INEQUALITIES AND UNIFORM ASYMPTOTIC FORMULAE FOR SPT-CRANK OF PARTITIONS

Abstract

We establish some inequalities that arise from truncating Lerch sums and derive uniform asymptotic formulae for the spt-crank of ordinary partitions. The uniform asymptotic formulae improve upon a result of Mao [‘Asymptotic formulas for spt-crank of partitions’, J. Math. Anal. Appl. 460(1) (2018), 121–139].

1. Introduction

A partition of a positive integer n is a sequence of nonincreasing positive integers whose sum equals n. Let $p(n)$ denote the number of partitions of n. Let $\mathrm {spt}(n)$ denote the total number of appearances of the smallest parts in all the partitions of n. This smallest part function, or spt-function, was introduced by Andrews [1]. In the same paper, Andrews proved

$$ \begin{align*}\mathrm{spt}(5n+4)\equiv 0~(\hspace{-1pt}\bmod\,5),\quad \mathrm{spt}(7n+5)\equiv 0~(\hspace{-1pt}\bmod\,7),\quad \mathrm{spt}(13n+6)\equiv 0~(\hspace{-1pt}\bmod\,{13}),\end{align*} $$

for all $n\in \mathbb N_0$ , analogous to the famous congruences for $p(n)$ which were found and later proved by Ramanujan [8].

To provide combinatorial explanations for these congruences for $\mathrm {spt}(n)$ , Andrews et al. [3] introduced the spt-crank of an S-partition. Let $\mathcal {P}$ denote the set of partitions and $\mathcal {D}$ denote the set of partitions into distinct parts. For a partition $\lambda \in \mathcal {P}$ , define $\#(\lambda )$ as the number of parts in $\lambda $ , $|\lambda |$ as the sum of the parts of $\lambda $ and $s(\lambda )$ as the smallest part in the partition $\lambda $ with $s(\varnothing )=+\infty $ for the empty partition $\varnothing $ . Define

$$ \begin{align*}\mathcal{S}=\{(\mu_1,\mu_2,\mu_3)\in \mathcal{D}\times\mathcal{P}\times\mathcal{P}: \mu_1\neq \varnothing,\; s(\mu_1)\le \min(s(\mu_2), s(\mu_3)) \}.\end{align*} $$

A triple $\mu =(\mu _1, \mu _2, \mu _3)$ of partitions in $\mathcal {S}$ is called an S-partition. Further, define a weight $w(\mu )=(-1)^{\#(\mu _1)-1}$ and the spt-crank by $\mathrm {scrk}(\mu )=\#(\mu _2)-\#(\mu _3)$ . Finally, define $|\mu |:=|\mu _1|+|\mu _2|+|\mu _3|$ , where $|\mu _j|$ is the sum of the parts of $\mu _j$ . If $|\mu |=n$ , we call $\mu $ an S-partition of n. Denote the set of all S-partitions of n by $\mathcal {S}(n)$ . Then $\mathrm { spt}(n)$ is equal to the net number of S-partitions of n counted according to the weight. The number of S-partitions of n with spt-crank m counted according to the weight w is denoted by $N_S(m,n)$ , so that

$$ \begin{align*}N_S(m,n)=\sum_{\mu \in \mathcal{S}(n),\;\mathrm{scrk}(\mu)=m}w(\mu).\end{align*} $$

Chen et al. [4] showed that $N_S(m, n)$ counts the number of doubly marked partitions of n with spt-crank m.

Andrews et al. [3, Theorem 2.1] established the generating function for the spt-crank:

$$ \begin{align*}S(z,q):=\sum_{n\ge 0}\sum_{m\in\mathbb Z}N_S(m,n)z^mq^n=\sum_{n\ge 1}\frac{(q^{n+1})_\infty q^n}{(zq^n)_\infty(z^{-1}q^n)_\infty},\end{align*} $$

where $(a)_{\infty }=\prod _{j\ge 0}(1-aq^j)$ for any $a\in {\mathbb C}$ and $|q|<1$ . Note that $S(1,q)$ is the generating function for $\mathrm {spt}(n)$ (see [3, Corollary 2.2]). Using [3, Corollary 2.5] gives the one variable generating function:

(1.1) $$ \begin{align} \sum_{n\ge 0}N_S(m,n)q^n=\frac{1}{(q)_\infty}\sum_{\ell\ge 1}(-1)^{\ell-1}\bigg(\frac{q^{{\ell(\ell+1)}/{2}+|m|\ell}}{1-q^\ell}-\frac{q^{{\ell(3\ell+1)}/{2}+|m|\ell}}{1-q^\ell}\bigg) \end{align} $$

for any $m\in \mathbb Z$ . This identity relates $N_S(m,n)$ to the difference of two half Lerch sums.

