,\\[9pt]
&h_{3} \gt z^{24}x^{36}i^{12}\beta_1(w)^2\beta_1(y)^2,\ h_{4} \lt y^{12}z^{12}x^{24}w^{12}i^{12}\alpha_2(w)\alpha(x)\alpha_2(y)\alpha_2(z),\\[9pt]
&h_{5} \lt z^{24}x^{24}w^{24}\alpha_2(i)\alpha(x)\alpha_2(y)^2,
\end{aligned}
\end{equation} and denote the right-hand sides of (3.16) by 
$\tilde{h_1},\tilde{h_2},\tilde{h_3},\tilde{h_4},\tilde{h_5}$, respectively.
Now our task is to estimate 
$e^{2z+i-3x},e^{2w+y-3x},e^{z+w-2x}$ and 
$e^{y+i-2x}$. By (3.13), it is obvious that for 
$x\geq 18$,
\begin{align}
&2z_{1}+i_{1}-3x \lt 2z+i-3x \lt 2z_{2}+i_{2}-3x,
\end{align}
\begin{align*}
&2w_{1}+y_{1}-3x \lt 2w+y-3x \lt 2w_{2}+y_{2}-3x,
\end{align*}
\begin{align*}
&z_{1}+w_{1}-2x \lt z+w-2x \lt z_{2}+w_{2}-2x,
\end{align*}
\begin{align*}
&y_{1}+i_{1}-2x \lt y+i-2x \lt y_{2}+i_{2}-2x,
\end{align*}which implies that
\begin{align}
&e^{2z_{1}+i_{1}-3x} \lt e^{2z+i-3x} \lt e^{2z_{2}+i_{2}-3x},
\end{align}
\begin{align*}
&e^{2w_{1}+y_{1}-3x} \lt e^{2w+y-3x} \lt e^{2w_{2}+y_{2}-3x},
\end{align*}
\begin{align*}
&e^{z_{1}+w_{1}-2x} \lt e^{z+w-2x} \lt e^{z_{2}+w_{2}-2x},
\end{align*}
\begin{align*}
&e^{y_{1}+i_{1}-2x} \lt e^{y+i-2x} \lt e^{y_{2}+i_{2}-2x}.
\end{align*} In order to give feasible bounds for 
$e^{2z+i-3x},e^{2w+y-3x},e^{z+w-2x}$ and 
$e^{j+i-2x}$, we define
\begin{equation}
\Phi(t)=1+t +\frac{t ^2}{2}+\frac{t ^3}{6}+\frac{t ^4}{24},
\end{equation}and
\begin{equation}
\phi(t)=1+t +\frac{t ^2}{2}+\frac{t ^3}{6}+\frac{t ^4}{24}+\frac{t ^5}{120}.
\end{equation}Then, for t < 0,
\begin{equation}
\phi(t) \lt e^{t} \lt \Phi(t).
\end{equation} To apply (3.21) to (3.18), it is necessary to check that 
$2z_{2}+i_{2}-3x,2w_{2}+y_{2}-3x,z_{2}+w_{2}-2x$ and 
$w_{2}+i_{2}-2x$ are negative. By a straightforward calculation, one can get that for 
$x\geq 3$,
\begin{align*}
&2z_{2}+i_{2}-3x=-\frac{\pi^4\left(7\pi^8-210\pi^6x^2+270\pi^4x^4-216\pi^2x^6+648x^8\right)}{1944x^{11}} \lt 0 ,\\[9pt]
&2w_{2}+y_{2}-3x=-\frac{\pi^4\left(112\pi^8+105\pi^6x^2+135\pi^4x^4+108\pi^2x^6+324x^8\right)}{972x^{11}} \lt 0 ,\\[9pt]
&z_{2}+w_{2}-2x=-\frac{\pi^{4}\left(7\pi^{8}+60\pi^{4}x^{4}+432x^{8}\right)}{3888x^{11}} \lt 0,\\[9pt]
&y_{2}+i_{2}-2x=-\frac{4\pi^{4}\left(7\pi^{8}+15\pi^{4}x^{4}+27x^{8}\right)}{243x^{11}} \lt 0.
\end{align*} Thus, we can apply (3.21) to (3.18) and obtain that for 
$x\geq 18$,
\begin{align}
&\phi (2z_{1}+i_{1}-3x) \lt e^{2z+i-3x} \lt \Phi (2z_{2}+i_{2}-3x),
\end{align}
\begin{align*}
&\phi (2w_{1}+y_{1}-3x) \lt e^{2w+y-3x} \lt \Phi (2w_{2}+y_{2}-3x),
\end{align*}
\begin{align*}
&\phi (z_{1}+w_{1}-2x) \lt e^{z+w-2x} \lt \Phi (z_{2}+w_{2}-2x),\\
&\phi (y_{1}+i_{1}-2x) \lt e^{y+i-2x} \lt \Phi (y_{2}+i_{2}-2x).
\end{align*}Now we proceed to prove (3.11). For convenience, we denote
\begin{equation}
A(x)=h_1+e^{2z+i-3x}h_2+e^{2w+y-3x}h_3-2e^{z+w-2x}h_4-e^{y+i-2x}h_5.
