On the Largest Prime Factor of Integers in Short Intervals

1. Introduction

The Legendre’s conjecture, which states that there is always a prime number between consecutive squares, is one of Landau’s problems on prime numbers. Clearly this means that there is always a prime number in the interval $[x,x+x^{{\displaystyle \frac{1}{2}}}]$. However, we cannot prove it even on the Riemann Hypothesis. Assuming RH, one can only show that there is always a prime number in the interval $[x,x+x^{{\displaystyle \frac{1}{2}}}logx]$. The best unconditional result is due to Li [1], where he showed the interval $[x,x+x^{0.52}]$ contains primes.

Instead of relaxing the length of the short interval, one can attack this conjecture by relaxing our restriction of primes. A number with a large prime factor is a good approximation of prime numbers. Thus, we can try to find numbers with a large prime factor in three intervals $[x,x+x^{{\displaystyle \frac{1}{2}}}]$, $[x,x+x^{{\displaystyle \frac{1}{2}}}{(logx)}^A]$ and $[x,x+x^{{\displaystyle \frac{1}{2}}+\epsilon }]$.

For the first interval, Ramachandra [2] showed in 1969 that this interval contains a number with a prime factor larger than $x^{0.576}$. The exponent 0.576 has been improved to

$$0.625,0.662,0.675225,0.692,0.7,0.71,0.723,0.728,0.732,0.738,0.74\mathrm{and}0.7428$$

by Ramachandra [3], Graham [4], Zhu [5], Jia [6], Baker [7], Jia [8], Jia [9] (and Liu [10]), Jia [11], Baker and Harman [12], Liu and Wu [13], Harman [14] Chapter 6, and Baker and Harman [15] respectively. For the second interval, Balog, Harman and Pintz [16] showed that this interval contains a number with a prime factor larger than $x^{0.712}$, and the exponent 0.712 has been improved to ${\displaystyle \frac{5}{6}}$ by Lou [17] and ${\displaystyle \frac{18}{19}}$ by Merikoski [18].

In this paper we shall focus on the third interval. In 1973, Jutila [19] showed that this interval contains a number with a prime factor larger than $x^{{\displaystyle \frac{2}{3}}-\epsilon }$. The exponent ${\displaystyle \frac{2}{3}}$ has been improved to

$$0.73,0.7338,0.772,0.82,{\displaystyle \frac{11}{12}},{\displaystyle \frac{17}{18}},{\displaystyle \frac{19}{20}},{\displaystyle \frac{24}{25}}\mathrm{and}{\displaystyle \frac{25}{26}}$$

by Balog [20,21], Balog, Harman and Pintz [22], Heath-Brown [23], Heath-Brown and Jia [24], Harman [14] Chapter 5, Haugland [25] and Jia and Liu [26] respectively. In his monograph, Harman [14] Exercise 5.1, encouraged us to reduce this exponent as much as we can. In this paper, we obtain the following result.

Theorem 1.1.

For sufficiently large x, the interval $[x,x+x^{{\displaystyle \frac{1}{2}}+\epsilon }]$ contains an integer with a prime factor larger than $x^{{\displaystyle \frac{51}{53}}-\epsilon }$.

Of course, our proof is much simpler than the similar arguments used in [24,25,26]. Throughout this paper, we always suppose that $\epsilon$ is a sufficiently small positive constant and $B=B\left(\epsilon \right)$ is a sufficiently large positive constant. We choose $\epsilon$ such that $K={\displaystyle \frac{8}{\epsilon }}({\displaystyle \frac{1}{26.5}}+{\displaystyle \frac{\epsilon }{2}})$ is an integer. The letter p, with or without subscript, is reserved for prime numbers. Let $v=x^{{\displaystyle \frac{51}{53}}-{\displaystyle \frac{\epsilon }{2}}}$, $P=x^{{\displaystyle \frac{\epsilon }{8}}}$ and $T_0=x^{{\displaystyle \frac{1}{2}}-{\displaystyle \frac{\epsilon }{6}}}$. Let $c_0$, $c_1$ and $c_2$ denote positive constants which may have different values at different places, and we write $m\sim M$ to mean that $c_{1}M<m⩽c_{2}M$. We use $M\left(s\right)$, $N\left(s\right)$ and some other capital letters to denote the Dirichlet polynomials

$$M\left(s\right)=\sum _{m\sim M}a\left(m\right)m^{-s},N\left(s\right)=\sum _{n\sim N}b\left(n\right)n^{-s}$$

where $a\left(m\right)$, $b\left(n\right)$ are complex numbers with $a\left(m\right)=O\left(1\right)$ and $b\left(n\right)=O\left(1\right)$. We also use $P\left(s\right)$ to denote

$$P\left(s\right)=\sum _{P<p⩽2P}{p}^{-s}.$$

2. Arithmetic Information

In this section we provide some arithmetic information (i.e. mean value bounds for some Dirichlet polynomials) which will help us prove the asymptotic formulas for sieve functions.

Lemma 2.1.

Suppose that $MN=v$ where $M\left(s\right),N\left(s\right)$ are Dirichlet polynomials and $v^{{\displaystyle \frac{49}{102}}}\ll M\ll v^{{\displaystyle \frac{53}{102}}}$. Let $b=1+{\displaystyle \frac{1}{logx}}$, $T_1={(logx)}^{2B}$, then for $T_1⩽T⩽T_0$ we have

$${\int }_T^{2T}\left|M(b+it)N(b+it)P^K(b+it)\right|dt\ll {(logx)}^{-B}.$$

Proof. 

The proof is similar to that of [26], Lemma 1. □

Lemma 2.2.

Suppose that $MNL=v$ where $M\left(s\right),N\left(s\right)$ are Dirichlet polynomials and $L\left(s\right)={\sum }_{l\sim L}{l}^{-s}$. Let $b=1+{\displaystyle \frac{1}{logx}}$, $T_2=\sqrt{L}$. Assume that $M\ll v^{{\displaystyle \frac{53}{102}}}$ and $N\ll v^{{\displaystyle \frac{53}{204}}}$, then for $T_2⩽T⩽T_0$ we have

$${\int }_T^{2T}\left|M(b+it)N(b+it)L(b+it)P^K(b+it)\right|dt\ll {(logx)}^{-B}.$$

Proof. 

