Autocorrelations of l-sequences with prime connection integer

Abstract

In this paper, the autocorrelations of l-sequences with prime connection integer are discussed. Let $\underline{a}$ be an l-sequence with connection integer p and period T = p − 1, we show that the autocorrelation $C_{\underline{a}}(\tau )$ of $\underline{a}$ with shift τ satisfies:

$$ \left\vert C_{\underline{a}}(\tau )-\frac{p-1}{p^{2}}\cdot \underset{c=1}{ \overset{p-1}{\sum }}\tan \left( \frac{\pi c2^{-\tau }}{p}\right) \tan \left( \frac{\pi c}{p}\right) \right\vert =O(\ln ^{2}p). $$

Thus by calculating this triangular sum, an estimate of $C_{\underline{a} }(\tau )$ can be obtained. Particularly, for any shift τ with $2^{-\tau }(\mbox{mod}\ p)=(p-3)/2$ or $(p+3)/2$, the autocorrelation $C_{ \underline{a}}(\tau )$ of $\underline{a}$ with shift τ satisfies $C_{ \underline{a}}(\tau )=O(\ln ^{2}p)$, thus when p is sufficiently large, the autocorrelation is low. Such result also holds for the decimations of l-sequences.

Appendix

The proof of Lemma 3 is presented in this Appendix.

Before giving the proof, we first show some necessary lemmas.

Lemma 4

For any real x , if $0<x<\frac{\pi }{2}$ , then

$$ \cos x>\frac{2}{\pi }\left( \frac{\pi }{2}-x\right) . $$

Proof

The result follows from $\cos x=\sin \left( \frac{\pi }{2}-x\right)$ and $\frac{\pi }{2}\sin x>x$ for $0<x<\frac{\pi }{2}$. □

Lemma 5

For any real x , if $0<x<\frac{\pi }{6}$ , then

$$ 0<\cos \left( \frac{\pi }{3}+x\right) <\frac{1-\sqrt[2]{3}x}{2}<\frac{1}{2}. $$

and if $0<x<\frac{\pi }{3}$ , then

$$ \frac{1}{2}<\frac{1}{2}+\frac{\sqrt[2]{3}x}{4}<\cos \left( \frac{\pi }{3} -x\right) <1, $$

Proof

It is clear that $0<\cos \left( \frac{\pi }{3}+x\right) <\frac{1}{2}$ for $0<x<\frac{\pi }{6}$, and $\frac{1}{2}<\cos \left( \frac{\pi }{3}-x\right) <1$ for $0<x<\frac{\pi }{3}$. Next we need only to show $\cos \left( \frac{\pi }{ 3}+x\right) <\frac{1-\sqrt[2]{3}x}{2}$ for $0<x<\frac{\pi }{6}$, and $\cos \left( \frac{\pi }{3}-x\right) >\frac{1}{2}+\frac{\sqrt[2]{3}x}{4}$ for $0<x< \frac{\pi }{3}$.

Since $\cos x\!=\!\frac{1}{2}\!-\!\frac{\sqrt[2]{3}}{2}\left( x\!-\!\frac{\pi }{3} \right) \!-\!\frac{1}{2}\cdot \frac{\left( x\!-\!\frac{\pi }{3}\right) ^{2}}{2!}\!+\! \frac{\sqrt[2]{3}}{2}\!\cdot\! \frac{\left( x-\frac{\pi }{3}\right) ^{3}}{3!}+ \frac{1}{2}\!\cdot\! \frac{\left( x-\frac{\pi }{3}\right) ^{4}}{4!}-\frac{\sqrt[2 ]{3}}{2}\!\cdot\! \frac{\left( x-\frac{\pi }{3}\right) ^{5}}{5!}-...$, if $0<x< \frac{\pi }{6}$, then we have

$$ \begin{array}{rcl} \cos \left( \frac{\pi }{3}+x\right) &=&\frac{1}{2}-\frac{\sqrt[2]{3}x}{2}- \frac{x^{2}}{4}+\frac{\sqrt[2]{3}x^{3}}{12}+... \\ &<&\frac{1}{2}-\frac{\sqrt[2]{3}x}{2}, \end{array} $$

and if $0<x<\frac{\pi }{3}$, then we have

$$ \begin{array}{rcl} \cos \left( \frac{\pi }{3}-x\right) &=&\frac{1}{2}+\frac{\sqrt[2]{3}x}{2}- \frac{x^{2}}{4}-\frac{\sqrt[2]{3}x^{3}}{12}+\frac{1}{2}\cdot \frac{x^{4}}{4!} +.... \\ &>&\frac{1}{2}+\frac{\sqrt[2]{3}x}{2}-\frac{x^{2}}{4}-\frac{\sqrt[2]{3}x^{3} }{12} \\ &=&\frac{1}{2}+\frac{\sqrt[2]{3}x}{12}\left( 6-\sqrt[2]{3}x-x^{2}\right) \\ &>&\frac{1}{2}+\frac{\sqrt[2]{3}x}{12}\left( 6-\sqrt[2]{3}\cdot \frac{\pi }{3 }-\left( \frac{\pi }{3}\right) ^{2}\right) \\ &>&\frac{1}{2}+\frac{\sqrt[2]{3}x}{4}. \end{array} $$

