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\begin{document}
\begin{CJK*}{GBK}{kai}
\CJKtilde
\title{The spt-function of Andrews}

\author[William Chen]{William Y.C. Chen}
\institute[CFC]
{
Center for Applied Mathematics\\[5pt]
Tianjin University,  Tianjin 300072, P. R. China\\[15pt]
}
\date{ July 7, 2017}
\begin{frame}
  \titlepage
\end{frame}
\begin{frame}
\frametitle{Goals for this talk}
My goals for this talk include the following:



\begin{itemize}

\item[{\rm (1)}] The introduction to spt-functions;
\vskip 0.7cm

\item[{\rm (2)}] The spt-crank;
\vskip 0.7cm

\item[{\rm (3)}] More spt-congruences on spt-functions;
\vskip 0.7cm

\item[{\rm (4)}] Generalizations and variations on spt-functions;
\vskip 0.7cm

\item[{\rm (5)}] Asymptotic properties on spt-functions;
\vskip 0.7cm

\item[{\rm (6)}] Conjecture on inequalities on spt-functions.
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{The definition of the spt-function}
The {\blue spt-function} was introduced by {\red Andrews} as the
weighted counting of partitions with respect to the
number of occurrences of {\red the smallest part}. To be specific, for a partition $\lambda$, let $n_s(\lambda)$ denote the number of appearances of the smallest of $\lambda$, then
$$\spt(n)=\sum_{|\lambda|=n}n_s(\lambda).$$



\begin{exam}
For $n=4$, we see that {\red $spt(4)=10$}. The five partitions of 4 and the values of $n_s(\lambda)$ are
listed below:
\[\begin{array}{c|ccccc}
\lambda&({\red 4})&(3,{\red 1})&({\red 2},{\red 2})&(2,{\red 1},{\red 1}) &
({\red 1},{\red 1},{\red 1},{\red 1})\\[3pt]
\hline
n_s(\lambda)&1&1&2&2&4
\end{array}
\]
\end{exam}
\end{frame}

\begin{frame}\frametitle{Dyson's rank}
Recall that {\blue the  rank of a partition} was introduced by {\red Dyson} as given below.

\begin{defi}[Dyson, 1944]
 Let $\lambda$ be a  partition.  The rank of   $\lambda$   is defined to be the largest part of $\lambda$ minus the number of parts of $\lambda$.
\end{defi}


For example, the rank of $(5,4,2,1,1,1)$ is equal to $5-6=-1$.


\footnotesize
\begin{thebibliography}{4}
{%\beamertemplatetextbibitems
\bibitem{Dyson-1944} F. Dyson, Some guesses in the theory of
partitions, Eureka (Cambridge)   8 (1944) 10--15.}
\end{thebibliography}
\end{frame}
\begin{frame}
\frametitle{The definition of crank}
In 1988, {\red Andrews and Garvan} gave the definition of crank for an ordinary partition as follows:
\begin{defi}[Andrews, Garvan, 1988]
For a partition $\lambda$, the crank of $\lambda$ is defined as follows:
\[crank(\lambda)=\begin{cases}
\lambda_1,&\text{if }n_1(\lambda)=0;\\
\mu(\lambda)-n_1(\lambda),&\text{if }n_1(\lambda)>0,
\end{cases}\]
where $n_1(\lambda)$ denotes the number of $1$'s in $\lambda$ and $\mu(\lambda)$ denotes the number of parts of $\lambda$ larger than $n_1(\lambda)$.
\end{defi}
\begin{exam}
For  example, let $\lambda=(\textcolor{blue}{7,7,6},4,3,\textcolor{red}{1,1,1,1})$, then $n_1(\lambda)=\textcolor{red}{4}$ and $\mu(\lambda)=\textcolor{blue}{3}$. This implies $crank(\lambda)=3-4=-1$.
\end{exam}
\end{frame}


\begin{frame}
\frametitle{The relation between spt-function and rank, crank}
{\red Andrews} proved that
\begin{thm}[Andrews, 2008]
For $n\geq 1$,
\[\spt(n)=\sum_{m=-\infty}^\infty m^2M(m,n)-\sum_{m=-\infty}^\infty m^2N(m,n),\]
where {\blue $M(m,n)$} denote the number of partitions of $n$ with crank $m$, and {\blue $N(m,n)$} denote the number of partitions of $n$ with rank $m$.
\end{thm}

\vskip 1cm
\footnotesize
\begin{thebibliography}{4}
{%\beamertemplatetextbibitems
\bibitem{Andrews-2008}   G.E. Andrews, The number of smallest parts in the partitions of $n$, J. Reine Angew. Math. 624
(2008) 133--142.
\bibitem{Andrews-2008}   George E. Andrews and F.G. Garvan, Dyson's crank of a partition,
Bull. Amer. Math. Soc. 18 (1988), 167--171.
}
\end{thebibliography}
\end{frame}

\begin{frame}
\frametitle{The congruences on the spt-function}
{\red Andrews} also deduced
 the following congruences on the spt-function
\begin{eqnarray*}
\spt(5n+4)&\equiv& 0\pmod{5};\\[3pt]
\spt(7n+5)&\equiv& 0\pmod{7};\\[3pt]
\spt({ 13}n+6)&\equiv& 0\pmod{{ 13}},
\end{eqnarray*}
which is a  striking resemblance to the Ramanujan's congruences.


\end{frame}

\begin{frame}\frametitle{$S$-partitions}
To give a combinatorial interpretation of these three spt-congruences, {\red Andrews, Garvan and Liang} introduced  the {\blue spt-crank} which is defined on {\blue $S$-partitions}.

\begin{defi}[Andrews, Garvan and Liang, 2012]
Let $\mathcal{D}$ denote  the set of partitions into distinct parts and $\mathcal{P}$ denote the set of partitions.
For   $\pi \in \mathcal{P}$, we use $s(\pi)$ to denote  the smallest part  of $\pi$ with the convention that $s(\emptyset)=+\infty$.    Define
\[S=\{(\pi_1, \pi_2,\pi_3)\in \mathcal{D} \times \mathcal{P} \times \mathcal{P} \colon {\red \pi_1 \neq \emptyset \text{ and }s(\pi_1)\leq \min\{s(\pi_2),s(\pi_3)\}}\}.\]
{\blue The triplet $\pi=(\pi_1,\,\pi_2,\,\pi_3) \in S$ is called  to be an   $S$-partitions of $n$} with weight $\omega(\pi)=(-1)^{\ell(\pi_1){\red -1}}$ if  $|\pi|=|\pi_1|+|\pi_2|+|\pi_3|=n$.
\end{defi}

\begin{exam}
The partition triplet
$((3,{\red 2}),(4,4,3,{\red 2}),(3,3,{\red 3}))$
is an $S$-partition of 27.
\end{exam}