The proof is similar to that of [26], Lemma 2. □

Lemma 2.3.

Suppose that $MNHL=v$ where $M\left(s\right),N\left(s\right),H\left(s\right)$ are Dirichlet polynomials and $L\left(s\right)={\sum }_{l\sim L}{l}^{-s}$. Let $b=1+{\displaystyle \frac{1}{logx}}$, $T_2=\sqrt{L}$. Assume that M, N and H satisfy the following conditions:

$M\ll v^{{\displaystyle \frac{53}{102}}},N\gg H,N^{{\displaystyle \frac{3}{4}}}H\ll v^{{\displaystyle \frac{53}{204}}},N{H}^{{\displaystyle \frac{1}{2}}}\ll v^{{\displaystyle \frac{53}{204}}},N^{{\displaystyle \frac{7}{4}}}{H}^{{\displaystyle \frac{3}{2}}}\ll v^{{\displaystyle \frac{53}{102}}}$,

Then for $T_2⩽T⩽T_0$ we have

$${\int }_T^{2T}\left|M(b+it)N(b+it)H(b+it)L(b+it)P^K(b+it)\right|dt\ll {(logx)}^{-B}.$$

Proof. 

The proof is similar to that of [26] Lemma 3, where [27] Theorem 2, is used. □

3. The Final Decomposition

Now we follow the discussion in [24,26]. Let $p_j=v^{t_j}$ and put

$$N\left(d\right)=\sum _{\begin{array}{c}x<p{p}_1\dots p_K⩽x+x^{{\displaystyle \frac{1}{2}}}\\ P<p_i⩽2P\end{array}}1,\mathcal{A}=\{n:2^{-K}v<n⩽2v,n\mathrm{repeats}N\left(n\right)\mathrm{times}\},$$

$$\mathcal{B}=\{n:v<n⩽2v\},{\mathcal{A}}_d=\{a:a\in \mathcal{A},d\mid a\},P\left(z\right)=\prod _{p<z}p,S(\mathcal{A},z)=\sum _{\begin{array}{c}a\in \mathcal{A}\\ (a,P(z\left)\right)=1\end{array}}1.$$

Then we only need to show that $S\left(\mathcal{A},{\left(2v\right)}^{{\displaystyle \frac{1}{2}}}\right)>0$. Our aim is to show that the sparser set $\mathcal{A}$ contains the expected proportion of primes compared to the bigger set $\mathcal{B}$, which requires us to decompose $S\left(\mathcal{A},{\left(2v\right)}^{{\displaystyle \frac{1}{2}}}\right)$ and prove asymptotic formulas of the form

$$S\left(\mathcal{A},z\right)=v^{-1}{x}^{{\displaystyle \frac{1}{2}}+\epsilon }{\left(\sum _{P<p⩽2P}{\displaystyle \frac{1}{p}}\right)}^K(1+o\left(1\right))S\left(\mathcal{B},z\right)$$

(1)

for some parts of it, and drop the other positive parts.

Let $\omega \left(u\right)$ denote the Buchstab function determined by the following differential-difference equation

$$\begin{array}{c}\hfill \left\{\begin{array}{cc}\omega \left(u\right)={\displaystyle \frac{1}{u}},\hfill & 1⩽u⩽2,\hfill \\ {\left(u\omega \left(u\right)\right)}^{\prime }=\omega (u-1),\hfill & u⩾2.\hfill \end{array}\right.\end{array}$$

Moreover, we have the upper and lower bounds for $\omega \left(u\right)$:

$$\begin{array}{c}\hfill \omega \left(u\right)⩾{\omega }_0\left(u\right)=\left\{\begin{array}{cc}{\displaystyle \frac{1}{u}},\hfill & 1⩽u<2,\hfill \\ {\displaystyle \frac{1+log(u-1)}{u}},\hfill & 2⩽u<3,\hfill \\ {\displaystyle \frac{1+log(u-1)}{u}}+{\displaystyle \frac{1}{u}}{\int }_2^{u-1}{\displaystyle \frac{log(t-1)}{t}}dt⩾0.5607,\hfill & 3⩽u<4,\hfill \\ 0.5612,\hfill & u⩾4,\hfill \end{array}\right.\end{array}$$

$$\begin{array}{c}\hfill \omega \left(u\right)⩽{\omega }_1\left(u\right)=\left\{\begin{array}{cc}{\displaystyle \frac{1}{u}},\hfill & 1⩽u<2,\hfill \\ {\displaystyle \frac{1+log(u-1)}{u}},\hfill & 2⩽u<3,\hfill \\ {\displaystyle \frac{1+log(u-1)}{u}}+{\displaystyle \frac{1}{u}}{\int }_2^{u-1}{\displaystyle \frac{log(t-1)}{t}}dt⩽0.5644,\hfill & 3⩽u<4,\hfill \\ 0.5617,\hfill & u⩾4.\hfill \end{array}\right.\end{array}$$

We shall use ${\omega }_0\left(u\right)$ and ${\omega }_1\left(u\right)$ to give numerical bounds for some sieve functions discussed below.