We study the inequalities that arise from truncating the Lerch sums in (1.1) and use them to investigate the uniform asymptotic behaviour of the spt-crank function $N_S(m,n)$ . For any integer $k\ge 1$ , we consider the truncation

(1.2) $$ \begin{align} \sum_{n\ge 0}\text{T}_S^k(m,n)q^n:=\frac{1}{(q)_\infty}\sum_{\ell\ge k}(-1)^{\ell-1}\frac{q^{{\ell(\ell+1)}/{2}+|m|\ell}(1-q^{\ell^2})}{1-q^\ell}. \end{align} $$

It is clear that $\text {T}_S^1(m,n)=N_S(m,n)$ ,

$$ \begin{align*} \text{T}_S^k(m,n)=N_S(m,n)+\sum_{1\le \ell< k}(-1)^{\ell}\sum_{0\le j<\ell}p(n-{\ell(\ell+1)}/{2}-(|m|+j)\ell) \end{align*} $$

for any integer $k\ge 2$ , and $\text {T}_S^k(m,n)=0$ for all $k>(\sqrt {8n+1}-1)/2$ . We can now state our main results.

Theorem 1.1. For any integers $m\in \mathbb Z, n\ge 0$ and $k\ge 1$ ,

$$ \begin{align*} (-1)^{k-1} \mathrm{T}_S^k(m,n)\ge 0. \end{align*} $$

Based on a Hardy–Ramanujan asymptotic result (see Lemma 3.1) for $p(n)$ , we establish the following uniform asymptotic formula for $N_S(m,n)$ .

Theorem 1.2. Set $\beta _n=\pi /\sqrt {6(n-1/24)}$ and assume that $m=o(n^{3/4})$ . As $n\to \infty $ ,

$$ \begin{align*} N_S(m,n)=\frac{\beta_n}{4}\mathrm{sech}^2\bigg(\frac{m\beta_n}{2}\bigg)\mathrm{spt}(n)(1+O(\beta_n+|m|^{2}\beta_n^{3})). \end{align*} $$

Theorem 1.2 improves the main result of Mao [7, Theorem 1.2], who established the following asymptotic formula for $N_S(m, n)$ using Wright’s circle method: if $\beta =\pi /\sqrt {6n}$ and $0<|m|\le n^{3/8}$ , then as $n\to \infty $ ,

$$ \begin{align*} N_S(m,n)=\frac{\beta}{4}\mathrm{sech}^2\bigg(\frac{m\beta}{2}\bigg)\mathrm{spt}(n)(1+O(|m|^{1/3}\beta^{1/2})). \end{align*} $$

In particular, Theorem 1.2 holds for a wider range of m and has a smaller error term than the result of Mao.

2. The proof of Theorem 1.1

We first establish the following lemma. This is an extended and rewritten version of Andrews et al. [2, Theorem 12], which only deals with the case when j is an odd integer.

Lemma 2.1. For all integers $m, j\ge 0$ , define

$$ \begin{align*}L_{j,m}(q)=\sum_{\ell\ge j}(-1)^{j+\ell}q^{Q(j+m;\ell)}+\sum_{\ell\ge j+1}(-1)^{j+\ell}q^{Q(j+1+m;\ell)},\end{align*} $$

where $Q(x;\ell )=x\ell +\binom {\ell }{2}$ . Then,

$$ \begin{align*} L_{j,m}(q)=&\sum_{\mu\in\{0,1\}}\sum_{\substack{\ell\ge j\\ \ell\equiv j~(\!\hspace{-1pt}\bmod\,2)}}q^{Q(m+j+\mu; \ell)}(1-q^{j+\mu(j+ m)})(1-q^{(m+j+\mu)+\ell}). \end{align*} $$

Proof. Note that $Q(m+j+1; j)=Q(m+j; j)+j$ . For any integer $j\ge 0$ ,

$$ \begin{align*} L_{j,m}(q) &=\sum_{\ell\ge j}(-1)^{\ell+j}q^{Q(m+j; \ell)}+\sum_{\ell\ge j}(-1)^{\ell+j}q^{Q(m+j+1; \ell)}-q^{Q(m+j+1; j)}\\ & =\sum_{\mu\in\{0,1\}}\sum_{\substack{\ell\ge j\\ \ell\equiv j~(\!\hspace{-1pt}\bmod\,2)}}(q^{Q(m+j+\mu; \ell)}-q^{Q(m+j+\mu; \ell+1)})-q^{Q(m+j; j)+j}\\ &=\sum_{\substack{\ell\ge j\\ \ell\equiv j~(\!\hspace{-1pt}\bmod\,2)}}q^{Q(m+j; \ell)}((1-q^{m+j+\ell})(1-q^j)+q^{\ell}(1-q^{m+j+\ell+1})(1-q^{2j+m}))\\ &\quad +\sum_{\substack{\ell\ge j\\ \ell\equiv j~(\!\hspace{-1pt}\bmod\,2)}}q^{Q(m+j; \ell)+j}-\sum_{\substack{\ell\ge j\\ \ell\equiv j~(\!\hspace{-1pt}\bmod\,2)}}q^{Q(m+j; \ell)+m+\ell+2j+m+j+\ell+1}-q^{Q(m+j,j)+j}\\ &=\sum_{\mu\in\{0,1\}}\sum_{\substack{\ell\ge j\\ \ell\equiv j~(\!\hspace{-1pt}\bmod\,2)}}q^{Q(m+j+\mu; \ell)}(1-q^{j+\mu(j+ m)})(1-q^{(m+j+\mu)+\ell})\\[4pt] &\quad +\sum_{\substack{\ell\ge j\\ \ell\equiv j~(\!\hspace{-1pt}\bmod\,2)}}q^{Q(m+j; \ell+2)+j}-\sum_{\substack{\ell\ge j\\ \ell\equiv j~(\!\hspace{-1pt}\bmod\,2)}}q^{Q(m+j; \ell)+m+\ell+2j+m+j+\ell+1}. \end{align*} $$