\end{frame}



\begin{frame}\frametitle{spt-crank}
The spt-crank is defined as follows.

\begin{defi}[Andrews, Garvan and Liang, 2012]Let $\pi$ be an $S$-partition, {\blue the spt-crank of $\pi$,}  denoted  $r(\pi)$, is defined to be the difference between the number of parts of $\pi_2$ and   $\pi_3$, that is,
\[ r(\pi)=\ell(\pi_2)-\ell(\pi_3).\]
\end{defi}


\begin{exam}
The $S$-partition
$\pi=((3, 2),({\red 4,4,3, 2}),({\blue 3,3, 3}))$
is an $S$-partition of 27 with weight $\omega(\pi)=-1$ and spt-crank ${\red 4}-{\blue 3}=1$.
\end{exam}
\footnotesize
\begin{thebibliography}{4}
{%\beamertemplatetextbibitems
 \bibitem{Andrews-Garvan-Liang-2011}G.E. Andrews, F.G. Garvan and J.L. Liang, Combinatorial interpretations of congruences for the spt-function,  Ramanujan J. 29 (2012) 321--338.
}


\end{thebibliography}
\end{frame}


\begin{frame}
\frametitle{Notation}
 Let {\blue $N_S(m,n)$ denote the
net number of $S$-partitions of $n$ with spt-crank $m$,}
that is,
\[
N_S(m,n)=\sum_{\stackrel{|\pi|=n}{r(\pi)=m}}\omega(\pi)
\]
and
\[N_S(k,t,n)=\sum_{m \equiv k \pmod{t}}N_S(m,n).\]
\end{frame}

\begin{frame}
\frametitle{Combinatorial interpretations}
{\red Andrews, Garvan and Liang} showed the following relations:
\begin{thm}[Andrews, Garvan and Liang, 2013]
\begin{eqnarray*}
N_S(k,5,5n+4)&=& \frac{\spt(5n+4)}{5}, \quad \text{for} \quad 0\leq k\leq 4, \\[3pt]
N_S(k,7,7n+5)&=& \frac{\spt(7n+5)}{7}, \quad \text{for} \quad 0\leq k\leq 6,
\end{eqnarray*}
\end{thm}
which imply $spt(5n+4)\equiv 0 \pmod{5}$ and $spt(7n+5)\equiv 0 \pmod{7}$.
\end{frame}

\begin{frame}\frametitle{Generating function for $N_S(m,n)$}
 {\red Andrews, Garvan and Liang} also showed that
{\small \begin{thm}[Andrews, Garvan and Liang, 2013]
\begin{align*}
\sum_{m=-\infty}^{+\infty}\sum_{n\geq 0}N_S(m,n)z^mq^n&=1+\sum_{m=0}^{\infty}z^m
\sum_{j=0}^{\infty}\frac{q^{j^2+mj+2j+m+1}}{(q;q)_{j+m}}\times\\
&\hskip 2cm \sum_{h=0}^{j}{j
\brack h}\frac{q^{h^2+h}}{(q;q)_h(1-q^{m+1+j+h})}\\[2pt]
&+\sum_{m=1}^{\infty}z^{-m}\sum_{j=m}^{\infty}\frac{q^{j^2-mj+2j-m+1}}{(q;q)_{j-m}}\times \\
&\hskip 2cm \sum_{h=0}^{j}{j
\brack h}\frac{q^{h^2+h}}{(q;q)_h(1-q^{j-m+1+h})}.
\end{align*}
\end{thm}}
\end{frame}

\begin{frame}
\frametitle{Nonnegativity of $N_S(m,n)$}
From the generating function of $N_S(m,n)$, {\red Andrews, Garvan and Liang} derived the nonnegativity of $N_S(m,n)$.

\begin{thm}[Andrews, Garvan and Liang, 2013]
For all $n\geq 0$,
\[N_S(m,n)\geq 0.\]
\end{thm}
\end{frame}

\begin{frame}
\frametitle{Marked partitions}
{\red Andrews, Dyson and Rhoades}  introduced marked partitions.
\begin{defi}[Andrews, Dyson and Rhoades, 2013]
A marked partition of $n$ is a pair $(\lambda,k)$, where $\lambda$ is an ordinary partition of $n$ and  $k$ is an integer identifying one of its minimum parts.
\end{defi}
{\blue Here we set $\lambda_k=s(\lambda)$ where $s(\lambda)$ denotes the minimum part of $\lambda$.}


\footnotesize
\begin{thebibliography}{4}
{%\beamertemplatetextbibitems
 \bibitem{Andrews-Garvan-Liang-2011}G.E. Andrews, F.J. Dyson and R.C. Rhoades, On the distribution of the spt-crank, Mathematics 1  (2013) 76--88.
}
\end{thebibliography}
\end{frame}
\begin{frame}
\frametitle{Example}

For example, there are ten marked partitions of $4$.
\[\begin{array}{c|c|c|c}
&\lambda \in \mathcal{P}(4)& n_s(\lambda) & (\lambda,k)\\[2pt] \hline
&(4) & 1 & {\blue ((4),1)}\\[5pt] \hline
&(3,1)& 1 &{\blue ((3,1),2)}\\[5pt] \hline
&(2,2)&2 &{\blue ((2,2),1), ((2,2),2)}\\[5pt] \hline
&(2,1,1)&2 & {\blue((2,1,1),2), ((2,1,1),3)}\\[5pt]\hline
&(1,1,1,1)& 4 & {\blue \begin{array}{cc}
 ((1,1,1,1),1), &((1,1,1,1),2)\\[2pt]
 ((1,1,1,1),3), &((1,1,1,1),4)
\end{array}}\\[5pt] \hline
 \text{Total} &5&10&10
\end{array}\]

\vskip 0.2cm
From the definition, it is easy to see that {\red the number of marked partitions of $n$ is equal to $spt(n)$.}
\end{frame}


\begin{frame}\frametitle{Problem}
The next problem is that

\begin{prob}
How to divide the set of marked partitions of $5n+4$ {\rm(} or $7n+5${\rm )} into five {\rm (} or seven{\rm )} equinumerous classes?
\end{prob}
\vskip 0.2cm

Here we first introduce a combinatorial structure called doubly marked partition. We then define $N_S(m,n)$ on doubly marked partitions and build a  connection between marked partitions and doubly marked partitions. We then divide these marked partitions according to the spt-crank of doubly marked partitions.