Before decomposing, we define the asymptotic regions $T_1$–$T_3$ and L as

$$\begin{array}{cc}\hfill T_1(m,n):=& \left\{m⩽{\displaystyle \frac{53}{102}},n⩽{\displaystyle \frac{53}{204}}\right\}\hfill \\ \hfill T_2(m,n,h):=& \left\{m⩽{\displaystyle \frac{53}{102}},n⩾h,{\displaystyle \frac{3}{4}}n+h⩽{\displaystyle \frac{53}{204}},n+{\displaystyle \frac{1}{2}}h⩽{\displaystyle \frac{53}{204}},{\displaystyle \frac{7}{4}}n+{\displaystyle \frac{3}{2}}h⩽{\displaystyle \frac{53}{102}}\right\},\hfill \\ \hfill T_3(m,n):=& \left\{{\displaystyle \frac{49}{102}}⩽m⩽{\displaystyle \frac{53}{102}}\mathrm{or}{\displaystyle \frac{49}{102}}⩽m+n⩽{\displaystyle \frac{53}{102}}\right\},\hfill \\ \hfill L(m,n):=& \left\{(m,n)\notin T_3,(m,n,n)\mathrm{cannot}\mathrm{be}\mathrm{partitioned}\mathrm{into}(\alpha ,\eta )\in T_1\mathrm{or}(\alpha ,\eta ,\gamma )\in T_2,\right.\hfill \\ \hfill & \left.n⩾{\displaystyle \frac{53}{255}}\mathrm{or}m⩾{\displaystyle \frac{1129}{2448}}\mathrm{or}{\displaystyle \frac{1}{2}}m+n⩾{\displaystyle \frac{9361}{24480}}\right\}.\hfill \end{array}$$

Lemma 3.1.

We can give an asymptotic formula for

$$\sum _{t_1\dots t_n}S\left({\mathcal{A}}_{p_1\dots p_n},v^{{\displaystyle \frac{2}{51}}}\right)$$

if we can group $(t_1,\dots ,t_n)$ into $(m,n)\in T_1$ or $(m,n,h)\in T_2$.

Lemma 3.2.

We can give an asymptotic formula for

$$\sum _{t_1\dots t_n}S\left({\mathcal{A}}_{p_1\dots p_n},p_n\right)$$

if we can group $(t_1,\dots ,t_n)$ into $(m,n)\in T_3$.

By Buchstab’s identity, we have

$$\begin{array}{cc}\hfill S\left(\mathcal{A},{\left(2v\right)}^{{\displaystyle \frac{1}{2}}}\right)=& S\left(\mathcal{A},v^{{\displaystyle \frac{2}{51}}}\right)-\sum _{{\displaystyle \frac{2}{51}}⩽t_1<{\displaystyle \frac{49}{102}}}S\left({\mathcal{A}}_{p_1},p_1\right)-\sum _{{\displaystyle \frac{49}{102}}⩽t_1<{\displaystyle \frac{1}{2}}}S\left({\mathcal{A}}_{p_1},p_1\right)\hfill \\ \hfill =& S\left(\mathcal{A},v^{{\displaystyle \frac{2}{51}}}\right)-\sum _{{\displaystyle \frac{2}{51}}⩽t_1<{\displaystyle \frac{49}{102}}}S\left({\mathcal{A}}_{p_1},v^{{\displaystyle \frac{2}{51}}}\right)-\sum _{{\displaystyle \frac{49}{102}}⩽t_1<{\displaystyle \frac{1}{2}}}S\left({\mathcal{A}}_{p_1},p_1\right)\hfill \\ \hfill & +\sum _{\begin{array}{c}{\displaystyle \frac{2}{51}}⩽t_1<{\displaystyle \frac{49}{102}}\\ {\displaystyle \frac{2}{51}}⩽t_2<min\left(t_1,{\displaystyle \frac{1}{2}}(1-t_1)\right)\end{array}}S\left({\mathcal{A}}_{p_1{p}_2},p_2\right)\hfill \\ \hfill =& S_1-S_2-S_3+S_4.\hfill \end{array}$$

(2)

By Lemma 2.1 and Lemma 2.2, we can give asymptotic formulas for $S_1$, $S_2$ and $S_3$. Before estimating $S_4$, we first split it into three parts:

$$\begin{array}{cc}\hfill S_4=& \sum _{\begin{array}{c}{\displaystyle \frac{2}{51}}⩽t_1<{\displaystyle \frac{49}{102}}\\ {\displaystyle \frac{2}{51}}⩽t_2<min\left(t_1,{\displaystyle \frac{1}{2}}(1-t_1)\right)\end{array}}S\left({\mathcal{A}}_{p_1{p}_2},p_2\right)\hfill \\ \hfill =& \sum _{\begin{array}{c}{\displaystyle \frac{2}{51}}⩽t_1<{\displaystyle \frac{49}{102}}\\ {\displaystyle \frac{2}{51}}⩽t_2<min\left(t_1,{\displaystyle \frac{1}{2}}(1-t_1)\right)\\ (t_1,t_2)\in T_3\end{array}}S\left({\mathcal{A}}_{p_1{p}_2},p_2\right)+\sum _{\begin{array}{c}{\displaystyle \frac{2}{51}}⩽t_1<{\displaystyle \frac{49}{102}}\\ {\displaystyle \frac{2}{51}}⩽t_2<min\left(t_1,{\displaystyle \frac{1}{2}}(1-t_1)\right)\\ (t_1,t_2)\in L\end{array}}S\left({\mathcal{A}}_{p_1{p}_2},p_2\right)\hfill \\ \hfill & +\sum _{\begin{array}{c}{\displaystyle \frac{2}{51}}⩽t_1<{\displaystyle \frac{49}{102}}\\ {\displaystyle \frac{2}{51}}⩽t_2<min\left(t_1,{\displaystyle \frac{1}{2}}(1-t_1)\right)\\ (t_1,t_2)\notin T_3\\ (t_1,t_2,t_2)\mathrm{can}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_1\mathrm{or}(m,n,h)\in T_2\end{array}}S\left({\mathcal{A}}_{p_1{p}_2},p_2\right)\hfill \\ \hfill & +\sum _{\begin{array}{c}{\displaystyle \frac{2}{51}}⩽t_1<{\displaystyle \frac{49}{102}}\\ {\displaystyle \frac{2}{51}}⩽t_2<min\left(t_1,{\displaystyle \frac{1}{2}}(1-t_1)\right)\\ (t_1,t_2)\notin T_3\\ (t_1,t_2)\notin L\\ (t_1,t_2,t_2)\mathrm{cannot}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_1\mathrm{or}(m,n,h)\in T_2\end{array}}S\left({\mathcal{A}}_{p_1{p}_2},p_2\right)\hfill \\ \hfill =& S_{41}+S_{42}+S_{43}+S_{44}.\hfill \end{array}$$