That is,

$$ \begin{align*} L_{j,m}(q)=\sum_{\mu\in\{0,1\}}\sum_{\substack{\ell\ge j\\ \ell\equiv j~(\!\hspace{-1pt}\bmod\,2)}}q^{Q(m+j+\mu; \ell)}(1-q^{j+\mu(j+ m)})(1-q^{(m+j+\mu)+\ell}), \end{align*} $$

which completes the proof.

Theorem 2.2. Let $Q(x; \ell )$ be defined in Lemma 2.1. For any integer $k\ge 1$ ,

$$ \begin{align*} \sum_{n\ge 0}(-1)^{k-1}\mathrm{T}_S^k(m,n)q^n & =\sum_{1\le j<k}\sum_{\substack{\ell\ge k\\ \ell\equiv k~(\!\hspace{-1pt}\bmod\,2)}}\frac{q^{Q(j+m;\ell)}(1-q^{j+m+\ell})}{(q)_\infty}\\[4pt] &\quad +\sum_{\mu\in\{0,1\}}\sum_{\substack{\ell\ge j\ge k\\ \ell\equiv j\equiv k~(\!\hspace{-1pt}\bmod\,2)}}\frac{q^{Q(m+j+\mu; \ell)}(1-q^{j+\mu(j+ m)})(1-q^{(m+j+\mu)+\ell})}{(q)_\infty}. \end{align*} $$

In particular,

$$ \begin{align*} \sum_{n\ge 0}N_S(m,n)q^n=\frac{1}{(q)_\infty}\sum_{\mu\in\{0,1\}}\sum_{\substack{\ell\ge j\ge 1\\ \ell\equiv j\equiv 1~(\!\hspace{-1pt}\bmod\,2)}}q^{Q(m+j+\mu; \ell)}(1-q^{j+\mu(j+ m)})(1-q^{(m+j+\mu)+\ell}). \end{align*} $$

Proof. From (1.2),

$$ \begin{align*} \sum_{n\ge 0}\mathrm{T}_S^k(m,n)q^n & =\frac{1}{(q)_\infty}\sum_{\ell\ge k}(-1)^{\ell-1}\frac{q^{{\ell(\ell+1)}/{2}+m\ell}(1-q^{\ell^2})}{1-q^\ell}\\[4pt] & =\frac{1}{(q)_\infty}\sum_{\ell\ge k}(-1)^{\ell-1}q^{{\ell(\ell+1)}/{2}+m\ell}\sum_{1\le j\le \ell}q^{(j-1)\ell}\\[4pt] & =\frac{1}{(q)_\infty}\sum_{j\ge 1}\sum_{\ell\ge \max(j, k)}(-1)^{\ell-1}q^{\binom{\ell}{2}+(m+j)\ell}. \end{align*} $$

Thus,

$$ \begin{align*} \sum_{n\ge 0}(-1)^{k-1}\mathrm{T}_S^k(m,n)q^n &=\frac{1}{(q)_\infty}\sum_{1\le j<k}\sum_{\ell\ge k}(-1)^{\ell-k}q^{\binom{\ell}{2}+(m+j)\ell}\\ &\quad +\frac{1}{(q)_\infty}\sum_{j\ge k}\sum_{\ell\ge j}(-1)^{\ell-k}q^{\binom{\ell}{2}+(m+j)\ell}\\ & =\sum_{1\le j<k}\sum_{\substack{\ell\ge k\\ \ell\equiv k~(\!\hspace{-1pt}\bmod\,2)}}\frac{q^{Q(j+m;\ell)}(1-q^{j+m+\ell})}{(q)_\infty}+\sum_{\substack{j\ge k\\ j\equiv k~(\!\hspace{-1pt}\bmod\,2)}}\frac{L_{j,m}(q)}{(q)_\infty}, \end{align*} $$

by using Lemma 2.1, which completes the proof.

Proof of Theorem 1.1

Since $(m+j+\mu )+\ell \neq j+\mu (j+ m)$ for $\mu \in \{0,1\}$ , $m\ge 0$ with $\ell \ge j\ge 1$ , the factors $(1-q^{j+m+\ell })$ , $(1-q^{j+\mu (j+ m)})$ and $(1-q^{(m+j+\mu )+\ell })$ in Theorem 2.2 merely cancel the corresponding factors in $1/(q)_\infty $ . It follows immediately from Theorem 2.2 that $(-1)^{k-1}\mathrm {T}_S^k(m,n)$ is nonnegative.