\end{equation} The remaining part of our proof aims to show the positivity of A(x). Using (3.16) and (3.22), we get that for 
$x\geq 18$,
\begin{eqnarray}
\begin{aligned}
A(x) \gt \tilde{h}_1+\phi (2z_{1}+i_{1}-3x)\tilde{h}_2+\phi (2w_{1}+y_{1}-3x)\tilde{h}_3\\-2\Phi (z_{2}+w_{2}-2x)\tilde{h}_4-\Phi (y_{2}+i_{2}-2x)\tilde{h}_5.
\end{aligned}
\end{eqnarray} We denote the right-hand side of (3.24) by 
$A_1(x)$ and rewrite 
$A_1(x)$ as a polynomial of x by using (3.12) and (3.14) to 
$A_1(x)$. 
$A_1(x)$ can be simplified as
\begin{equation}
A_1(x)={\sum_{k=0}^{164} a_k x^{k}\over 2^{25}3^{80} 5 x^{61}},
\end{equation} where ak are known real numbers, and the first few terms 
$a_{164}, a_{163}, a_{162}$ are given below.
\begin{eqnarray*}
&a_{164}=2^{26} 3^{74} 5\pi^{12} ,~ a_{163}=-2^{26}3^{74} 5(8748
+19\pi^{12}),\\ &a_{162}=2^{28} 3^{73} 5 (19683+81\pi^{12}-5\pi^{14}).
\end{eqnarray*} Thus, for 
$x\geq 18$, we have
\begin{equation}
A(x) \gt {\sum_{k=0}^{164} a_k x^{k}\over 2^{25}3^{80} 5 x^{61}}.
\end{equation} Since x is positive for 
$n\geq 1$, we have
\begin{equation}
\sum_{k=0}^{164} a_k x^{k} \gt \sum_{k=0}^{162} -|a_k| x^{k}+a_{163}x^{163}+a_{164} x^{164}.
\end{equation} For 
$0\leq k \leq 161$, we find that for 
$x\geq 18$,
\begin{equation}
-|a_k|x^{k} \gt -a_{162}x^{162}.
\end{equation} It follows that for 
$x\geq 18$,
\begin{equation}
\sum_{k=0}^{164} a_k x^{k} \gt \sum_{k=0}^{162} -|a_k| x^{k}+a_{163}x^{163}+a_{164} x^{164} \gt (-163a_{162}+a_{163}x+a_{164}x^2)x^{162}.
\end{equation}Combining (3.26) and (3.29), we get that A(x) is positive provided that
\begin{equation}
-163a_{162}+a_{163}x+a_{164}x^2 \gt 0,
\end{equation} which is true for 
$x\geq 94$, or equivalently, 
$n\geq 1343$. Thus, we arrive at that (3.3) is true for 
$n\geq 1520$. It can be checked that (3.3) is also true for 
$220\leq n \leq 1520$ by a straightforward computation.
We turn to considering the companion version of this theorem. For recent works on the companion version of the Turán type inequality and the Laguerre inequalities, see [Reference Banerjee3, Reference Chen4, Reference Dou, Wang and Yang10]. These results motivate us to consider the coefficient of 
$O(n^{-\frac{3}{2}})$ in the companion inequality. For details, we get the following result.
Theorem 3.3. For 
$n\geq 243$, we have
\begin{equation}
\begin{aligned}
&\left(1+\frac{\pi}{8n^{\frac{3}{2}}}\right)\left(p(n+1)^2-p(n)p(n+2)\right)\left(p(n+2)^2-p(n)p(n+4)\right)\\
&\quad \gt \left(p(n+1)p(n+2)-p(n)p(n+3)\right)^2.
\end{aligned}
\end{equation}Proof. To prove (3.31), we only need to verify
\begin{equation}
\begin{aligned}
&(1+\frac{\sqrt{6}\pi^4}{36\mu(n+2)^3})\left(p(n+1)^2-p(n)p(n+2)\right)\left(p(n+2)^2-p(n)p(n+4)\right)\\
&\quad \gt \left(p(n+1)p(n+2)-p(n)p(n+3)\right)^2,
\end{aligned}
\end{equation} where 
$\mu(n)$ is defined as (3.1).