(3)

$S_{41}$ has an asymptotic formula. For $S_{42}$, we cannot decompose further but have to discard the whole region giving the loss

$${\int }_{{\displaystyle \frac{2}{51}}}^{{\displaystyle \frac{49}{102}}}{\int }_{{\displaystyle \frac{2}{51}}}^{min\left(t_1,{\displaystyle \frac{1-t_1}{2}}\right)}{\mathbb{1}}_{(t_1,t_2)\in L}{\displaystyle \frac{\omega \left(\frac{1-t_1-t_2}{t_2}\right)}{t_1{t}_2^2}}d{t}_{2}d{t}_1<0.687415.$$

(4)

For $S_{43}$ we can use Buchstab’s identity to get

$$\begin{array}{cc}\hfill S_{43}=& \sum _{\begin{array}{c}{\displaystyle \frac{2}{51}}⩽t_1<{\displaystyle \frac{49}{102}}\\ {\displaystyle \frac{2}{51}}⩽t_2<min\left(t_1,{\displaystyle \frac{1}{2}}(1-t_1)\right)\\ (t_1,t_2)\notin T_3\\ (t_1,t_2,t_2)\mathrm{can}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_1\mathrm{or}(m,n,h)\in T_2\end{array}}S\left({\mathcal{A}}_{p_1{p}_2},p_2\right)\hfill \\ \hfill =& \sum _{\begin{array}{c}{\displaystyle \frac{2}{51}}⩽t_1<{\displaystyle \frac{49}{102}}\\ {\displaystyle \frac{2}{51}}⩽t_2<min\left(t_1,{\displaystyle \frac{1}{2}}(1-t_1)\right)\\ (t_1,t_2)\notin T_3\\ (t_1,t_2,t_2)\mathrm{can}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_1\mathrm{or}(m,n,h)\in T_2\end{array}}S\left({\mathcal{A}}_{p_1{p}_2},v^{{\displaystyle \frac{2}{51}}}\right)\hfill \\ \hfill & -\sum _{\begin{array}{c}{\displaystyle \frac{2}{51}}⩽t_1<{\displaystyle \frac{49}{102}}\\ {\displaystyle \frac{2}{51}}⩽t_2<min\left(t_1,{\displaystyle \frac{1}{2}}(1-t_1)\right)\\ (t_1,t_2)\notin T_3\\ (t_1,t_2,t_2)\mathrm{can}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_1\mathrm{or}(m,n,h)\in T_2\\ {\displaystyle \frac{2}{51}}⩽t_3<min\left(t_2,{\displaystyle \frac{1}{2}}(1-t_1-t_2)\right)\\ (t_1,t_2,t_3)\mathrm{can}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_3\end{array}}S\left({\mathcal{A}}_{p_1{p}_2{p}_3},p_3\right)\hfill \\ \hfill & -\sum _{\begin{array}{c}{\displaystyle \frac{2}{51}}⩽t_1<{\displaystyle \frac{49}{102}}\\ {\displaystyle \frac{2}{51}}⩽t_2<min\left(t_1,{\displaystyle \frac{1}{2}}(1-t_1)\right)\\ (t_1,t_2)\notin T_3\\ (t_1,t_2,t_2)\mathrm{can}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_1\mathrm{or}(m,n,h)\in T_2\\ {\displaystyle \frac{2}{51}}⩽t_3<min\left(t_2,{\displaystyle \frac{1}{2}}(1-t_1-t_2)\right)\\ (t_1,t_2,t_3)\mathrm{cannot}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_3\end{array}}S\left({\mathcal{A}}_{p_1{p}_2{p}_3},v^{{\displaystyle \frac{2}{51}}}\right)\hfill \\ \hfill & +\sum _{\begin{array}{c}{\displaystyle \frac{2}{51}}⩽t_1<{\displaystyle \frac{49}{102}}\\ {\displaystyle \frac{2}{51}}⩽t_2<min\left(t_1,{\displaystyle \frac{1}{2}}(1-t_1)\right)\\ (t_1,t_2)\notin T_3\\ (t_1,t_2,t_2)\mathrm{can}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_1\mathrm{or}(m,n,h)\in T_2\\ {\displaystyle \frac{2}{51}}⩽t_3<min\left(t_2,{\displaystyle \frac{1}{2}}(1-t_1-t_2)\right)\\ (t_1,t_2,t_3)\mathrm{cannot}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_3\\ {\displaystyle \frac{2}{51}}⩽t_4<min\left(t_3,{\displaystyle \frac{1}{2}}(1-t_1-t_2-t_3)\right)\\ (t_1,t_2,t_3,t_4)\mathrm{can}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_3\end{array}}S\left({\mathcal{A}}_{p_1{p}_2{p}_3{p}_4},p_4\right)\hfill \\ \hfill & +\sum _{\begin{array}{c}{\displaystyle \frac{2}{51}}⩽t_1<{\displaystyle \frac{49}{102}}\\ {\displaystyle \frac{2}{51}}⩽t_2<min\left(t_1,{\displaystyle \frac{1}{2}}(1-t_1)\right)\\ (t_1,t_2)\notin T_3\\ (t_1,t_2,t_2)\mathrm{can}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_1\mathrm{or}(m,n,h)\in T_2\\ {\displaystyle \frac{2}{51}}⩽t_3<min\left(t_2,{\displaystyle \frac{1}{2}}(1-t_1-t_2)\right)\\ (t_1,t_2,t_3)\mathrm{cannot}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_3\\ {\displaystyle \frac{2}{51}}⩽t_4<min\left(t_3,{\displaystyle \frac{1}{2}}(1-t_1-t_2-t_3)\right)\\ (t_1,t_2,t_3,t_4)\mathrm{cannot}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_3\end{array}}S\left({\mathcal{A}}_{p_1{p}_2{p}_3{p}_4},p_4\right)\hfill \\ \hfill =& S_{431}-S_{432}-S_{433}+S_{434}+S_{435}.\hfill \end{array}$$

(5)