3. Lemmas for Theorem 1.2

For $r_0\in \mathbb N_0$ and $r_1, r_2\in \mathbb N$ , define

(3.1) $$ \begin{align} \sum_{n\ge 0}T_n(r_0,r_1,r_2)q^n=\frac{q^{r_0}}{(q)_\infty}(1-q^{r_1})(1-q^{r_2}). \end{align} $$

Then by Theorem 2.2,

(3.2) $$ \begin{align} N_S(m,n)=&\sum_{\mu\in\{0,1\}}\sum_{\substack{\ell\ge j\ge 1\\ \ell\equiv j\equiv 1~(\!\hspace{-1pt}\bmod\,2)}}T_{m,n; j, \ell}^\mu, \end{align} $$

where

$$ \begin{align*}T_{m,n; j,\ell}^\mu=T_n(Q(m+j+\mu;\ell), (1+\mu)j+\mu m, m+j+\mu+\ell).\end{align*} $$

Since there is no cancellation in the summation in (3.2), to establish the asymptotics for $N_S(m,n)$ , it is sufficient to establish an asymptotic formula for $T_n(r_0,r_1,r_2)$ . We need the following Hardy–Ramanujan asymptotic result for $p(n)$ , which is an immediate consequence of [5, (1.61)].

Lemma 3.1. For $n\in \mathbb N$ and with $w=n-1/24\rightarrow \infty $ ,

$$ \begin{align*} p(n)=\frac{1}{\sqrt{3}B^2}\partial_{w}^2 e^{B\sqrt{w}}+O(e^{({1}/{2})B\sqrt{w}}), \end{align*} $$

where $B=2\pi /\sqrt {6}$ and $\partial _w^k:={d^k}/{d w^k}$ for any $k\in \mathbb N$ .

Next, using (3.1) and Lemma 3.1, we prove the following proposition.

Proposition 3.2. Define $T_n(r_0,r_1,r_2)$ as in (3.1). Then, for $0\le r_0, r_1, r_2=O(n^{3/4})$ with $r_0\gg r_1\gg r_2$ ,

$$ \begin{align*}\frac{1}{p(n)}T_n(r_0,r_1,r_2)-\widehat{T}_n(r_0,r_1,r_2)\ll \beta_ne^{-\beta_n r_0}(1+(\beta_nr_0)^2)\min(1, (\beta_nr_1)^2) \quad\mbox{as}\ n\to\infty,\end{align*} $$

where $\beta _n=\pi /\sqrt {6(n-1/24)}$ and $\widehat {T}_n(r_0,r_1,r_2)=e^{-\beta _nr_0}(1-e^{-\beta _nr_1})(1-e^{-\beta _nr_2}).$

Proof. From (3.1) and Lemma 3.1,

(3.3) $$ \begin{align} T_n(r_0,r_1,r_2) & =p(n-r_0)-p(n-r_0-r_1)-p(n-r_0-r_2)+p(n-r_0-r_1-r_2)\nonumber\\ & =\frac{1}{\sqrt{3}B^2}S_n(r_0,r_1,r_2)+O(e^{({1}/{2})B\sqrt{w}}), \end{align} $$

where

$$ \begin{align*} S_n(r_0,r_1,r_2) :\!\!&=\partial_{w}^2 (e^{B\sqrt{w-r_0}}-e^{B\sqrt{w-r_0-r_1}}-e^{B\sqrt{w-r_0-r_2}}+e^{B\sqrt{w-r_0-r_1-r_2}})\\ & =\partial_{w}^2 \int_{0}^{r_1}\partial_w (e^{B\sqrt{w-r_0-t}}-e^{B\sqrt{w-r_0-r_2-t}}) \,d t\\ & =\partial_{w}^2 \int_{0}^{r_1}\partial_w \int_{0}^{r_2} \partial_w e^{B\sqrt{w-r_0-t-u}}\,d u\, d t\\ & =r_1r_2\int_{0}^{1}\int_{0}^{1}(\partial_w^4 e^{B\sqrt{w-r_0-r_1t-r_2u}})\,d u\, d t. \end{align*} $$

Note that as $w\to +\infty $ ,

$$ \begin{align*}\partial_w^4 e^{B\sqrt{w}}=\frac{B^4}{16 w^2}e^{B\sqrt{w}}(1+O(w^{-1/2}))\end{align*} $$

and

$$ \begin{align*}e^{B\sqrt{w-r}}=e^{B\sqrt{w}-{Br}/{2\sqrt{w}}}(1+O(r^2/w^{3/2}))\end{align*} $$

hold for $r=O(w^{3/4})$ . For $0\le r_0, r_1,r_2=O(n^{3/4})$ with $r_0\gg r_1\gg r_2$ ,