\vskip 0.2cm
\footnotesize
\begin{thebibliography}{4}
{%\beamertemplatetextbibitems
\bibitem{Andrews-2008}   William Y.C. Chen, Kathy Q. Ji, and Wenston J.T. Zang, The spt-crank for ordinary partitions, J. Reine Angew. Math. 711 (2016)  231--249.
}
\end{thebibliography}
\end{frame}


\begin{frame}\frametitle{Doubly marked partitions}
We find that  $N_S(m,n)$ can be interpreted in terms of  the doubly marked partitions of $n$ with spt-crank $m$.
\vskip 0.2cm
  \parbox[c]{5.5cm}{\framebox{\parbox[t]{2in}{{\blue \bf Definition.}
{\red A doubly marked partition of $n$} is a triplet $(\lambda,s,t)$, where
\begin{itemize}
\item $\lambda$ is an ordinary partition of $n${\rm ;}
 \item $1\leq s\leq d(\lambda)$ {\rm ;}
  \item $s\leq t\leq \lambda_1${\rm ;}
  \item $\lambda'_s=\lambda'_t$.
  \end{itemize}}}
 }
  \parbox[c]{3cm}{ \input{double.tpx}}




\end{frame}

\begin{frame}\frametitle{The definition of spt-crank}
{\small We give the definition of spt-crank in terms of doubly marked partitions.
\begin{defi}[Chen, Ji and Zang, 2016]  Let  $(\mu,s,t)$ be a doubly marked partition.  Then the spt-crank of $(\mu,s,t)$, denoted $d(\mu,s,t)$, is defined by
\begin{equation*}
d(\mu,s,t)=\mu'_s-\mu_{\mu'_s-s+1}+t-2s+1.
\end{equation*}
\end{defi}



Let {\blue $((4,4,1,1),2,3)$} be a doubly marked partition  of $10$, its spt-crank is equal to}

  \parbox[c]{5.5cm}{$d((4,4,1,1),2,3)$\\[3pt]
  $={\red \mu'_2}-{\blue \mu_{\mu'_2-2+1}}+3-4+1$\\[3pt]
 $={\red 2}- {\blue 4}+3-4+1$\\[3pt]
 $=-2$.}
 \parbox[c]{3cm}{ \input{scexa.TpX}}



\end{frame}

\begin{frame}\frametitle{Example}

\begin{table}
\[\begin{array}{ccc|cc}
S\text{-partition}& \text{weight}&\text{ crank}&\text{doubly marked partition}&\text{ crank}\\[2pt]
((1),(1,1,1),\emptyset)&+1&3& ((1,1,1,1),1,1)&3\\[2pt]\hline
((1),(2,1),\emptyset)&+1&2&((2,1,1),1,1)&2\\[2pt]\hline
((1),(1,1),(1))&+1&1&((3,1),1,1)&1\\[2pt]
((1),(3),\emptyset)&+1&1&((2,2),1,2)&1\\[2pt]
((2,1),(1),\emptyset)&-1&1&&\\[2pt]
((2),(2),\emptyset)&+1&1&&\\[2pt]\hline
((1),(2),(1))&+1&0&((2,2),1,1)&0\\[2pt]
((1),(1),(2))&+1&0&((4),1,4)&0\\[2pt]
((3,1),\emptyset,\emptyset)&-1&0&&\\[2pt]
((4),\emptyset,\emptyset)&+1&0&&\\[2pt]\hline
\end{array}
\]
\end{table}\end{frame}

\begin{frame}\frametitle{Example}
\begin{table}
\[\begin{array}{ccc|cc}
S\text{-partition}& \text{weight}&\text{crank}&\text{doubly marked partition}&\text{crank}\\[2pt]
((1),(1),(1,1))&+1&-1&((2,2),2,2)&-1\\[2pt]
((1),\emptyset,(3))&+1&-1&((4),1,3)&-1\\[2pt]
((2,1),\emptyset,(1))&-1&-1&&\\[2pt]
((2),\emptyset,(2))&+1&-1&&\\[2pt]\hline
((1),\emptyset,(2,1))&+1&-2&((4),1,2)&-2\\[2pt]\hline
((1),\emptyset,(1,1,1))&+1&-3&((4),1,1)&-3
\end{array}
\]
\end{table}
\end{frame}




\begin{frame}\frametitle{Connection}
We have the following result.
\begin{thm}[Chen, Ji and Zang, 2016]
There is a bijection $\triangle$ between the set of marked partitions of $n$ and the set of doubly marked partitions of $n$.
\end{thm}



\end{frame}
\begin{frame}\frametitle{The bijection $\triangle$}
The bijection $\triangle$ is constructed as shown in the following figure.  Let $(\lambda,k)$ be a marked partition and $(\mu, a,b)=\triangle(\lambda,k)$.
\input{triangle.TpX}
\end{frame}

\begin{frame}\frametitle{The map $\phi$}
The map $\phi \colon ((6,5,3,1),2,6)\rightarrow ((4,3,3,3,1,1),3,4).$
\input{phi.TpX}
\end{frame}

\begin{frame}\frametitle{ An example of the bijection $\triangle$}
We give an example to explain {\blue how to get a double marked partition $((2,2,1,1),2,2)$ from a marked partition $((2,1,1,1,1),5)$.}
 \input{triexa.TpX}
We have shown that this process  is well-defined and reversible.
\end{frame}


\begin{frame}
\frametitle{Example-1}
For example, for $n=4$,  we list  ten marked partitions
of $4$, the corresponding doubly marked partitions, and the spt-crank modulo $5$.
{\small
\begin{table}\label{table-1}
 \[\begin{array}{c|c|c|c}
 (\lambda,k)& (\mu,s,t)=\triangle(\lambda,k) & d(\mu,s,t) & d(\mu,s,t) \text{mod } 5\\[2pt] \hline
  ((1,1,1,1),4)& ((4),1,4) & 0 &0\\[2pt]
  ((2,2),1)& ((2,2),1,1)& 0& 0\\[2pt]\hline
 ((3,1),2)& ((3,1),1,1)& 1& 1\\[2pt]
 ((2,2),2)& ((2,2),1,2)& 1& 1\\[2pt]\hline
 ((1,1,1,1),1)& ((4),1,1) & -3 &2\\[2pt]
 ((2,1,1),2)& ((2,1,1),1,1)& 2 &2\\[2pt]\hline
  ((1,1,1,1),2)& ((4),1,2) & -2 &3\\[2pt]
  ((4),1) & ((1,1,1,1),1,1)& 3& 3\\[2pt]\hline
 ((2,1,1),3)& ((2,2),2,2)& -1 &4\\[2pt]
 ((1,1,1,1),3)& ((4),1,3) & -1 &4
\end{array}\]
\end{table}}