We also set t = 10 in Lemma 3.1 and use the same Taylor expansion of 
$y,z,w,i$ as that in the proof of Theorem 3.2 to estimate their upper and lower bounds. We denote the left-hand side of (3.32) by B(x) and find 
$B_1(x)$ as a lower bound of B(x). Using the definition of 
$\alpha(t),\beta(t),\alpha_2(t),\beta_1(t)$ in (3.5) and (3.15), we only need to verify the positivity of
\begin{equation}
\begin{aligned}
B_1(x):=&-\Phi(y_2-x)(\sqrt{6}\pi^4+36x^3)h_1'-\Phi(i_2+2z_2+y_2-4x)(\sqrt{6}\pi^4+36x^3)h_2'\\ &-36\Phi(2w_2+2y_2-4x)h_3'+72\phi(w_1+z_1+y_1-3x)h_4'\\ &+\phi(2y_1+i_1-3x)(\sqrt{6}\pi^4+36x^3)h_5'+\sqrt{6}\pi^4\phi(2z_1-2x)h_6',
\end{aligned}
\end{equation} where 
$\Phi(t)$ and 
$\phi(t)$ are the first 9 and 10 terms in the Taylor expansion of et, and
\begin{align}
&h'_{1}=i^{12}w^{24}y^{12}z^{24}\alpha(x)^3\alpha_2(y), \ h'_{2}=w^{24}x^{36}y^{12}\alpha_2(i)\alpha_2(y)\alpha_2(z)^2,
\end{align}
\begin{align*}
&h'_{3}=i^{12}x^{39}z^{24}\alpha_2(w)^2\alpha_2(y)^2, \
h'_{4}=i^{12}w^{12}x^{27}y^{12}z^{12}\beta_1(w)\beta(x)\beta_1(y)\beta_1(z),\\
&h'_{5}=w^{24}x^{24}z^{24}\beta_1(i)\beta(x)\beta_1(y)^2, \
h'_{6}=i^{12}w^{24}x^{12}y^{24}\beta(x)^2\beta_1(z)^2.
\end{align*} We simplify 
$B_1(x)$ as
\begin{equation}
B_1(x)=\frac{\sum_{k=0}^{215}b_kx^k}{2^{47}3^{100}\cdot5 \cdot7x^{109}}
\end{equation}where the coefficient of the first term of the numerator is
\begin{equation*}
b_{215}=2^{49}3^{90}\pi^{12}(25515\sqrt{6}-51030-4\pi^6)
\end{equation*} is positive. With similar arguments as in the proof of Theorem 3.2, we have that (3.31) holds for 
$x\geq 461$, or equivalently, 
$n\geq 32300$. Moreover, we can verify that (3.31) holds for 
$243\leq n\leq 32300$. This completes the proof.
4. Overpartition
In this section, we will give thresholds for the iterated Laguerre inequality of order 2 and its companion version for the overpartition function. An overpartition of n is an ordinary partition of n with the added condition that the first occurrence of any part may be overlined or not. The number of overpartitions of n is denoted by 
$\bar{p}(n)$. One can use the Rademacher-type estimation to obtain a bound for 
$\bar{p}(n)$. Let
\begin{equation}
\bar{\mu}(n):=\pi\sqrt{n}.
\end{equation} The approach presented in the proof of Theorem 3.2 is still available. Thus, we will omit some tedious formulae and statements in our remaining proofs. Wang and Yang [Reference Wang and Yang31, Lemma 5.1] also gave the following estimate of 
$\bar{p}(n)$.
Lemma 4.1. For any given integer t, there exists N(t) such that for all 
$n\geq N(t)$,
\begin{equation}
\begin{aligned}\frac{\pi^2e^{\bar\mu}}{8\bar\mu^2}\left(1-\frac{1}{\bar\mu}-\frac{1}{\bar\mu^t}\right) \lt \bar p(n) \lt \frac{\pi^2e^{\bar\mu}}{8\bar\mu^2}\left(1-\frac{1}{\bar\mu}+\frac{1}{\bar\mu^t}\right),\end{aligned}
\end{equation} where 
$\bar{\mu}$ is the abbreviation for 
$\bar{\mu}(n)$.
By Lemma 4.1, we obtain the access to prove the iterated Laguerre inequality of order 2 for 
$\bar{p}(n)$. In view of the proof is similar to those in 
$\S$ 3, we omit some details.
Theorem 4.2. For 
$n\geq 40$, 
$\bar{p}(n)$ satisfies the iterated Laguerre inequality of order 2, i.e.,
\begin{equation}
\mathcal{L}_1^{(2)}\left(\bar{p}(n)\right) \gt 0.
\end{equation}Proof. Specifically, setting t = 10 in Lemma 4.1 gives that for 
$n\geq 821$,
\begin{equation}
\begin{aligned}\frac{\pi^2e^{\bar{\mu}}}{8\bar{\mu}^2}\left(1-\frac1{\bar{\mu}}-\frac1{\bar{\mu}^{10}}\right) \lt &\bar{p}(n) \lt \frac{\pi^2e^{\bar{\mu}}}{8\bar{\mu}^2}\left(1-\frac1{\bar{\mu}}+\frac1{\bar{\mu}^{10}}\right).\end{aligned}
\end{equation}For convenience, we denote
\begin{equation}
\bar{y}=\bar{\mu}(n),\bar z=\bar{\mu}(n+1),\bar{x}=\bar{\mu}(n+2), \bar w=\bar{\mu}(n+3),\bar i=\bar{\mu}(n+4).
\end{equation} We can also express 
$\bar{y},\bar z,\bar w,\bar i$ as functions of 
$\bar{x}$, i.e.,
\begin{equation*}
\bar{y}=\sqrt{\bar{x}^2-2\pi ^2},\bar z=\sqrt{\bar{x}^2-\pi ^2},\bar w=\sqrt{\bar{x}^2+\pi^2}, \bar i=\sqrt{\bar{x}^2+2\pi ^2}.