We have asymptotic formulas for $S_{431}$–$S_{434}$. For the remaining $S_{435}$, we have two ways to get more possible savings: One way is to use Buchstab’s identity twice more for some parts if we can group $(t_1,t_2,t_3,t_4,t_4)$ into $(m,n)\in T_1$ or $(m,n,h)\in T_2$. Another way is to use Buchstab’s identity in reverse to make almost-primes visible. The details of further decompositions are similar to those in [1]. Combining the cases above we get a loss from $S_{43}$ of

$$\begin{array}{cc}\hfill & \left({\int }_{(t_1,t_2,t_3,t_4)\in U_1}{\displaystyle \frac{\omega \left(\frac{1-t_1-t_2-t_3-t_4}{t_4}\right)}{t_1{t}_2{t}_3{t}_4^2}}d{t}_{4}d{t}_{3}d{t}_{2}d{t}_1\right)\hfill \\ \hfill +& \left({\int }_{(t_1,t_2,t_3,t_4,t_5,t_6)\in U_2}{\displaystyle \frac{{\omega }_1\left(\frac{1-t_1-t_2-t_3-t_4-t_5-t_6}{t_6}\right)}{t_1{t}_2{t}_3{t}_4{t}_5{t}_6^2}}d{t}_{6}d{t}_{5}d{t}_{4}d{t}_{3}d{t}_{2}d{t}_1\right)\hfill \\ \hfill -& \left({\int }_{(t_1,t_2,t_3,t_4,t_5)\in U_3}{\displaystyle \frac{\omega \left(\frac{1-t_1-t_2-t_3-t_4-t_5}{t_5}\right)}{t_1{t}_2{t}_3{t}_4{t}_5^2}}d{t}_{5}d{t}_{4}d{t}_{3}d{t}_{2}d{t}_1\right)\hfill \\ \hfill ⩽& (0.161005+0.073993-0.009022)=0.225976,\hfill \end{array}$$

(6)

where

$$\begin{array}{cc}\hfill U_1(t_1,t_2,t_3,t_4):=& \left\{(t_1,t_2)\notin T_3,(t_1,t_2,t_2)\mathrm{can}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_1\mathrm{or}(m,n,h)\in T_2,\right.\hfill \\ \hfill & {\displaystyle \frac{2}{51}}⩽t_3<min\left(t_2,{\displaystyle \frac{1}{2}}(1-t_1-t_2)\right),\hfill \\ \hfill & (t_1,t_2,t_3)\mathrm{cannot}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_3,\hfill \\ \hfill & {\displaystyle \frac{2}{51}}⩽t_4<min\left(t_3,{\displaystyle \frac{1}{2}}(1-t_1-t_2-t_3)\right),\hfill \\ \hfill & (t_1,t_2,t_3,t_4)\mathrm{cannot}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_3,\hfill \\ \hfill & (t_1,t_2,t_3,t_4,t_4)\mathrm{cannot}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_1\mathrm{or}(m,n,h)\in T_2,\hfill \\ \hfill & \left.{\displaystyle \frac{2}{51}}⩽t_1<{\displaystyle \frac{49}{102}},{\displaystyle \frac{2}{51}}⩽t_2<min\left(t_1,{\displaystyle \frac{1}{2}}(1-t_1)\right)\right\},\hfill \\ \hfill U_2(t_1,t_2,t_3,t_4,t_5,t_6):=& \left\{(t_1,t_2)\notin T_3,(t_1,t_2,t_2)\mathrm{can}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_1\mathrm{or}(m,n,h)\in T_2,\right.\hfill \\ \hfill & {\displaystyle \frac{2}{51}}⩽t_3<min\left(t_2,{\displaystyle \frac{1}{2}}(1-t_1-t_2)\right),\hfill \\ \hfill & (t_1,t_2,t_3)\mathrm{cannot}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_3,\hfill \\ \hfill & {\displaystyle \frac{2}{51}}⩽t_4<min\left(t_3,{\displaystyle \frac{1}{2}}(1-t_1-t_2-t_3)\right),\hfill \\ \hfill & (t_1,t_2,t_3,t_4)\mathrm{cannot}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_3,\hfill \\ \hfill & (t_1,t_2,t_3,t_4,t_4)\mathrm{can}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_1\mathrm{or}(m,n,h)\in T_2,\hfill \\ \hfill & {\displaystyle \frac{2}{51}}⩽t_5<min\left(t_4,{\displaystyle \frac{1}{2}}(1-t_1-t_2-t_3-t_4)\right),\hfill \\ \hfill & (t_1,t_2,t_3,t_4,t_5)\mathrm{cannot}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_3,\hfill \\ \hfill & {\displaystyle \frac{2}{51}}⩽t_6<min\left(t_5,{\displaystyle \frac{1}{2}}(1-t_1-t_2-t_3-t_4-t_5)\right),\hfill \\ \hfill & (t_1,t_2,t_3,t_4,t_5,t_6)\mathrm{cannot}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_3,\hfill \\ \hfill & \left.{\displaystyle \frac{2}{51}}⩽t_1<{\displaystyle \frac{49}{102}},{\displaystyle \frac{2}{51}}⩽t_2<min\left(t_1,{\displaystyle \frac{1}{2}}(1-t_1)\right)\right\},\hfill \\ \hfill U_3(t_1,t_2,t_3,t_4,t_5):=& \left\{(t_1,t_2)\notin T_3,(t_1,t_2,t_2)\mathrm{can}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_1\mathrm{or}(m,n,h)\in T_2,\right.\hfill \\ \hfill & {\displaystyle \frac{2}{51}}⩽t_3<min\left(t_2,{\displaystyle \frac{1}{2}}(1-t_1-t_2)\right),\hfill \\ \hfill & (t_1,t_2,t_3)\mathrm{cannot}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_3,\hfill \\ \hfill & {\displaystyle \frac{2}{51}}⩽t_4<min\left(t_3,{\displaystyle \frac{1}{2}}(1-t_1-t_2-t_3)\right),\hfill \\ \hfill & (t_1,t_2,t_3,t_4)\mathrm{cannot}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_3,\hfill \\ \hfill & (t_1,t_2,t_3,t_4,t_4)\mathrm{cannot}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_1\mathrm{or}(m,n,h)\in T_2,\hfill \\ \hfill & t_4<t_5<{\displaystyle \frac{1}{2}}(1-t_1-t_2-t_3-t_4),\hfill \\ \hfill & (t_1,t_2,t_3,t_4,t_5)\mathrm{can}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_3,\hfill \\ \hfill & \left.{\displaystyle \frac{2}{51}}⩽t_1<{\displaystyle \frac{49}{102}},{\displaystyle \frac{2}{51}}⩽t_2<min\left(t_1,{\displaystyle \frac{1}{2}}(1-t_1)\right)\right\}.\hfill \end{array}$$