\end{frame}
\begin{frame}\frametitle{Example-2}
   For another example,    for $n=5$, we list fourteen marked partitions
of $5$, the corresponding doubly marked partitions, and the spt-crank modulo $7$.



\begin{table}[h]
\[\begin{array}{ c|c|c|c}
 (\lambda,k)& (\mu,s,t)=\triangle(\lambda,k) & d(\mu,s,t) & d(\mu,s,t) \mod 7\\[2pt] \hline
  ((3,1,1),2)& ((3,2),1,1)& 0& 0\\[2pt]
 ((1,1,1,1,1),5)& ((5),1,5) & 0 &0\\[2pt]\hline
  ((3,1,1),3)& ((3,2),1,2)& 1& 1\\[2pt]
  ((4,1),2)& ((4,1),1,1)& 1& 1\\[2pt]\hline
 ((3,2),2)& ((3,1,1),1,1)& 2& 2\\[2pt]
 ((2,2,1),3)& ((2,2,1),1,1)& 2& 2\\[2pt]\hline
 ((2,1,1,1),2)& ((2,1,1,1),1,1)& 3 &3\\[2pt]
  ((1,1,1,1,1),1)& ((5),1,1) & -4 &3\\[2pt]\hline
  ((5),1) & ((1,1,1,1,1),1,1)& 4& 4\\[2pt]
   ((1,1,1,1,1),2)& ((5),1,2) & -3 &4
\end{array}\]


\end{table}

\end{frame}
\begin{frame}\frametitle{Example-2}

\[\begin{array}{c|c|c|c}
(\lambda,k)& (\mu,a,b)=\triangle(\lambda,k) & d(\mu,a,b) & d(\mu,a,b)  \mod 7\\[2pt] \hline
 ((2,1,1,1),3)& ((3,2),2,2)& -2 &5\\[2pt]
 ((1,1,1,1,1),3)& ((5),1,3) & -2 &5\\[2pt]\hline
 ((1,1,1,1,1),4)& ((5),1,4) & -1 &6\\[2pt]
 ((2,1,1,1),4)& ((2,2,1),2,2)& -1 &6
\end{array}\]
\end{frame}


\begin{frame}
\frametitle{The conjecture on spt-crank}
In 2013, {\red Andrews, Dyson and Rhoades} raised the following conjecture on $N_S(m,n)$:

\begin{conj}[Andrews, Dyson and Rhoades, 2013]
For $m\geq 0$ and $n\geq 1$:
$$N_S(m,n)\geq N_S(m+1,n).$$
\end{conj}
They proved this conjecture for sufficiently large $n$:
\begin{thm}[Andrews, Dyson and Rhoades, 2013]
For fixed $m\geq 0$,
\[N_S(m,n)-N_S(m+1,n)\sim \frac{(2m+1)\pi^2}{384\sqrt{3}n^2}\exp\left(\pi\sqrt{\frac{2n}{3}}\right)
\quad\text{as }n\rightarrow\infty .
\]
\end{thm}
\end{frame}
\begin{frame}
They also showed that this conjecture is equivalent to the following inequality:
\begin{conj}[Andrews, Dyson and Rhoades, 2013]
For $m\geq 0$ and $n\geq 1$, we have
\[M(\leq m,n)\leq N(\leq m,n),\]
\end{conj}
where  \textcolor{blue}{$N(\leq m,n)$} denote the number of partitions of $n$ with rank less than or equal to $m$, and \textcolor{blue}{$M(\leq m,n)$} denote the number of partitions of $n$ with crank less than or equal to $m$.

\vskip 0.2cm

When $m=0$, this inequality was conjectured by {\red Kaavya}.


\footnotesize
\begin{thebibliography}{4}
{%\beamertemplatetextbibitems
\bibitem{Andrews, Dyson, Rhoades, 2013}   G.E. Andrews, F.J. Dyson, R.C. Rhoades, On the distribution of the spt-crank, Mathematics 1
(2013) 76--98.
\bibitem{Kaavya}S.J. Kaavya, Crank $0$ partitions and the parity of the partition function, Int. J. Number Theory 7 (2011) 793--801.
}
\end{thebibliography}
\end{frame}

\begin{frame}
\frametitle{Confirmation of the conjecture}
We build an injection from the set of partitions of $n$ with crank less than or equal to $m$ to the set of partitions of $n$ with rank less than or equal to $m$. Thus we give a combinatorial proof of the following theorem.
\begin{thm}[Chen, Ji and Zang, 2015]
For all $m\geq 0$ and $n\geq 1$, we have
\[M(\leq m,n)\leq N(\leq m,n).\]
\end{thm}
\footnotesize
\begin{thebibliography}{4}
{%\beamertemplatetextbibitems
\bibitem{Andrews, Dyson, Rhoades, 2013}   William Y.C. Chen, Kathy Q. Ji and Wenston J.T. Zang, Proof of the Andrews-Dyson-Rhoades conjecture on the spt-crank, Adv. Math. 270(2015) 60--96.
}
\end{thebibliography}
\end{frame}

\begin{frame}
\frametitle{The inequality}
Actually, the inequality of rank and crank mentioned above has already been conjectured in 2009, by {\red Bringmann and Mahlburg}, as follows:

\begin{conj}[Bringmann and Mahlburg]
For all $n\geq 1$ and $m\geq 0$, we have
\[\textcolor{red}{N(\leq m-1,n)\leq M(\leq m,n)}\leq N(\leq m,n).\]
\end{conj}

We give a combinatorial proof of the first half of the inequality. Thus the conjecture has been confirmed.
\footnotesize
\begin{thebibliography}{4}
{%\beamertemplatetextbibitems
\bibitem{Bringmann-2009} K. Bringmann, K. Mahlburg, Inequalities between ranks and cranks, Proc. Amer. Math.
Soc. 137 (2009), no. 8, 2567--2574.

\bibitem{Chen-Ji-Zang}W.Y.C. Chen, K.Q. Ji and W.J.T. Zang, Nearly equal distributions of the rank and the crank of partitions, to appear.