\end{equation*} By the Taylor expansions of 
$\bar{y},\bar z,\bar w,\bar i$, we obtain that for 
$x\geq 19$,
\begin{align}
&\bar{y}_1 \lt \bar{y} \lt \bar{y}_2,\ \bar z_1 \lt \bar z \lt \bar z_2,\ \bar w_1 \lt \bar w \lt \bar w_2,\ \bar i_1 \lt \bar i \lt \bar i_2.
\end{align}where
\begin{equation}
\begin{aligned}
&\bar{y}_1=\bar{x}-\frac{\pi ^2}{\bar{x}}-\frac{\pi ^4}{2\bar{x}^3}-\frac{\pi ^6}{2\bar{x}^5}-\frac{5\pi ^8}{8\bar{x}^7}-\frac{7\pi ^{10}}{8\bar{x}^9}-\frac{22\pi^{12}}{16x^{11}},\\[9pt]
&\bar{y}_2=\bar{x}-\frac{\pi ^2}{\bar{x}}-\frac{\pi ^4}{2\bar{x}^3}-\frac{\pi ^6}{2\bar{x}^5}-\frac{5\pi ^8}{8\bar{x}^7}-\frac{7\pi ^{10}}{8\bar{x}^9}-\frac{21\pi^{12}}{16x^{11}},\\[9pt]
&\bar z_1=\bar{x}-\frac{\pi ^2}{2 \bar{x}}-\frac{\pi ^4}{8 \bar{x}^3}-\frac{\pi ^6}{16 \bar{x}^5}-\frac{5 \pi ^8}{128 \bar{x}^7}-\frac{7 \pi ^{10}}{256 \bar{x}^9}-\frac{22\pi^{12}}{1024x^{11}},\\[9pt]
&\bar z_2=\bar{x}-\frac{\pi ^2}{2 \bar{x}}-\frac{\pi ^4}{8 \bar{x}^3}-\frac{\pi ^6}{16 \bar{x}^5}-\frac{5 \pi ^8}{128 \bar{x}^7}-\frac{7 \pi ^{10}}{256 \bar{x}^9}-\frac{21\pi^{12}}{1024x^{11}},\\[9pt]
&\bar w_1=\bar{x}+\frac{\pi ^2}{2 \bar{x}}-\frac{\pi ^4}{8 \bar{x}^3}+\frac{\pi ^6}{16 \bar{x}^5}-\frac{5 \pi ^8}{128 \bar{x}^7}+\frac{7 \pi ^{10}}{256 \bar{x}^9}-\frac{21\pi^{12}}{1024x^{11}},\\[9pt]
&\bar w_2=\bar{x}+\frac{\pi ^2}{2 \bar{x}}-\frac{\pi ^4}{8 \bar{x}^3}+\frac{\pi ^6}{16 \bar{x}^5}-\frac{5 \pi ^8}{128 \bar{x}^7}+\frac{7 \pi ^{10}}{256 \bar{x}^9},\\[9pt]
&\bar i_1=\bar{x}+\frac{\pi ^2}{\bar{x}}-\frac{\pi ^4}{2\bar{x}^3}+\frac{\pi ^6}{2\bar{x}^5}-\frac{5\pi ^8}{8\bar{x}^7}+\frac{7\pi ^{10}}{8 \bar{x}^9}-\frac{21\pi^{12}}{16x^{11}},\\[9pt]
&\bar i_2=\bar{x}+\frac{\pi ^2}{\bar{x}}-\frac{\pi ^4}{2\bar{x}^3}+\frac{\pi ^6}{2\bar{x}^5}-\frac{5\pi ^8}{8\bar{x}^7}+\frac{7\pi ^{10}}{8\bar{x}^9}.
\end{aligned}
\end{equation} Similarly, we replace 
$x,y,z,w,i,y_1,y_2,z_1,z_2,w_1,w_2,i_1,i_2$ in (3.16) and (3.24) by 
$\bar x,\bar y,\bar z,\bar w,\bar i,\bar y_1,\bar y_2,\bar z_1,\bar z_2,\bar w_1,\bar w_2,\bar i_1,\bar i_2$ in the definition (4.5) and (4.7) and denote the result by 
$\bar{A}(\bar{x})$ and 
$\bar{A}_1(\bar{x})$, respectively. With similar arguments as in the proof of Theorem 3.2, we have the numerator of 
$\bar{A}_1(\bar{x})$ is a certain polynomial of degree 164 with positive coefficient of the first term. It can be verified that 
$\bar{A}_1(\bar{x}) \gt 0$ holds for 
$n\geq 244$, thus (4.3) holds for 
$n\geq 821$. Moreover, a direct calculation shows that (4.3) holds for 
$40\leq n \leq 821$. This completes the proof.
The companion Laguerre inequality of order 2 also holds for the overpartition function as follows.
Theorem 4.3. For 
$n\geq 44$, 
$\bar{p}(n)$ satisfies the companion Laguerre inequality of order 2, i.e.,
\begin{equation}
\begin{aligned}
&(1+\frac{\sqrt{6}\pi}{18n^{\frac{3}{2}}})\left(\bar{p}(n+1)^2-\bar{p}(n)\bar{p}(n+2)\right)\left(\bar{p}(n+2)^2-\bar{p}(n)\bar{p}(n+4)\right)\\
& \gt \left(\bar{p}(n+1)\bar{p}(n+2)-\bar{p}(n)\bar{p}(n+3)\right)^2.