$$ \begin{align*} S_n(r_0,r_1,r_2) & =r_1r_2\int_{0}^{1}\int_{0}^{1}\frac{B^4e^{B\sqrt{w-r_0-r_1t-r_2u}}}{16 (w-r_0-r_1t-r_2u)^2}(1+O(w^{-1/2}))\,d u\, d t\\ & = r_1r_2\int_{0}^{1}\int_{0}^{1}\frac{B^4e^{B\sqrt{w}-B{(r_0+r_1t+r_2u)}/{2\sqrt{w}}}}{16 w^2}\bigg(1+O\bigg(w^{-1/2}+\frac{r_0^2}{w^{3/2}}+\frac{r_0}{w}\bigg)\bigg)\,d u\, d t\\ & = \frac{B^2e^{B\sqrt{w}}}{4 w}e^{-{Br_0}/{2\sqrt{w}}}(1-e^{-{Br_1}/{2\sqrt{w}}})(1-e^{-{Br_2}/{2\sqrt{w}}})+E_n(r_0,r_1,r_2), \end{align*} $$

where

$$ \begin{align*} E_n(r_0,r_1,r_2)&\ll \frac{e^{B\sqrt{w}}}{ w^2}e^{-{Br_0}/{2\sqrt{w}}}\bigg(w^{-1/2}+\frac{r_0^2}{w^{3/2}}+\frac{r_0}{w}\bigg)r_1r_2\int_{0}^{1}\int_{0}^{1}e^{-B{(r_1t+r_2u)}/{2\sqrt{w}}}\,d u\, d t\\ &\ll p(n)w^{-1/2}e^{-{Br_0}/{2\sqrt{w}}}(1+w^{-1}r_0^2)\min(1, w^{-1}r_1^2). \end{align*} $$

In the last line, we used

$$ \begin{align*}p(n)=\frac{1}{4\sqrt{3}n}e^{2\pi\sqrt{n/6}}(1+O(n^{-1/2})).\end{align*} $$

Combining the above estimates and (3.3) and using $\beta _n=\pi /\sqrt {6(n-1/24)}$ completes the proof.

We use the following version of the Euler–Maclaurin summation formula.

Lemma 3.3. Let $a,b\in \mathbb Z$ with $a\le b$ and let $f\in \mathcal {C}^1([a,b])$ . Then

$$ \begin{align*}\sum_{a\le\ell\le b}f(\ell \varepsilon)=\frac{1}{\varepsilon}\int_{a\varepsilon}^{b\varepsilon}f(u)\,du+\frac{f(a\varepsilon)+f(b\varepsilon)}{2}+O\bigg(\int_{a\varepsilon}^{b\varepsilon}|f'(u)|\,du\bigg) \end{align*} $$

for any $\varepsilon \in (0,1)$ , where the implied constant is absolute.

We also need the following lemmas.

Lemma 3.4. Let $m\ge 0$ and define

$$ \begin{align*}F_m(q)=\sum_{\ell\ge 1}(-1)^{\ell-1} q^{{\ell(\ell+1)}/{2}+m\ell}\frac{1-q^{\ell^2}}{1-q^\ell}.\end{align*} $$

Then, as $t\to 0^+$ ,

$$ \begin{align*} F_m(e^{-t})=\frac{1}{4}\mathrm{sech}^2\bigg(\frac{(m+1/2)t}{2}\bigg)(1+O(t)). \end{align*} $$

Proof. From Liu and Zhou [6, Theorem 2.7], with $\ell =0$ and $p=2$ ,

$$ \begin{align*}\sum_{n\ge 1}(-1)^{n-1}e^{-n^2x-n\alpha}=\sum_{0\le k<2}\frac{(-x)^{k}}{k !} \partial_{\alpha}^{2k}\bigg(\frac{1}{1+e^{\alpha}}\bigg)+O(x^2e^{-\alpha}),\end{align*} $$

as $x\rightarrow 0^+$ , uniformly for all $\alpha \ge 0$ . Note that

$$ \begin{align*} F_m(q)-F_{m+1}(q)=\sum_{\ell\ge 1}(-1)^{\ell-1}q^{{\ell(\ell+1)}/{2}+m\ell}(1-q^{\ell^2}). \end{align*} $$

Therefore, for any $h\ge 0$ , as $t\rightarrow 0^+$ ,

$$ \begin{align*} F_h(e^{-t})-F_{h+1}(e^{-t})&=\sum_{\ell\ge 1}(-1)^{\ell-1}(e^{-t({\ell^2}/{2}+(h+1/2)\ell)}-e^{-t({3\ell^2}/{2}+(h+1/2)\ell)})\\ &=\sum_{0\le k<2}\frac{(-t/2)^{k}-(-3t/2)^k}{k !} \partial_{\alpha}^{2k}\big|_{\alpha=(h+1/2)t}\bigg(\frac{1}{1+e^{\alpha}}\bigg)+O(t^2e^{-(h+1/2)t})\\ &=t\partial_{\alpha}^{2}\big|_{\alpha=(h+1/2)t}\bigg(\frac{1}{1+e^{\alpha}}\bigg)+O(t^2e^{-(h+1/2)t}). \end{align*} $$