}
\end{thebibliography}
\end{frame}
\begin{frame}
\frametitle{The property of $\tau_n$}
This inequality implies a bijection $\tau_n:P(n)\rightarrow P(n)$, which has the following interesting property:
\begin{thm}[Bringmann and Mahlburg, 2009]
For any $|\lambda|=n$, we have:
\[|crank(\lambda)|-|rank(\tau_n(\lambda))|=0\text{ or }1.\]
\end{thm}


{\red Bringmann and Mahlburg} pointed that the above property of $\tau_n$ leads to an upper bound of spt-function:
\begin{thm}[Bringmann and Mahlburg, 2009]
\[spt(n)\leq \sqrt{2n}p(n).\]
\end{thm}

\end{frame}

\begin{frame}
\frametitle{More congruences}
{\red Garvan} found the following congruences of the spt-function

\begin{thm}[Garvan, 2010]
For $n\geq 0$,
\begin{eqnarray*}
\spt(11\cdot19^4\cdot n+22006)&\equiv& 0\pmod{11},\\[3pt]
 \spt(17\cdot7^4\cdot n+243)&\equiv& 0\pmod{17},\\[3pt]
 \spt(19\cdot5^4\cdot n+99)&\equiv& 0\pmod{19},\\[3pt]
 \spt(29\cdot13^4\cdot n+18583)&\equiv& 0\pmod{29},\\[3pt]
 \spt(31\cdot29^4\cdot n+409532)&\equiv& 0\pmod{31},\\[3pt]
 \spt(37\cdot5^4\cdot n+1349)&\equiv& 0\pmod{37}.
\end{eqnarray*}
\end{thm}
\footnotesize
\begin{thebibliography}{4}
{%\beamertemplatetextbibitems
\bibitem{Garvan-2010}F.G. Garvan, Congruences for Andrews' smallest parts partition function and new congruences for Dyson's rank, Int. J. Number Theory 6 (2) (2010) 281--309.
}
\end{thebibliography}



\end{frame}

\begin{frame}
\frametitle{General case}
{\red Bringmann}  proved the following assertion.

\begin{thm}[Bringmann, 2008]
For any prime $\ell\geq 5$, there are infinitely many
arithmetic progressions $an+b$ such that
\[\spt(an+b)\equiv 0\pmod{\ell}.\]
\end{thm}

This assertion is reminiscent to the following result obtained by {\red Ono}.

\begin{thm}[Ono, 2000]
 For any prime $\ell\geq 5$,  there are infinitely  many arithmetic progressions $an+b$ such that
  \begin{equation*}
  p(an+b)\equiv 0\pmod{\ell}.
  \end{equation*}
  \end{thm}

\footnotesize
\begin{thebibliography}{4}
{%\beamertemplatetextbibitems
\bibitem{Bringmann-2008} K. Bringmann, On the explicit construction of higher deformations of partition statistics, Duke Math. J.
144 (2) (2008)  195--233.
\bibitem{Ono-2000} K. Ono, Distribution of the partition function modulo $m$, Ann. of Math. (2)  151 (1) (2000) 293--307.
}
\end{thebibliography}

\end{frame}

\begin{frame}
\frametitle{Explicit Ramanujan-type congruence}
{\red Ono} derived the following  Ramanujan-type congruences of $\spt(n)$ modulo $\ell$ for any prime $\ell\geq 5$.

\begin{thm}[Ono, 2011]
Let $\ell\geq 5$ be a prime and let $\left(\frac{\bullet}{\circ}\right)$ denote the Legendre symbol.
 \begin{itemize}
 \item[{\rm(i)}] For $n\geq 1$, if $\left(\frac{-n}{\ell}\right)=1$,
\[\spt\left((\ell^2 n+1)/24\right)\equiv 0\pmod{\ell}.
\]
\item[{\rm(ii)}] For $n\geq 0$,
\[\spt\left((\ell^{3}n+1)/24\right)\equiv \left(\frac{3}{\ell}\right)\spt\left((\ell n+1)/24\right)\pmod{\ell}.\]
\end{itemize}
\end{thm}

\footnotesize
\begin{thebibliography}{4}
{%\beamertemplatetextbibitems

\bibitem{Ono-2011}K. Ono, Congruences for the Andrews spt-function, Proc. Natl. Acad. Sci. USA 108 (2) (2011) 473--476.
}
\end{thebibliography}

\end{frame}


\begin{frame}
\frametitle{Explicit Ramanujan-type congruence on prime powers}
{\red Ahlgren,  Bringmann and  Lovejoy}  extended the above result to any prime power. 

\begin{thm}[Ahlgren, Bringmann and Lovejoy, 2011]\label{thm-ell-adic}
Let $\ell\geq 5$ be a prime and let $m\geq 1$.
\begin{itemize}
\item[{\rm(i)}] For $n\geq 1$, if $\left(\frac{-n}{\ell}\right)=1$,
\[\spt\left((\ell^{2m}n+1)/24\right)\equiv 0\pmod{\ell^m}.\]

\item[{\rm(ii)}] For $n\geq 0$,
\[\spt\left((\ell^{2m+1}n+1)/24\right)\equiv\left(\frac{3}{\ell}\right) \spt\left((\ell^{2m-1}n+1)/24\right)\pmod{\ell^m}.\]
\end{itemize}
\end{thm}
\footnotesize
\begin{thebibliography}{4}
{%\beamertemplatetextbibitems

\bibitem{Ahlgren-Bringmann-Lovejoy-2011}
S. Ahlgren, K. Bringmann and J. Lovejoy, $\ell$-adic properties of smallest parts functions, Adv. Math. 228 (1) (2011) 629--645.
}
\end{thebibliography}


\end{frame}



\begin{frame}
\frametitle{The congruences of powers of $5,7$ and $13$.}
{\red Garvan} gave the following congruences of powers of $5,7$ and $13$.

\begin{thm}[Garvan, 2012]\label{thm-Garvan-5713p}
Let $a,b,c$ be positive integers, and let $\delta_a$, $\lambda_b$ and $\gamma_c$ be the least nonnegative residues of the reciprocals of $24$ mod $5^a$, $7^b$ and $13^c$ respectively. Then
\[\spt(5^an+\delta_a)\equiv 0\pmod{5^{\lfloor\frac{a+1}{2}\rfloor}},\]
\[\spt(7^bn+\lambda_b)\equiv 0\pmod{7^{\lfloor\frac{b+1}{2}\rfloor}},\]
\[\spt(13^cn+\gamma_c)\equiv 0\pmod{13^{\lfloor\frac{c+1}{2}\rfloor}}.\]
\end{thm}

\footnotesize
\begin{thebibliography}{4}
{%\beamertemplatetextbibitems
\bibitem{Garvan-2012} F.G. Garvan, Congruences for Andrews' spt-function modulo powers of 5, 7 and 13, Trans. Amer. Math. Soc. 364 (9) (2012) 4847--4873.
}
\end{thebibliography}