\end{aligned}
\end{equation}Proof. To prove (4.8), it is equivalent to demonstrating the validity of the following inequality,
\begin{equation}
\begin{aligned}
&(1+\frac{\sqrt{6}\pi^4}{18\bar{\mu}(n+2)^3})\left(\bar{p}(n+1)^2-\bar{p}(n)\bar{p}(n+2)\right)\left(\bar{p}(n+2)^2-\bar{p}(n)\bar{p}(n+4)\right)\\
& \gt \left(\bar{p}(n+1)\bar{p}(n+2)-\bar{p}(n)\bar{p}(n+3)\right)^2,
\end{aligned}
\end{equation} where 
$\bar{\mu}(n)$ is defined as (4.1).
We still set t = 10 in Lemma 4.1 and use (4.5) to give the definition of 
$\bar y,\bar z,\bar x,\bar w,\bar i$ and use the first 6 and 7 terms of the Taylor expansion to approximate 
$\bar y,\bar z,\bar w,\bar i$. We denote the left-hand side of (4.9) by 
$\bar{B}(\bar x)$ and look for a lower bound 
$\bar{B}_1(\bar x)$. Thus, it is sufficient to prove
\begin{equation}
\begin{aligned}
\bar{B}_1(\bar x):=&-\Phi(\bar y_2-\bar x)(\sqrt{6}\pi^4+18\bar x^3)\bar h_1'-\Phi(\bar i_2+2\bar z_2+\bar y_2-4\bar x)(\sqrt{6}\pi^4+18\bar x^3)\bar h_2'\\ &-18\Phi(2\bar w_2+2\bar y_2-4\bar x)\bar h_3'+36\phi(\bar w_1+\bar z_1+\bar y_1-3\bar x)\bar h_4'\\ &+\phi(2\bar y_1+\bar i_1-3\bar x)(\sqrt{6}\pi^4+18\bar x^3)\bar h_5'+\sqrt{6}\pi^4\phi(2\bar z_1-2\bar x)\bar h_6' \gt 0,
\end{aligned}
\end{equation} where 
$\phi(t)$ and 
$\Phi(t)$ are the first 10 and 11 terms in the Taylor expansion of et and 
$\bar h'_1,\bar h'_2,\bar h'_3,\bar h'_4,\bar h'_5,\bar h'_6$ are the results obtained by replacing 
$y,z,x,w,i$ in (3.34) by 
$\bar y,\bar z,\bar x,\bar w,\bar i$ in the definition (4.5). With the similar argument as in the proof of Theorem 3.2, we have the numerator of 
$\bar{B}_1(\bar{x})$ is a certain polynomial of degree 227 with positive coefficient of the first term. It is easily seen that (4.8) holds for 
$\bar x\geq 687$, or equivalently, 
$n\geq 47821$. Moreover, it can be checked that (4.8) also holds for 
$44\leq n \leq 47821$ by a direct calculation. This completes the proof.
5. Spt function
In this section, we will give the thresholds for the iterated Laguerre inequality of order 2 and its companion version for the spt function. The smallest part function 
$\mathop{\mathrm{spt}}\nolimits(n)$ is defined by Andrews [Reference Andrews2] to be the total number of smallest parts in all partitions of n. More precisely, for a partition λ of n, denote 
$n_s(\lambda)$ the number of occurrences of the smallest part in λ. Let P(n) denote the set of partitions of n, then
\begin{equation}
\mathop{\mathrm{spt}}\nolimits(n)=\sum_{\lambda \in P(n)}n_s(\lambda).
\end{equation}For the background for spt-function, see [Reference Ahlgren, Bringmann and Lovejoy1, Reference Chen4, Reference Dixit and Yee9, Reference Eichhorn and Hirschhorn11, Reference Garvan12]. First, we introduce a lemma proposed by Dou, Wang and Yang [Reference Dou, Wang and Yang10, Lemma 2.6] which will be used in our proofs.
Lemma 5.1 (Dou, Wang and Yang)
For any positive t, we have that for 
$n \gt N(t)$
\begin{equation}
\mathop{\mathrm{spt}}\nolimits(n)=2\sqrt3e^{\mu(n)}\left(\frac{1}{\mu(n)}+R(n)\right),
\end{equation}where
\begin{equation}
N(t)=\frac{3}{2\pi^2}(1+4t\log(2t))^2
\end{equation}and
\begin{equation}
|R(n)| \lt \frac{1}{\mu(n)^t}.
\end{equation}Using this upper bound, we will prove Theorem 5.2 and 5.3. For convenience, we denote
\begin{equation}
x_0=\mu(n),\quad x_i=\mu(n+i)=\sqrt{x_0^2+\frac{2\pi^2}{3}i},\quad s=\sqrt{x_0^2+\frac{\pi^2}{3}},\quad R_i=x_0R(n+i),
\end{equation} where 
$i \gt -n$. Then we have
\begin{equation}
{\rm spt}(n+i)=\frac{2\sqrt3e^s}{x_0}e^{x_i-s}\left(\frac{x_0}{x_i}+R_i\right).