Hence, for any $m\ge 0$ , the Euler–Maclaurin summation formula implies

$$ \begin{align*} t^{-1}F_m(e^{-t})&=\sum_{h\ge m}\partial_{\alpha}^{2}\big|_{\alpha=(h+1/2)t}\bigg(\frac{1}{1+e^{\alpha}}\bigg)+O\bigg(t\sum_{h\ge m}e^{-(h+1/2)t}\bigg)\\ & =\int_{m}^{\infty}\partial_{\alpha}^{2}\big|_{\alpha=(x+1/2)t}\bigg(\frac{1}{1+e^{\alpha}}\bigg)\,d x\\ &\quad +O\bigg(\int_{m}^{\infty}\bigg|\partial_x\partial_{\alpha}^{2}\big|_{\alpha=(x+1/2)t}\bigg(\frac{1}{1+e^{\alpha}}\bigg)\bigg|\,d x+e^{-(m+1/2)t}\bigg)\\ & =-t^{-1}\partial_{\alpha}\big|_{\alpha=(m+1/2)t}\bigg(\frac{1}{1+e^{\alpha}}\bigg)+O(e^{-(m+1/2)t}), \end{align*} $$

which completes the proof.

Finally, we need the following uniform upper bound estimate for partial theta functions.

Lemma 3.5. Uniformly for all $\alpha \ge 0$ and any fixed $k\ge 0$ , as $t\to 0^+$ ,

$$ \begin{align*}\sum_{\ell\ge 1}e^{-t(\ell^2+\alpha\ell)}\ell^k\ll t^{-(k+1)/2}\min(1, (\alpha\sqrt{t})^{-k-3}).\end{align*} $$

Proof. Using integration by parts for a Riemann–Stieltjes integral,

$$ \begin{align*} \sum_{\ell\ge 1}e^{-t(\ell^2+\alpha\ell)}\ell^k&=\int_{0}^\infty e^{-t(x^2+\alpha x) }\,d \sum_{1\le \ell\le x} \ell^k\\ &=\int_{0}^{\infty}\bigg(\sum_{1\le \ell\le x} \ell^k\bigg)(2x+\alpha)t e^{-t(x^2+\alpha x) }\, d x. \end{align*} $$

Using the fact that $\sum _{1\le \ell \le x}\ell ^k\ll x^{k+1}$ for any fixed $k\ge 0$ yields

$$ \begin{align*} &\sum_{\ell\ge 1}e^{-t(\ell^2+\alpha\ell)} \ell^k \ll t\int_{0}^\infty x^{k+2}e^{-t(x^2+\alpha x) }\,d x+\alpha t\int_{0}^{\infty} x^{k+1} e^{-t(x^2+\alpha x)}\,d x\\ &\quad = t^{-(k+1)/2}\int_{0}^\infty x^{k+2}e^{-x^2-(\alpha\sqrt{t}) x }\,d x+(\alpha \sqrt{t})t^{-(k+1)/2}\int_{0}^{\infty} x^{k+1} e^{-x^2-(\alpha\sqrt{t}) x}\,d x. \end{align*} $$

However, for any fixed $v\ge 0$ and any $\lambda \ge 0$ ,

$$ \begin{align*}\int_{0}^\infty x^{v}e^{-x^2-\lambda x }\,d x\ll \mathbf{1}_{\lambda\le 1}\int_{0}^\infty x^{v}e^{-x^2}\,d x+\mathbf{1}_{\lambda\ge 1} \int_{0}^\infty x^{v}e^{-\lambda x }\,d x\ll \min(1, \lambda^{-v-1}).\end{align*} $$

Here $\mathbf {1}_{A}$ denotes the indicator function of A. Thus,

$$ \begin{align*} t^{(k+1)/2}\sum_{\ell\ge 1}e^{-t(\ell^2+\alpha\ell)}\ell^k\ll \min(1, (\alpha\sqrt{t})^{-k-3})+(\alpha \sqrt{t})\min(1, (\alpha\sqrt{t})^{-k-2}), \end{align*} $$

which completes the proof.

4. The proof of Theorem 1.2

In this section, we give the proof of Theorem 1.2. We begin with two lemmas.

Lemma 4.1. For any $m\ge 0$ with $m=o(n^{3/4})$ , as $n\to \infty $ ,

$$ \begin{align*}N_S(m,n)=p(n)(F_m(e^{-\beta_n})+E_S(m,n)),\end{align*} $$

where

$$ \begin{align*}E_S(m,n)\ll e^{-n^{1/4}}+\sum_{\ell\ge 1}\beta_n\ell e^{-\beta_n(\ell^2+m\ell)}(1+\beta_n^2(m+\ell)^2\ell^2)\min(1,\beta_n^2(\ell+m)^2).\end{align*} $$