\end{frame}
\begin{frame}
\frametitle{The parity of the spt-function}
For $n\geq 1$, the parity of $\spt(n)$ is determined by {\red Folsom and Ono}. They obtained the following characterization of the  parity of $\spt(n)$.

\begin{thm}[Folsom and Ono, 2008]\label{thm-spt-parity}
 The function $\spt(n)$ is odd if and only if $24n-1=pm^2$, where $m$ is an integer and $p\equiv 23\pmod{24}$ is  prime.
\end{thm}

As pointed out by {\red Andrews, Garvan and Liang},
 the above theorem contains an error.

 \begin{exam}
   For $n=507$, it is clear that $507\times 24-1=12167=23\times 23^2=pm^2$, where $p=m=23$. Obviously, $507$ satisfies the condition of the above theorem. But  $\spt(507)=60470327737556285225064$ is even.
   \end{exam}
\footnotesize
\begin{thebibliography}{4}
{%\beamertemplatetextbibitems
\bibitem{Folsom-Ono-2008}A. Folsom and K. Ono, The spt-function of Andrews, Proc. Natl. Acad. Sci. USA 105 (51) (2008) 20152--20156.
}
\end{thebibliography}


\end{frame}

\begin{frame}
\frametitle{The correct parity of the spt-function}
{\red Andrews, Garvan and Liang} corrected Folosom and Ono's result as given below.

\begin{thm}[Andrews, Garvan and Liang, 2013]\label{thm-pir-spt-co}
The function spt$(n)$ is odd if and only if $24n-1 = p^{4a+1}m^2$ for some prime $p \equiv 23  \pmod{24}$
and some integers $a, m$ with $(p,m) = 1$.
\end{thm}

\footnotesize
\begin{thebibliography}{4}
{%\beamertemplatetextbibitems
\bibitem{Andrews-Garvan-Liang-2013} G.E. Andrews, F.G. Garvan and J. Liang, Self-conjugate vector partitions and the parity of the spt-function, Acta Arith. 158 (3) (2013) 199--218.
}
\end{thebibliography}
\end{frame}

\begin{frame}
\frametitle{The congruence of modulo 3}
On the congruences of modulo $3$, {\red Folsom and Ono} gave the following result:
\begin{thm}[Folsom and Ono, 2008]
Let $\ell\geq 5$ be a prime such that $\ell\equiv 2\pmod{3}$. If $0 < k < \ell-1,$
then for   $n\geq 0$,
\begin{eqnarray*}
&&\spt(\ell^4n+\ell^3k-(\ell^4-1)/24)\equiv
 0 \pmod{3}.\end{eqnarray*}
\end{thm}

For example, for $\ell=5$, we may deduce the following congruences:
\begin{eqnarray*}
\spt(625n + 99) &\equiv& \spt(625n + 224)
\equiv \spt(625n + 349)\\[3pt]
 &\equiv& \spt(625n + 474)
  \equiv 0 \pmod {3}.
  \end{eqnarray*}

\end{frame}

\begin{frame}
\frametitle{The variations of the spt-function}
{\blue The  $k$-th symmetrized rank function $\eta_k(n)$} was defined by  {\red Andrews}  as
\begin{equation*}\label{equ-def-etak}
\eta_{k}(n) =\sum_{m=-n}^n{m+\lfloor \frac{k-1}{2}\rfloor\choose k}N(m,n).
\end{equation*}

{\red Garvan} introduced {\blue the $k$-th symmetrized crank function $\mu_k(n)$}
as follows:
\begin{equation*}\label{equ-def-muk}
\mu_{k}(n) =\sum_{m=-n}^n{m+\lfloor \frac{k-1}{2}\rfloor\choose k}M(m,n).
\end{equation*}
\footnotesize
\begin{thebibliography}{4}
{%\beamertemplatetextbibitems
\bibitem{Andrews-2007}G.E. Andrews, Partitions, Durfee symbols, and the Atkin-Garvan moments of ranks, Invent. Math. 169 (1) (2007) 37--73.
\bibitem{Garvan-2011}F.G. Garvan, Higher order spt-functions, Adv. Math. 228 (1) (2011) 241--265.
}
\end{thebibliography}


\end{frame}

\begin{frame}
\frametitle{The higher order spt-function}
{\red Garvan}  introduced {\blue the higher order $\spt$-function $\spt_k(n)$} in terms of $\mu_k(n)$ and $\eta_k(n)$.

\begin{defi}[Garvan, 2011]
For  $k\geq 1$, define
\begin{equation*}\label{def-sptk}
\spt_k(n)=\mu_{2k}(n)-\eta_{2k}(n).
\end{equation*}
\end{defi}

Garvan proved that
\[\mu_{2k}(n)\geq \eta_{2k}(n),\]
which implies that $\spt_k(n)\geq 0$.
\end{frame}

\begin{frame}
\frametitle{The congruences of the higher spt-function}
Garvan  obtained     congruences of $\spt_2(n)$, $\spt_3(n)$ and $\spt_4(n)$.

\begin{thm}[Garvan, 2011]
For $n\geq 1$,
\begin{eqnarray*}
\spt_2(n)\equiv 0\pmod{5},&&\text{if } n\equiv 0,1,4\pmod{5},\\[3pt]
\spt_2(n)\equiv 0\pmod{7},&&\text{if } n\equiv 0,1,5\pmod{7},\\[3pt]
\spt_2(n)\equiv 0\pmod{11},&&\text{if } n\equiv 0\pmod{11},\\[3pt]
\spt_3(n)\equiv 0\pmod{7},&&\text{if } n\not\equiv 3,6\pmod{7},\\[3pt]
\spt_3(n)\equiv 0\pmod{2},&&\text{if } n\equiv 1\pmod{4},\\[3pt]
\spt_4(n)\equiv 0\pmod{3}, &&\text{if } n\equiv 0\pmod{3}.
\end{eqnarray*}
\end{thm}

Garvan also provided a combinatorial interpretation of $\spt_k(n)$.