\end{equation} Since the approach in the remaining are different from those in the proofs above, we give the detailed process for the positivity of 
$\mathcal{L}_1^{(2)}\left(\mathop{\mathrm{spt}}\nolimits(n-1)\right) \gt 0$.
Theorem 5.2. For 
$n\geq 241$, 
$\mathop{\mathrm{spt}}\nolimits(n)$ satisfies the iterated Laguerre inequality of order 2, i.e.,
\begin{equation}
\mathcal{L}_1^{(2)}\left(\mathop{\mathrm{spt}}\nolimits(n-1)\right) \gt 0.
\end{equation}Proof. Let 
${y_i}$ be the first 13 terms of the Laurent expansion of 
$x_0/x_i$, that is,
\begin{equation}
{y_i}=1-\frac{i\pi^2}{3x_0^2}+\frac{i^2\pi^4}{6x_0^4}-\frac{5i^3\pi^6}{54x_0^6}+\frac{35i^4\pi^8}{648x_0^8}-\frac{7i^5\pi^{10}}{216x_0^{10}}
+\frac{77i^6\pi^{12}}{3888x_0^{12}},
\end{equation} Setting t = 13 in Lemma 5.1, we obtain that 
$|R_i| \lt 2x_0^{-12}$ for 
$x_0 \gt 74$ and 
$x_0/x_i+R_i$ can be rewritten as 
${y_i}+{r_i}$ with
\begin{equation}
|{r_i}| \lt \frac{143i^7\pi^{14}}{5832x_0^{14}}+|R_i| \lt \frac{2288\pi^{14}}{729}x_0^{-14}+2x_0^{-12} \lt 500x_0^{-12}
\end{equation} for 
$x_0\geq171$. Denote
\begin{equation}
\begin{aligned}
{\phi}(x)&=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\dots+\frac{x^{10}}{3628800}+\frac{x^{11}}{39916800}+\frac{x^{12}}{479001600},\\
{j_i}&=x_0+\frac{i\pi^2}{3x_0}-\frac{i^2\pi^4}{18x_0^3}+\frac{i^3\pi^6}{54x_0^5}-\frac{5i^4\pi^8}{648x_0^7}+\frac{7i^5\pi^{10}}{1944x_0^9}-\frac{7i^6\pi^{12}}{3888x_0^{11}},\\
{j}&=x_0+\frac{\pi^2}{6x_0}-\frac{\pi^4}{72x_0^3}+\frac{\pi^6}{432x_0^5}-\frac{5\pi^8}{10368x_0^7}+\frac{7\pi^{10}}{62208x_0^9}-\frac{7\pi^{12}}{248832x_0^{11}},
\end{aligned}
\end{equation}then
\begin{equation}
\begin{aligned}
&\left|e^{x_i-s}-{\phi}({j_i}-{j})\right|=\left|\left(e^{x_i-s}-e^{{j_i}-{j}}\right)+(e^{{j_i}-{j}}-{\phi}({j_i}-{j}))\right|\\ \lt &\frac{8}{7}\left|(x_i-s)-({j_i}-{j})\right|+\left|\frac{13}{12}\frac{({j_i}-{j})^{13}}{13!}\right|\\ \lt &\frac{11i^7\pi^{14}}{5832x_0^{13}}+\frac{\pi^{26}}{5748019200x_0^{13}} \lt 5000x_0^{-12}.
\end{aligned}
\end{equation} Thus, 
$\mathop{\mathrm{spt}}\nolimits(n+i)$ can be rewritten as
\begin{equation}
\mathop{\mathrm{spt}}\nolimits(n+i)=\frac{2\sqrt3e^x}{x_0}({\phi}({j_i}-{j})+{w_i})({y_i}+{r_i}),
\end{equation} with 
$|{r_i}|,|{w_i}| \lt {r}=5000x_0^{-12}$. It means that the positivity of 
$\mathcal{L}_1^{(2)}\left(\mathop{\mathrm{spt}}\nolimits(n-1)\right)$ is equivalent to
\begin{equation}
\begin{aligned}
&(\phi(j_0-j)+w_0)^2(\phi(j_{-1}-j)+w_{-1})(\phi(j_3-j)+w_3)(y_0+r_0)^2(y_{-1}+r_{-1})(y_3+r_3)\\
&+(\phi(j_{-1}-j)+w_{-1})(\phi(j_1-j)+w_1)^3(y_{-1}+r_{-1})(y_1+r_1)^3\\
&-(\phi(j_{-1}-j)+w_{-1})^2(\phi(j_1-j)+w_1)(\phi(j_3-j)+w_3)(y_{-1}+r_{-1})^2(y_1+r_1)(y_3+r_3)\\
&+(\phi(j_{-1}-j)+w_{-1})^2(\phi(j_2-j)+w_2)^2(y_{-1}+r_{-1})^2(y_2+r_2)^2\\
&-2(\phi(j_{-1}-j)+w_{-1})(\phi(j_0-j)+w_0)(\phi(j_1-j)+w_1)(\phi(j_2-j)+w_2)\\
&\quad\quad(y_{-1}+r_{-1})(y_0+r_0)(y_1+r_1)(y_2+r_2) \gt 0.