Next we shall decompose $S_{44}$. By Buchstab’s identity, we have

$$\begin{array}{cc}\hfill S_{44}=& \sum _{\begin{array}{c}{\displaystyle \frac{2}{51}}⩽t_1<{\displaystyle \frac{49}{102}}\\ {\displaystyle \frac{2}{51}}⩽t_2<min\left(t_1,{\displaystyle \frac{1}{2}}(1-t_1)\right)\\ (t_1,t_2)\notin T_3\\ (t_1,t_2)\notin L\\ (t_1,t_2,t_2)\mathrm{cannot}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_1\mathrm{or}(m,n,h)\in T_2\end{array}}S\left({\mathcal{A}}_{p_1{p}_2},p_2\right)\hfill \\ \hfill =& \sum _{\begin{array}{c}{\displaystyle \frac{2}{51}}⩽t_1<{\displaystyle \frac{49}{102}}\\ {\displaystyle \frac{2}{51}}⩽t_2<min\left(t_1,{\displaystyle \frac{1}{2}}(1-t_1)\right)\\ (t_1,t_2)\notin T_3\\ (t_1,t_2)\notin L\\ (t_1,t_2,t_2)\mathrm{cannot}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_1\mathrm{or}(m,n,h)\in T_2\end{array}}S\left({\mathcal{A}}_{p_1{p}_2},v^{{\displaystyle \frac{2}{51}}}\right)\hfill \\ \hfill & -\sum _{\begin{array}{c}{\displaystyle \frac{2}{51}}⩽t_1<{\displaystyle \frac{49}{102}}\\ {\displaystyle \frac{2}{51}}⩽t_2<min\left(t_1,{\displaystyle \frac{1}{2}}(1-t_1)\right)\\ (t_1,t_2)\notin T_3\\ (t_1,t_2)\notin L\\ (t_1,t_2,t_2)\mathrm{cannot}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_1\mathrm{or}(m,n,h)\in T_2\\ {\displaystyle \frac{2}{51}}⩽t_3<min\left(t_2,{\displaystyle \frac{1}{2}}(1-t_1-t_2)\right)\end{array}}S\left({\mathcal{A}}_{p_1{p}_2{p}_3},p_3\right)\hfill \\ \hfill =& \sum _{\begin{array}{c}{\displaystyle \frac{2}{51}}⩽t_1<{\displaystyle \frac{49}{102}}\\ {\displaystyle \frac{2}{51}}⩽t_2<min\left(t_1,{\displaystyle \frac{1}{2}}(1-t_1)\right)\\ (t_1,t_2)\notin T_3\\ (t_1,t_2)\notin L\\ (t_1,t_2,t_2)\mathrm{cannot}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_1\mathrm{or}(m,n,h)\in T_2\end{array}}S\left({\mathcal{A}}_{p_1{p}_2},v^{{\displaystyle \frac{2}{51}}}\right)\hfill \\ \hfill & -\sum _{\begin{array}{c}{\displaystyle \frac{2}{51}}⩽t_1<{\displaystyle \frac{49}{102}}\\ {\displaystyle \frac{2}{51}}⩽t_2<min\left(t_1,{\displaystyle \frac{1}{2}}(1-t_1)\right)\\ (t_1,t_2)\notin T_3\\ (t_1,t_2)\notin L\\ (t_1,t_2,t_2)\mathrm{cannot}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_1\mathrm{or}(m,n,h)\in T_2\\ {\displaystyle \frac{2}{51}}⩽t_3<min\left(t_2,{\displaystyle \frac{1}{2}}(1-t_1-t_2)\right)\\ (t_1,t_2,t_3)\mathrm{can}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_1\mathrm{or}(m,n,h)\in T_2\end{array}}S\left({\mathcal{A}}_{p_1{p}_2{p}_3},p_3\right)\hfill \\ \hfill & -\sum _{\begin{array}{c}{\displaystyle \frac{2}{51}}⩽t_1<{\displaystyle \frac{49}{102}}\\ {\displaystyle \frac{2}{51}}⩽t_2<min\left(t_1,{\displaystyle \frac{1}{2}}(1-t_1)\right)\\ (t_1,t_2)\notin T_3\\ (t_1,t_2)\notin L\\ (t_1,t_2,t_2)\mathrm{cannot}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_1\mathrm{or}(m,n,h)\in T_2\\ {\displaystyle \frac{2}{51}}⩽t_3<min\left(t_2,{\displaystyle \frac{1}{2}}(1-t_1-t_2)\right)\\ (t_1,t_2,t_3)\mathrm{cannot}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_1\mathrm{or}(m,n,h)\in T_2\end{array}}S\left({\mathcal{A}}_{p_1{p}_2{p}_3},p_3\right)\hfill \\ \hfill =& S_{441}-S_{442}-S_{443}.\hfill \end{array}$$

(7)