\end{frame}

\begin{frame}
\frametitle{The ospt-function}
{\red Andrews, Chan and Kim} introduced the   modified rank and crank moments $N^+_j(n)$ and $M^+_j(n)$
by considering the following unilateral sums:
\[N^+_j(n)=\sum_{m\geq 0}m^jN(m,n)\]
and
\[M^+_j(n)=\sum_{m\geq 0}m^jM(m,n).\]



They defined the ospt-function $\ospt(n)$
as given below:

\begin{defi}[Andrews, Chan and Kim, 2013]
For $n\geq 1$,
\begin{equation*}\label{defi-ospt-eq}
\ospt(n)={M}^+_1(n)-{N}^+_1(n).
\end{equation*}
\end{defi}
\footnotesize
\begin{thebibliography}{4}
{%\beamertemplatetextbibitems
\bibitem{Andrews-Chan-Kim-2013}
G.E. Andrews, S.H. Chan and B. Kim, The odd moments of ranks and cranks, J. Combin. Theory Ser. A 120 (1) (2013) 77--91.
}
\end{thebibliography}
\end{frame}



\begin{frame}
\frametitle{The positivity of the ospt-function}
{\red Andrews, Chan and Kim} obtained  the following inequality.

\begin{thm}[Andrews, Chan and Kim, 2013]\label{thm-mk+n>nk+n}
For all positive integer $n$ and $j$,
\begin{equation*}\label{thm-mk+n>nk+neq}
M^+_j(n)>N^+_j(n).
\end{equation*}
\end{thm}

This yields that $\ospt(n)>0$. They also gave the combinatorial interpretation of $\ospt(n)$ in terms of even and odd strings of a partition. Using this combinatorial interpretation, {\red Bringmann and Mahlburg} proved    a monotone
 property of $\ospt(n)$ as given below.

\begin{thm}[Bringmann and Mahlburg, 2014]
For $n\geq 1$,
\[\ospt(n+1)\geq \ospt(n).\]
\end{thm}

\footnotesize
\begin{thebibliography}{4}
{%\beamertemplatetextbibitems
\bibitem{Bringmann-Mahlburg-2014} K. Bringmann and K. Mahlburg, Asymptotic inequalities for positive crank and rank moments, Trans.
Amer. Math. Soc. 366 (2) (2014) 1073--1094.
}
\end{thebibliography}
\end{frame}

\begin{frame}
\frametitle{The inequalities of the ospt-function}
{\red Chan and Mao} proved the following inequalities of $\ospt(n)$.

\begin{thm}[Chan and Mao, 2014]
We have
\begin{align}
&\ospt(n)>\frac{p(n)}{4}+\frac{N(0,n)}{2}-\frac{M(0,n)}{4}\quad \text{for } n\geq 8,\nonumber\\[15pt]
&\ospt(n)<\frac{p(n)}{4}+\frac{N(0,n)}{2}-\frac{M(0,n)}{4}+\frac{N(1,n)}{2}\quad \text{for } n\geq 7,\nonumber\\[15pt]
&\ospt(n)<\frac{p(n)}{2} \quad \text{for } n\geq 3.\nonumber
\end{align}
\end{thm}

\footnotesize
\begin{thebibliography}{4}
{%\beamertemplatetextbibitems

\bibitem{Chan-Mao-2014}S.H. Chan and R. Mao, Inequalities for ranks of partitions and the first moment of ranks and cranks of partitions, Adv. Math. 258 (2014) 414--437.
}
\end{thebibliography}
\end{frame}

\begin{frame}
\frametitle{The asymptotic properties of the spt-function}

By applying the circle method to  the second symmetrized rank
moment $\eta_2(n)$,  {\red Bringmann}  obtained an asymptotic expression of the spt-function $\spt(n)$.

\begin{thm}[Bringmann, 2008]\label{thm-asy-p-o}
As $n\rightarrow\infty$,
\begin{equation*}\label{thm-asy-p-o-eqn}
\spt(n)\sim\frac{\sqrt{6}}{\pi}\sqrt{n}p(n)\sim\frac{1}{2\sqrt{2}\pi\sqrt{n}}
e^{\pi\sqrt{\frac{2n}{3}}}.
\end{equation*}
\end{thm}
\footnotesize
\begin{thebibliography}{4}
{%\beamertemplatetextbibitems
\bibitem{Bringmann-2008} K. Bringmann, On the explicit construction of higher deformations of partition statistics, Duke Math. J.
144 (2) (2008)  195--233.
}
\end{thebibliography}


\end{frame}

\begin{frame}
\frametitle{The exact relation of the spt-function}
{\red Ahlgren and Andresen}  obtained
an exact expression for the spt-function.

\begin{thm}[Ahlgren and Andresen, 2016]
For $n\geq 1$,
\[\spt(n)=\frac{\pi}{6}(24n-1)^{\frac{1}{4}}\sum_{c=1}^\infty \frac{A_c(n)}{c}(I_{1/2}-I_{3/2})\left(\frac{\pi\sqrt{24n-1}}{6c}\right),\]
where $I_\nu$ is the I-Bessel function, $A_c(n)$ is the Kloosterman sum
\[A_c(n)=\sum_{d \mod c \atop (d,c)=1}e^{\pi i s(d,c)-2i\pi\frac{dn}{c}}.\]
\end{thm}

\footnotesize
\begin{thebibliography}{4}
{%\beamertemplatetextbibitems
\bibitem{Ahlgren-Scott-Andresen-2016}S. Ahlgren and N. Andersen, Algebraic and transcendental formulas for the smallest parts function, Adv. Math. 289 (2016) 411--437.
}
\end{thebibliography}

\end{frame}

\begin{frame}
\frametitle{The asymptotic property of $\spt_k(n)$}
The asymptotic property of $\spt_k(n)$ was first conjectured by { Bringmann and Mahlburg}, and then confirmed by {\red Bringmann, Mahlburg and Rhoades}.

\begin{thm}[Bringmann, Mahlburg and Rhoades, 2011]
As $n\rightarrow\infty$,
\[\spt_k(n)\sim\beta_{2k}n^{k-\frac{1}{2}}p(n),\]
where $\beta_{2k}\in\frac{\sqrt{6}}{\pi}\mathbb{Q}$ is positive.
\end{thm}



\footnotesize
\begin{thebibliography}{4}
{%\beamertemplatetextbibitems
\bibitem{Bringmann-Mahlburg-2009}K. Bringmann and K. Mahlburg, Inequalities between ranks and cranks, Proc. Amer. Math. Soc. 137 (8) (2009) 2567--2574.