\end{aligned}
\end{equation} For 
$x_0\geq171$ and 
$-1\leq i,j,k,l\leq3$, we have
\begin{equation}
\begin{aligned}
&\prod_{s=i,j,k,l}({\phi}({j}_s-{j})+{w}_s)({y}_s+{r}_s)\\ \lt &\prod_{s=i,j,k,l}{\phi}({j}_s-{j}){y}_s+4{r}(y_{-1}+{r})^3({\phi}({j}_3-{j})+{r})^4+4{r}(y_{-1}+{r})^4({\phi}({j}_3-{j})+{r})^3\\ \lt &\prod_{s=i,j,k,l}{\phi}({j}_s-{j}){y}_s+8{r}\times1.1^7\\ \lt &\prod_{s=i,j,k,l}{\phi}({j}_s-{j}){y}_s+16{r}.
\end{aligned}
\end{equation} On the other hand, with similar arguments, we get for 
$x_0\geq171$ and 
$-1\leq i,j,k,l\leq3$
\begin{equation}
\prod_{s=i,j,k,l}({\phi}({j}_s-{j})+{w}_s)({y}_s+{r}_s) \gt \prod_{s=i,j,k,l}{\phi}({j}_s-{j}){y}_s-16{r}.
\end{equation}It remains to show that
\begin{equation}
\begin{aligned}
&\phi(j_{-1}-j)^2\phi(j_2-j)^2y_{-1}^2y_2^2-2\phi(j_{-1}-j)\phi(j_0-j)\phi(j_1-j)\phi(j_2-j)y_{-1}y_0y_1y_2\\
&\quad+\phi(j_{-1}-j)\phi(j_1-j)^3y_{-1}y_1^3+\phi(j_0-j)^2\phi(j_{-1}-j)\phi(j_3-j)y_0^2y_{-1}y_3\\
&\quad \quad-\phi(j_{-1}-j)^2\phi(j_1-j)\phi(j_3-j)y_{-1}^2y_1y_3-96r \gt 0.
\end{aligned}
\end{equation} Denote the numerator of the left-hand side of the above inequality is 
$C(x_0)$. Note that 
$C(x_0)$ is a polynomial of degree 567 with positive coefficient of the first term. One can see that for n > 4446, (5.16) holds. It follows that (5.7) holds for n > 4446. A direct calculation reveals that for 
$241\leq n\leq 4446$, 
$\mathop{\mathrm{spt}}\nolimits(n)$ satisfies the iterated Laguerre inequality. The proof is completed.
Now we proceed to prove the companion iterated Laguerrre inequality for 
$\mathop{\mathrm{spt}}\nolimits(n)$.
Theorem 5.3. For 
$n\geq 222$, we have
\begin{equation}
\begin{aligned}
&\left(1+\frac{16\sqrt{2}}{3\pi^2n^{\frac{3}{2}}}\right)\left(\mathop{\mathrm{spt}}\nolimits(n)^2-\mathop{\mathrm{spt}}\nolimits(n-1)\mathop{\mathrm{spt}}\nolimits(n+1)\right)\left(\mathop{\mathrm{spt}}\nolimits(n+1)^2-\mathop{\mathrm{spt}}\nolimits(n-1)\mathop{\mathrm{spt}}\nolimits(n+3)\right)\\
& \gt (\mathop{\mathrm{spt}}\nolimits(n)\mathop{\mathrm{spt}}\nolimits(n+1)-\mathop{\mathrm{spt}}\nolimits(n-1)\mathop{\mathrm{spt}}\nolimits(n+2))^2.
\end{aligned}
\end{equation}Proof. To prove (5.17), we only need to show
\begin{equation}
\begin{aligned}
&\left(1+\frac{\sqrt{3}\pi}{x_0^{3}}\right)\left(\mathop{\mathrm{spt}}\nolimits(n)^2-\mathop{\mathrm{spt}}\nolimits(n-1)\mathop{\mathrm{spt}}\nolimits(n+1)\right)\left(\mathop{\mathrm{spt}}\nolimits(n+1)^2-\mathop{\mathrm{spt}}\nolimits(n-1)\mathop{\mathrm{spt}}\nolimits(n+3)\right)\\
& \gt (\mathop{\mathrm{spt}}\nolimits(n)\mathop{\mathrm{spt}}\nolimits(n+1)-\mathop{\mathrm{spt}}\nolimits(n-1)\mathop{\mathrm{spt}}\nolimits(n+2))^2.