We have an asymptotic formula for $S_{441}$. For $S_{442}$ we can use the same methods as above (i.e. using Buchstab’s identity twice more and making almost-primes visible) to get a loss of

$$\begin{array}{cc}\hfill & \left({\int }_{(t_1,t_2,t_3,t_4)\in U_4}{\displaystyle \frac{\omega \left(\frac{1-t_1-t_2-t_3-t_4}{t_4}\right)}{t_1{t}_2{t}_3{t}_4^2}}d{t}_{4}d{t}_{3}d{t}_{2}d{t}_1\right)\hfill \\ \hfill -& \left({\int }_{(t_1,t_2,t_3,t_4,t_5)\in U_5}{\displaystyle \frac{\omega \left(\frac{1-t_1-t_2-t_3-t_4-t_5}{t_5}\right)}{t_1{t}_2{t}_3{t}_4{t}_5^2}}d{t}_{5}d{t}_{4}d{t}_{3}d{t}_{2}d{t}_1\right)\hfill \\ \hfill ⩽& (0.038404-0.005445)=0.032959,\hfill \end{array}$$

(8)

where

$$\begin{array}{cc}\hfill U_4(t_1,t_2,t_3,t_4):=& \left\{(t_1,t_2)\notin T_3,(t_1,t_2,t_2)\mathrm{cannot}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_1\mathrm{or}(m,n,h)\in T_2,\right.\hfill \\ \hfill & {\displaystyle \frac{2}{51}}⩽t_3<min\left(t_2,{\displaystyle \frac{1}{2}}(1-t_1-t_2)\right),\hfill \\ \hfill & (t_1,t_2,t_3)\mathrm{can}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_1\mathrm{or}(m,n,h)\in T_2,\hfill \\ \hfill & (t_1,t_2,t_3)\mathrm{cannot}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_3,\hfill \\ \hfill & {\displaystyle \frac{2}{51}}⩽t_4<min\left(t_3,{\displaystyle \frac{1}{2}}(1-t_1-t_2-t_3)\right),\hfill \\ \hfill & (t_1,t_2,t_3,t_4)\mathrm{cannot}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_3,\hfill \\ \hfill & \left.{\displaystyle \frac{2}{51}}⩽t_1<{\displaystyle \frac{49}{102}},{\displaystyle \frac{2}{51}}⩽t_2<min\left(t_1,{\displaystyle \frac{1}{2}}(1-t_1)\right)\right\},\hfill \\ \hfill U_5(t_1,t_2,t_3,t_4,t_5):=& \left\{(t_1,t_2)\notin T_3,(t_1,t_2,t_2)\mathrm{cannot}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_1\mathrm{or}(m,n,h)\in T_2,\right.\hfill \\ \hfill & {\displaystyle \frac{2}{51}}⩽t_3<min\left(t_2,{\displaystyle \frac{1}{2}}(1-t_1-t_2)\right),\hfill \\ \hfill & (t_1,t_2,t_3)\mathrm{can}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_1\mathrm{or}(m,n,h)\in T_2,\hfill \\ \hfill & (t_1,t_2,t_3)\mathrm{cannot}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_3,\hfill \\ \hfill & {\displaystyle \frac{2}{51}}⩽t_4<min\left(t_3,{\displaystyle \frac{1}{2}}(1-t_1-t_2-t_3)\right),\hfill \\ \hfill & (t_1,t_2,t_3,t_4)\mathrm{cannot}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_3,\hfill \\ \hfill & t_4<t_5<{\displaystyle \frac{1}{2}}(1-t_1-t_2-t_3-t_4),\hfill \\ \hfill & (t_1,t_2,t_3,t_4,t_5)\mathrm{can}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_3,\hfill \\ \hfill & \left.{\displaystyle \frac{2}{51}}⩽t_1<{\displaystyle \frac{49}{102}},{\displaystyle \frac{2}{51}}⩽t_2<min\left(t_1,{\displaystyle \frac{1}{2}}(1-t_1)\right)\right\}.\hfill \end{array}$$

For $S_{443}$ we can perform a role-reversal to get a small saving. For the definition of a role-reversal one can see [28] or [14] Chapter 5, and we refer the readers to [29] and [1] for more applications of role-reversals. In this way we have

$$\begin{array}{cc}\hfill S_{443}=& \sum _{\begin{array}{c}{\displaystyle \frac{2}{51}}⩽t_1<{\displaystyle \frac{49}{102}}\\ {\displaystyle \frac{2}{51}}⩽t_2<min\left(t_1,{\displaystyle \frac{1}{2}}(1-t_1)\right)\\ (t_1,t_2)\notin T_3\\ (t_1,t_2)\notin L\\ (t_1,t_2,t_2)\mathrm{cannot}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_1\mathrm{or}(m,n,h)\in T_2\\ {\displaystyle \frac{2}{51}}⩽t_3<min\left(t_2,{\displaystyle \frac{1}{2}}(1-t_1-t_2)\right)\\ (t_1,t_2,t_3)\mathrm{cannot}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_1\mathrm{or}(m,n,h)\in T_2\end{array}}S\left({\mathcal{A}}_{p_1{p}_2{p}_3},p_3\right)\hfill \\ \hfill =& \sum _{\begin{array}{c}{\displaystyle \frac{2}{51}}⩽t_1<{\displaystyle \frac{49}{102}}\\ {\displaystyle \frac{2}{51}}⩽t_2<min\left(t_1,{\displaystyle \frac{1}{2}}(1-t_1)\right)\\ (t_1,t_2)\notin T_3\\ (t_1,t_2)\notin L\\ (t_1,t_2,t_2)\mathrm{cannot}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_1\mathrm{or}(m,n,h)\in T_2\\ {\displaystyle \frac{2}{51}}⩽t_3<min\left(t_2,{\displaystyle \frac{1}{2}}(1-t_1-t_2)\right)\\ (t_1,t_2,t_3)\mathrm{cannot}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_1\mathrm{or}(m,n,h)\in T_2\end{array}}S\left({\mathcal{A}}_{\beta p_2{p}_3},{\left({\displaystyle \frac{2v}{\beta p_2{p}_3}}\right)}^{{\displaystyle \frac{1}{2}}}\right)\hfill \\ \hfill =& \sum _{\begin{array}{c}{\displaystyle \frac{2}{51}}⩽t_1<{\displaystyle \frac{49}{102}}\\ {\displaystyle \frac{2}{51}}⩽t_2<min\left(t_1,{\displaystyle \frac{1}{2}}(1-t_1)\right)\\ (t_1,t_2)\notin T_3\\ (t_1,t_2)\notin L\\ (t_1,t_2,t_2)\mathrm{cannot}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_1\mathrm{or}(m,n,h)\in T_2\\ {\displaystyle \frac{2}{51}}⩽t_3<min\left(t_2,{\displaystyle \frac{1}{2}}(1-t_1-t_2)\right)\\ (t_1,t_2,t_3)\mathrm{cannot}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_1\mathrm{or}(m,n,h)\in T_2\end{array}}S\left({\mathcal{A}}_{\beta p_2{p}_3},v^{{\displaystyle \frac{2}{51}}}\right)\hfill \\ \hfill & -\sum _{\begin{array}{c}{\displaystyle \frac{2}{51}}⩽t_1<{\displaystyle \frac{49}{102}}\\ {\displaystyle \frac{2}{51}}⩽t_2<min\left(t_1,{\displaystyle \frac{1}{2}}(1-t_1)\right)\\ (t_1,t_2)\notin T_3\\ (t_1,t_2)\notin L\\ (t_1,t_2,t_2)\mathrm{cannot}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_1\mathrm{or}(m,n,h)\in T_2\\ {\displaystyle \frac{2}{51}}⩽t_3<min\left(t_2,{\displaystyle \frac{1}{2}}(1-t_1-t_2)\right)\\ (t_1,t_2,t_3)\mathrm{cannot}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_1\mathrm{or}(m,n,h)\in T_2\\ {\displaystyle \frac{2}{51}}⩽t_4<{\displaystyle \frac{1}{2}}{t}_1\end{array}}S\left({\mathcal{A}}_{\beta p_2{p}_3{p}_4},p_4\right),\hfill \end{array}$$