Proof. From (3.2) and Proposition 3.2,

$$ \begin{align*} \frac{1}{p(n)}T_{m,n; j, \ell}^\mu & = \widehat{T}_n(Q(m+j+\mu;\ell), (1+\mu)j+\mu m, m+j+\mu+\ell)\\ &\quad +O(\beta_ne^{-\beta_nQ(m+j+\mu;\ell)}(1+\beta_n^2Q(m+j+\mu;\ell)^2)\min(1, \beta_n^2(m+j+\ell)^2)) \end{align*} $$

for all $\ell \ge j\ge 1$ with $Q(m+j+\mu ;\ell )=O(n^{{3}/{4}})$ . Moreover,

$$ \begin{align*}\sum_{\mu\in\{0,1\}}\sum_{\substack{\ell\ge j\ge 1\\ \ell\equiv j\equiv 1~(\!\hspace{-1pt}\bmod\,2)\\ Q(m+j+\mu;\ell)\ge n^{3/4}}}T_{m,n; j, \ell}^\mu \ll n^2 p(n-\lfloor n^{3/4}\rfloor)\ll p(n)e^{-n^{1/4}},\end{align*} $$

since the inner summation has no more than $n^2$ terms not equal to $0$ , each term is less than $p(n-\lfloor n^{3/4}\rfloor )$ , and $B/2=\pi /\sqrt {6}>1$ . Setting

$$ \begin{align*}\widehat{T}_{m,n; j,\ell}^\mu = \widehat{T}_n(Q(m+j+\mu;\ell), (1+\mu)j+\mu m, m+j+\mu+\ell),\end{align*} $$

we see that for all $\ell \ge j\ge 1$ with $Q(m+j+\mu ;\ell )> n^{3/4}$ ,

$$ \begin{align*}\widehat{T}_{m,n; j,\ell}^\mu\ll e^{-\beta_nQ(m+j+\mu;\ell)}.\end{align*} $$

Therefore, noting that

$$ \begin{align*}\ell^2/2+m\ell \le Q(m+j;\ell)\le Q(m+j+\mu;\ell)\ll \ell^2+m\ell,\end{align*} $$

we have

$$ \begin{align*} \frac{N_S(m,n)}{p(n)} & = \sum_{\mu\in\{0,1\}}\sum_{\substack{\ell\ge j\ge 1\\ \ell\equiv j\equiv 1~(\!\hspace{-1pt}\bmod\,2)}}\widehat{T}_{m,n; j,\ell}^\mu+O\Bigg(e^{-n^{1/4}}+\sum_{\substack{\ell\ge j\ge 1\\ \ell\equiv j\equiv 1~(\!\hspace{-1pt}\bmod\,2)\\ Q(m+j+1;\ell)> n^{3/4}}}e^{-\beta_nQ(m+j;\ell)}\Bigg)\\ &\quad +O\Bigg(\sum_{\substack{\ell\ge j\ge 1\\ \ell\equiv j\equiv 1~(\!\hspace{-1pt}\bmod\,2)}}\beta_n e^{-\beta_n(\ell^2/2+m\ell)}(1+\beta_n^2(m+\ell)^2\ell^2)\min(1,\beta_n^2(\ell+m)^2)\Bigg). \end{align*} $$

Further simplification yields

$$ \begin{align*} p(n)^{-1}N_S(m,n)=\sum_{\mu\in\{0,1\}}\sum_{\substack{\ell\ge j\ge 1\\ \ell\equiv j\equiv 1~(\!\hspace{-1pt}\bmod\,2)}}\widehat{T}_{m,n; j,\ell}^\mu+E_S(m,n), \end{align*} $$

where

(4.1) $$ \begin{align} E_S(m,n)&\ll e^{-n^{1/4}}+\sum_{\ell\ge 1}\beta_n\ell e^{-\beta_n(\ell^2+m\ell)}(1+\beta_n^2(m+\ell)^2\ell^2)\min(1,\beta_n^2(\ell+m)^2). \end{align} $$

Using the definitions of $\widehat {T}_{m,n; j,\ell }^\mu $ in Proposition 3.2 and of $F_m(q)$ in Lemma 3.4, we complete the proof.

Lemma 4.2. Let $E_S(m,n)$ be defined as in Lemma 4.1. For $0\le m=o(n^{3/4})$ ,

$$ \begin{align*}E_S(m,n)\ll (\beta_n+m^2\beta_n^3)e^{-\beta_n m}.\end{align*} $$

Proof. From (4.1), for $1/\beta _n\le m=o(n^{3/4})$ ,

$$ \begin{align*} E_S(m,n)&\ll e^{-n^{1/4}}+\sum_{\ell\ge 1}\ell \beta_n e^{-\beta_n m\ell}\beta_n^2m^2\ell^2\ll e^{-n^{1/4}}+\beta_n^3m^2e^{-\beta_n m}\ll \beta_n^3m^2e^{-\beta_n m}, \end{align*} $$

by noting that $\beta _n m=o(n^{1/4})$ . For $0\le m\le 1/\beta _n$ ,

$$ \begin{align*} E_S(m,n)+O(e^{-n^{1/4}}) & \ll \beta_n^3\sum_{\ell\ge 1}e^{-\beta_n(\ell^2+m\ell)}(1+\beta_n^2(m+\ell)^2\ell^2)\ell (\ell+m)^2\\ & \ll \beta_n^3\sum_{\ell\ge 1}e^{-\beta_n(\ell^2+m\ell)}(\ell^3+\ell m^2+\beta_n^2(m^4+\ell^4)\ell^3)\\ & \ll \beta_n^{3-(3+1)/2}\min(1, (m\beta_n^{1/2})^{-6})+m^2\beta_n^{3-(1+1)/2}\min(1, (m\beta_n^{1/2})^{-4})\\ &\quad +m^4\beta_n^{5-(3+1)/2}\min(1, (m\beta_n^{1/2})^{-6})+\beta_n^{5-(7+1)/2}\min(1, (m\beta_n^{1/2})^{-10}). \end{align*} $$