\end{aligned}
\end{equation}Let
\begin{equation*}M:= \frac{\sqrt{3}\pi}{x_0^{3}}.\end{equation*}With similar arguments as in the proof of Theorem 5.2 and following the notation of Theorem 5.2, (5.17) is equivalent to
\begin{equation*}
\begin{aligned}
&M\phi(z_{0}-z)^2\phi(z_1-z)^2y_{0}^2y_1^2+(1+M)\Bigg(-\phi(z_{-1}-z)\phi(z_1-z)^3y_{-1}y_1^3\\
&-\phi(z_0-z)^2\phi(z_{-1}-z)\phi(z_3-z)y_0^2y_{-1}y_3+\phi(z_{-1}-z)^2\phi(z_1-z)\phi(z_3-z)y_{-1}^2y_1y_3\Bigg)\\
&-\phi(z_{-1}-z)^2\phi(z_2-z)^2y_{-1}^2y_2^2+2\phi(z_{-1}-z)\phi(z_0-z)\phi(z_1-z)\phi(z_2-z)y_{-1}y_0y_1y_2\\ &-16r(4M+6) \gt 0.
\end{aligned}
\end{equation*} With similar arguments, we have the numerator of the above inequality is a certain polynomial of degree 570 with positive coefficient of the first term. It can be checked that the inequality above holds for n > 12606. It means that the positivity of (5.17). A direct computation gives that for 
$222\leq n \leq 12606$, 
$\mathop{\mathrm{spt}}\nolimits(n)$ satisfies the companion iterated Laguerre inequality.
6. Further work
We will further consider the inequality 
$\mathcal{L}_{r}^{(2)}({\alpha(n)}) \gt 0$. For r = 2, following the approach given in Theorem 2.1, we conclude the following result. Since there are too many terms in 
$\mathcal{L}_{2}^{(2)}({\alpha(n)})$ and the estimate method is similar to those in Theorem 2.1, we omitted the tedious details.
Theorem 6.1. Assume that 
$\alpha(n)$ satisfies the conditions of Theorem 2.1, then for sufficiently large n,
\begin{equation}
\mathcal{L}_{2}^{(2)}({\alpha(n)}) \gt 0.
\end{equation}
\begin{align*}
&\mathcal{L}_{2}^{(2)}({\alpha(n)})\\
=& \left(\alpha(n)\alpha(n+4) - 4\alpha(n+1)\alpha(n+3)+3\alpha(n+2)^2\right)\left(\alpha(n)\alpha(n+8) - 4\alpha(n+2)\alpha(n+6)\right.\\ &+\left. 3\alpha(n+4)^2\right)\\
&-4\left(\alpha(n)\alpha(n+5) - 3\alpha(n+1)\alpha(n+4)+2\alpha(n+2)\alpha(n+3)\right)\\
&\quad\quad \left(\alpha(n)\alpha(n+7) - \alpha(n+1)\alpha(n+6)-3\alpha(n+2)\alpha(n+5)+3\alpha(n+3)\alpha(n+4)\right)\\
&+3\left(\alpha(n)\alpha(n+6) - 2\alpha(n+1)\alpha(n+5)+\alpha(n+2)\alpha(n+4)+2\alpha(n+3)^2\right)^2\\
=&55296A_2(n)^6+o(A_2(n)^6)\\
=&55296\delta(n)^{12}+o(\delta(n)^{12}) \gt 0,
\end{align*}as desired.
As a corollary, 
$\gamma(n)$ can be proved to satisfy 
$\mathcal{L}_{2}^{(2)}({\gamma(n)}) \gt 0$ for sufficiently large n.
Theorem 6.2. For sufficiently large n,
\begin{equation}
\mathcal{L}_{2}^{(2)}({\gamma(n)}) \gt 0.
\end{equation} For p(n), 
$\bar p(n)$ and 
$\mathop{\mathrm{spt}}\nolimits(n)$, employing similar arguments in Theorem 3.2, 4.2 and 5.2, we can prove 
$\mathcal{L}_{2}^{(2)}$ preserves the positivity for p(n), 
$\bar p(n)$ and 
$\mathop{\mathrm{spt}}\nolimits(n)$.
Theorem 6.3. For 
$ n\geq1230$,
\begin{equation}
\mathcal{L}_{2}^{(2)}({p(n)}) \gt 0.
\end{equation}Theorem 6.4. For 
$n\geq266$,
\begin{equation}
\mathcal{L}_{2}^{(2)}({\bar{p}(n)}) \gt 0.
\end{equation}Theorem 6.5. For 
$n\geq1276$,
\begin{equation}
\mathcal{L}_{2}^{(2)}(\mathop{\mathrm{spt}}\nolimits(n)) \gt 0.
\end{equation} We still intend to study the inequality arising from the higher order iterated Laguerre operator acting on functions. The first step is to write 
$\mathcal{L}_{r}^{(m)}({\alpha(n)})$ as an explicit expression. However, we found that there is no pattern in the expression of 
$\mathcal{L}_{r}^{(m)}({\alpha(n)})$ and it seems like it is difficult to represent it in a unified form. Moreover, as the order increases, the number of terms increases rapidly, and it is too complicated to calculate one by one. Therefore, we propose a conjecture on the general case.
Conjecture 6.6. Let 
$\alpha(n)$ be one of 
$\gamma(n),p(n),\bar{p}(n)$, or spt(n). Then, for sufficiently large n and any positive integers r and m,
\begin{equation}
\mathcal{L}_{r}^{(m)}({\alpha(n)}) \gt 0.
\end{equation}Acknowledgements
We thank the anonymous referee for helpful comments. This work was supported by the National Science Foundation of China (grant number 12171254).