where $\beta \sim v^{1-t_1-t_2-t_3}$ and $\left(\beta ,P\left(p_3\right)\right)=1$. Again, we can use Buchstab’s identity in reverse to gain a small saving on the last term. Altogether we get a loss from $S_{443}$ of

$$\begin{array}{cc}\hfill & \left({\int }_{(t_1,t_2,t_3,t_4)\in U_6}{\displaystyle \frac{\omega \left(\frac{t_1-t_4}{t_4}\right)\omega \left(\frac{1-t_1-t_2-t_3}{t_3}\right)}{t_2{t}_3^2{t}_4^2}}d{t}_{4}d{t}_{3}d{t}_{2}d{t}_1\right)\hfill \\ \hfill -& \left({\int }_{(t_1,t_2,t_3,t_4,t_5)\in U_7}{\displaystyle \frac{\omega \left(\frac{t_1-t_4-t_5}{t_5}\right)\omega \left(\frac{1-t_1-t_2-t_3}{t_3}\right)}{t_2{t}_3^2{t}_4{t}_5^2}}d{t}_{5}d{t}_{4}d{t}_{3}d{t}_{2}d{t}_1\right)\hfill \\ \hfill ⩽& (0.046566-0.007144)=0.039422,\hfill \end{array}$$

(9)

where

$$\begin{array}{cc}\hfill U_6(t_1,t_2,t_3,t_4):=& \left\{(t_1,t_2)\notin T_3,(t_1,t_2,t_2)\mathrm{cannot}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_1\mathrm{or}(m,n,h)\in T_2,\right.\hfill \\ \hfill & {\displaystyle \frac{2}{51}}⩽t_3<min\left(t_2,{\displaystyle \frac{1}{2}}(1-t_1-t_2)\right),\hfill \\ \hfill & (t_1,t_2,t_3)\mathrm{cannot}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_1\mathrm{or}(m,n,h)\in T_2,\hfill \\ \hfill & (t_1,t_2,t_3)\mathrm{cannot}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_3,\hfill \\ \hfill & {\displaystyle \frac{2}{51}}⩽t_4<{\displaystyle \frac{1}{2}}{t}_1,\hfill \\ \hfill & (1-t_1-t_2-t_3,t_2,t_3,t_4)\mathrm{cannot}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_3,\hfill \\ \hfill & \left.{\displaystyle \frac{2}{51}}⩽t_1<{\displaystyle \frac{49}{102}},{\displaystyle \frac{2}{51}}⩽t_2<min\left(t_1,{\displaystyle \frac{1}{2}}(1-t_1)\right)\right\},\hfill \\ \hfill U_7(t_1,t_2,t_3,t_4,t_5):=& \left\{(t_1,t_2)\notin T_3,(t_1,t_2,t_2)\mathrm{cannot}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_1\mathrm{or}(m,n,h)\in T_2,\right.\hfill \\ \hfill & {\displaystyle \frac{2}{51}}⩽t_3<min\left(t_2,{\displaystyle \frac{1}{2}}(1-t_1-t_2)\right),\hfill \\ \hfill & (t_1,t_2,t_3)\mathrm{can}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_1\mathrm{or}(m,n,h)\in T_2,\hfill \\ \hfill & (t_1,t_2,t_3)\mathrm{cannot}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_3,\hfill \\ \hfill & {\displaystyle \frac{2}{51}}⩽t_4<{\displaystyle \frac{1}{2}}{t}_1,\hfill \\ \hfill & (1-t_1-t_2-t_3,t_2,t_3,t_4)\mathrm{cannot}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_3,\hfill \\ \hfill & t_4<t_5<{\displaystyle \frac{1}{2}}(t_1-t_4),\hfill \\ \hfill & (1-t_1-t_2-t_3,t_2,t_3,t_4,t_5)\mathrm{can}\mathrm{be}\mathrm{partitioned}\mathrm{into}(m,n)\in T_3,\hfill \\ \hfill & \left.{\displaystyle \frac{2}{51}}⩽t_1<{\displaystyle \frac{49}{102}},{\displaystyle \frac{2}{51}}⩽t_2<min\left(t_1,{\displaystyle \frac{1}{2}}(1-t_1)\right)\right\}.\hfill \end{array}$$

Finally, by (2)–(9), the total loss is less than

$$0.687415+0.225976+0.032959+0.039422<0.986$$

and the proof of Theorem 1.1 is completed.