In the last step, we used Lemma 3.5. Further simplification yields

$$ \begin{align*}E_S(m,n)\ll e^{-n^{1/4}}+\beta_n+(m^2\beta_n^2+m^4\beta_n^{3})\mathbf{1}_{m\le \beta_n^{-1/2}}+m^{-2}\mathbf{1}_{m\ge \beta_n^{-1/2}}\ll \beta_n e^{-\beta_n m},\end{align*} $$

where $\mathbf {1}_{A}$ is the indicator function of A. This completes the proof.

Proof of Theorem 1.2

From Lemmas 3.4, 4.1 and 4.2,

(4.2) $$ \begin{align} N_S(m,n)&=\frac{p(n)}{4}\mathrm{sech}^2\bigg(\frac{(m+1/2)\beta_n}{2}\bigg)(1+O(\beta_n))+O(p(n)(\beta_n+m^2\beta_n^3)e^{-\beta_n m})\nonumber\\ &=\frac{p(n)}{4}\mathrm{sech}^2\bigg(\frac{m\beta_n}{2}\bigg)(1+O(\beta_n+m^2\beta_n^3)) \end{align} $$

for any $0\le m=o(n^{3/4})$ and as $n\to \infty $ . Theorem 1.2 immediately follows from the fact that $N_S(-m,n)=N_S(m,n)$ and the following Lemma 4.3.

Lemma 4.3. We have $\mathrm {spt}(n)=\beta _n^{-1}p(n)(1+O(\beta _n))$ .

Proof. From Theorem 1.1 with $k=2$ ,

$$ \begin{align*}N_S(m,n)\le p(n-|m|-1)\le p(n-|m|).\end{align*} $$

The asymptotic formula for $p(n)$ yields

$$ \begin{align*}N_S(m,n)\ll (n-|m|)^{-1}e^{B\sqrt{n-|m|}}\ll p(n)e^{-|m|\beta_n}\end{align*} $$

for all $n\ge |m|+1$ . In particular, one has

$$ \begin{align*}N_S(m,n)\ll p(n)e^{-|m|\beta_n}\end{align*} $$

for all $|m|\ge \beta _n^{-5/4}$ . Now, using (4.2) and the fact that $N_S(-m,n)=N_S(m,n)$ ,

$$ \begin{align*} \sum_{m\in \mathbb Z}\frac{N_S(m,n)}{p(n)}&=\sum_{|m|\le \beta_n^{-5/4}}\frac{N_S(m,n)}{p(n)}+\sum_{|m|> \beta_n^{-5/4}}\frac{N_S(m,n)}{p(n)}\\ & =\sum_{|m|\le \beta_n^{-5/4}}\frac{1}{4}\mathrm{sech}^2\bigg(\frac{m\beta_n}{2}\bigg)(1+O(\beta_n+m^2\beta_n^3))+\sum_{|m|>\beta_n^{-5/4}}O(e^{-|m|\beta_n})\\ & =\sum_{|m|\le \beta_n^{-5/4}}\frac{1}{4}\mathrm{sech}^2\bigg(\frac{m\beta_n}{2}\bigg)+R_S(n), \end{align*} $$

where

$$ \begin{align*} R_S(n)&\ll \beta_n\sum_{|m|\le \beta_n^{-5/4}}\mathrm{sech}^2\bigg(\frac{m\beta_n}{2}\bigg)\bigg(1+\frac{(m\beta_n)^2}{4}\bigg)+\sum_{|m|\ge \beta_n^{-5/4}}e^{-|m|\beta_n}\\ &\ll \int_{\mathbb R}((1+u^2)\mathrm{sech}^2(u)+\beta_n|\partial_u((1+u^2)\mathrm{sech}^2(u))|)\,d u+1\ll 1, \end{align*} $$

by using the Euler–Maclaurin summation formula. Using the Euler–Maclaurin summation formula again yields

$$ \begin{align*} \sum_{|m|\le \beta_n^{-5/4}}\mathrm{sech}^2\bigg(\frac{m\beta_n}{2}\bigg)&=\frac{1}{\beta_n}\int_{|u|\le \beta_n^{-1/4}}\mathrm{sech}^2(u/2)\,d u+O\bigg(1+\int_{\mathbb R}|\partial_u\mathrm{sech}^2(u/2)|\,d u\bigg)\\ &=\frac{1}{\beta_n}\int_{\mathbb R}\mathrm{sech}^2(u/2)\,d u+O(1)=\frac{4}{\beta_n}+O(1), \end{align*} $$

which completes the proof of the lemma.