\bibitem{Bringmann-Mahlburg-Rhoades-2011}  K. Bringmann, K. Mahlburg and R.C. Rhoades, Asymptotics for rank and crank moments, Bull. London Math. Soc.  43 (4) (2011) 661--672.
}
\end{thebibliography}

\end{frame}
\begin{frame}
\frametitle{The asymptotic property of $\ospt(n)$}
{\red Bringmann and Mahlburg} derived the asymptotic formula of $\ospt(n)$ as given below.

\begin{thm}[Bringmann and Mahlburg, 2014]
As $n\rightarrow \infty$,
\begin{equation*}\label{BriMaheva-for}
\ospt(n)\sim \frac{p(n)}{4}\sim\frac{1}{16\sqrt{3}n}e^{\pi\sqrt{\frac{2n}{3}}}.
\end{equation*}
\end{thm}

\footnotesize
\begin{thebibliography}{4}
{%\beamertemplatetextbibitems

\bibitem{Bringmann-Mahlburg-2014} K. Bringmann and K. Mahlburg, Asymptotic inequalities for positive crank and rank moments, Trans.
Amer. Math. Soc. 366 (2) (2014) 1073--1094.
}
\end{thebibliography}
\end{frame}

\begin{frame}
\frametitle{The log-concavity of $p(n)$}
 {\red DeSalvo and Pak}
proved that the partition function $p(n)$ satisfies the log-concave property for $n\geq 26$.


\begin{thm}[DeSalvo and Pak, 2015]\label{thm-p-log-c-thm}
For $n\geq 26$,
\begin{equation*} \label{thm-p-log-c}
p(n)^2>p(n-1)p(n+1).
\end{equation*}
\end{thm}

They also proved the following theorem.

\begin{thm}[DeSalvo and Pak, 2015]
For  $n\geq 2$,
\begin{equation*} \label{thm-p-log-chen1}
\frac{p(n-1)}{p(n)}\left(1+\frac{1}{n}\right)>\frac{p(n)}{p(n+1)}.
\end{equation*}
\end{thm}

\footnotesize
\begin{thebibliography}{4}
{%\beamertemplatetextbibitems
\bibitem{DeSalvo-Pak-2015}S. DeSalvo and I. Pak, Log-concavity of the partition function, Ramanujan J. 38 (1) (2015) 61--73.
}
\end{thebibliography}

\end{frame}

\begin{frame}
\frametitle{Results on $p(n)$}
{\red DeSalvo and Pak} further proved that the term $(1+1/n)$ in the above theorem  can be improved  to $(1+O(n^{-3/2}))$.

\begin{thm}[DeSalvo and Pak, 2015]
For   $n>6$,
\begin{equation*} \label{thm-p-log-chen3}
\frac{p(n-1)}{p(n)}\left(1+\frac{240}{(24n)^{3/2}}\right)>\frac{p(n)}{p(n+1)}.
\end{equation*}
\end{thm}

{\red DeSalvo and Pak}  conjectured that the coefficient of $1/n^{3/2}$  can be improved to $\pi/\sqrt{24}$, which was proved by {\red Chen, Wang and Xie}.

\begin{thm}[Chen, Wang and Xie, 2016]
For   $n\geq 45$,
\begin{equation*}
\frac{p(n-1)}{p(n)}\left(1+\frac{\pi}{\sqrt{24}n^{3/2}}\right)>\frac{p(n)}{p(n+1)}.
\end{equation*}
\end{thm}

\footnotesize
\begin{thebibliography}{4}
{%\beamertemplatetextbibitems
\bibitem{Chen-Wang-Xie-2016} W.Y.C. Chen, L.X.W. Wang and G.Y.B. Xie, Finite differences of the logarithm of the partition function, Math. Comp. 85 (298) (2016) 825--847.
}
\end{thebibliography}
\end{frame}

\begin{frame}
\frametitle{Similar inequalities on $\spt(n)$}
We now present some conjectures on $\spt(n)$.
{\small\begin{conj}
\begin{itemize}
\item[(1)]For $n\geq 36$,
\begin{equation*} \label{conj-logcav-spt}
\spt(n)^2>\spt(n-1)\spt(n+1).
\end{equation*}
\item[(2)]For   $n\geq 13$,
\begin{equation*}
\frac{\spt(n-1)}{\spt(n)}\left(1+\frac{1}{n}\right)>\frac{\spt(n)}{\spt(n+1)}.
\end{equation*}
\item[(3)]For   $n\geq 73$,
\begin{equation*}
\frac{\spt(n-1)}{\spt(n)}\left(1+\frac{\pi}{\sqrt{24}n^{3/2}}\right)>\frac{\spt(n)}{\spt(n+1)}.
\end{equation*}
\item[(4)]For  $n>m>1$,
\begin{equation*} \label{conj-chen2}
\spt(n)^2> \spt(n-m)\spt(n+m).
\end{equation*}
\item[(5)]If $a,b$ are integers with $a,b>1$  and $(a,b)\neq (2,2)$ or $(3,3)$,  then
\begin{equation*}
\spt(a)\spt(b)> \spt(a+b).
\end{equation*}
\end{itemize}
\end{conj}}
\end{frame}

\begin{frame}
\frametitle{Confirmation of the conjecture}
Recently {\red Madeline Locus}  gives the following upper-bound and lower-bound on $\spt(n)$.
\begin{thm}[Madeline Locus]
For each $a\in\mathbb{Z}^+$ and $k\in\mathbb{Z}^+$, there is a $B_k(a)$ such that for all $n\geq B_k(a)$,
\[\frac{\sqrt{3}}{\pi\sqrt{24n-1}} \left(1-\frac{1}{an^k}\right) e^{\mu(n)}<\spt(n)<\frac{\sqrt{3}}{\pi\sqrt{24n-1}}
\left(1+\frac{1}{an^k}\right)e^{\mu(n)},\]
where $\mu(n)=\frac{\pi}{6}\sqrt{24n-1}$.
\end{thm}
With the aid of the above estimate, Locus announces a proof of all the above conjectures on spt-function.
\footnotesize
\begin{thebibliography}{4}
{%\beamertemplatetextbibitems
\bibitem{Chen-Wang-Xie-2016} M. Locus, Inequalities satisfied by the Andrews spt-function, preprint.
}
\end{thebibliography}
\end{frame}


\begin{frame}
\frametitle{Higher order Tur$\acute{\text{a}}$n inequality}
Recall that a sequence $\{a_n\}_{ n\geq 0}$ satisfies the higher order Tur\'{a}n inequality   if for $ n\geq 1$,
\begin{equation*}\label{high-turan}
4(a_n^2-a_{n-1}a_{n+1})(a_{n+1}^2-a_{n}a_{n+2})-(a_na_{n+1}-a_{n-1}a_{n+2})^2> 0.
\end{equation*}

Numerical evidence indicates that
both  $p(n)$ and  $\spt(n)$ satisfy  the higher order Tur\'an inequality.

 \begin{conj}\label{conj-dis-3}
For $n\geq 95$,  $p(n)$ satisfies the higher order Tur\'an
 inequality,  whereas   $\spt(n)$
satisfies  the higher order Tur\'an
 inequality for $n\geq 108$.
\end{conj}
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