□

Lemma 6

For any real x , if $0<x<\frac{\pi }{3}$ , then

$$ \left\{ \begin{array}{c} 1<\frac{1}{\cos x(2\cos x-1)}<\frac{4\sqrt[2]{3}}{3\left( \frac{\pi }{3} -x\right) } \\[5pt] \frac{1}{3}<\frac{1}{\cos x(2\cos x+1)}<1 \end{array} \right. , $$

if $\frac{\pi }{3}<x<\frac{\pi }{2}$ , then

$$ \left\{ \begin{array}{c} -\frac{\sqrt[2]{3}\pi }{6\left( x-\frac{\pi }{3}\right) \left( \frac{\pi }{2} -x\right) }<\frac{1}{\cos x(2\cos x-1)}<-2 \\[5pt] 1<\frac{1}{\cos x(2\cos x+1)}<\frac{\pi }{2\left( \frac{\pi }{2}-x\right) } \end{array} \right. , $$

if $\frac{\pi }{2}<x<\frac{2\pi }{3}$ , then

$$ \left\{ \begin{array}{c} 1<\frac{1}{\cos x(2\cos x-1)}<\frac{\pi }{2\left( x-\frac{\pi }{2}\right) } \\[5pt] -\frac{\sqrt[2]{3}\pi }{6\left( x-\frac{\pi }{2}\right) \left( \frac{2\pi }{3 }-x\right) }<\frac{1}{\cos x(2\cos x+1)}<-2 \end{array} \right. , $$

and if $\frac{2\pi }{3}<x<\pi$ , then

$$ \left\{ \begin{array}{c} \frac{1}{3}<\frac{1}{\cos x(2\cos x-1)}<1 \\[5pt] 1<\frac{1}{\cos x(2\cos x+1)}<\frac{4\sqrt[2]{3}}{3\left( x-\frac{2\pi }{3} \right) } \end{array} \right. . $$

Proof

• (1) If $0<x<\frac{\pi }{3}$, then from Lemma 5 we know that $1{\kern-1pt} >{\kern-1pt}\cos x{\kern-1pt}={\kern-1pt}\cos \left( \frac{\pi }{3}-\left( \frac{\pi }{3}-x\right) \right)>$ $\frac{1}{2}+\frac{\sqrt[2]{3}}{4}\left( \frac{\pi }{3}-x\right) >\frac{1}{2}$ , thus we have

  $$ \left\{ \begin{array}{c} \frac{1}{2}<\frac{1}{2}+\frac{\sqrt[2]{3}}{4}\left( \frac{\pi }{3}-x\right) <\cos x<1 \\[5pt] \frac{\sqrt[2]{3}}{2}\left( \frac{\pi }{3}-x\right) <2\cos x-1<1 \\ 2<2\cos x+1<3 \end{array} \right. . $$

  Hence

  $$ \left\{ \begin{array}{c} \frac{\sqrt[2]{3}}{4}\left( \frac{\pi }{3}-x\right) <\cos x(2\cos x-1)<1 \\[5pt] 1<\cos x(2\cos x+1)<3 \end{array} \right. , $$

  that is,

  $$ \left\{ \begin{array}{c} 1<\frac{1}{\cos x(2\cos x-1)}<\frac{4\sqrt[2]{3}}{3\left( \frac{\pi }{3} -x\right) } \\[5pt] \frac{1}{3}<\frac{1}{\cos x(2\cos x+1)}<1 \end{array} \right. . $$

• (2) If $\frac{\pi }{3}<x<\frac{\pi }{2}$, then from Lemma 4 and 5 we have

  $$ \cos x>\frac{2}{\pi }\left( \frac{\pi }{2}-x\right) >0 $$

  and

  $$ \cos x=\cos \left( \frac{\pi }{3}+\left( x-\frac{\pi }{3}\right) \right) < \frac{1}{2}-\frac{\sqrt[2]{3}}{2}\left( x-\frac{\pi }{3}\right) <\frac{1}{2}. $$

  Thus we have

  $$ \left\{ \begin{array}{c} 0<\frac{2}{\pi }\left( \frac{\pi }{2}-x\right) <\cos x<\frac{1}{2}-\frac{ \sqrt[2]{3}}{2}\left( x-\frac{\pi }{3}\right) <\frac{1}{2} \\[5pt] -1<2\cos x-1<-\sqrt[2]{3}\left( x-\frac{\pi }{3}\right) \\[5pt] 1<2\cos x+1<2 \end{array} \right. . $$

  Hence

  $$ \left\{ \begin{array}{c} -\frac{\sqrt[2]{3}\pi }{6\left( x-\frac{\pi }{3}\right) \left( \frac{\pi }{2} -x\right) }<\frac{1}{\cos x(2\cos x-1)}<-2 \\[5pt] 1<\frac{1}{\cos x(2\cos x+1)}<\frac{\pi }{2\left( \frac{\pi }{2}-x\right) } \end{array} \right. . $$

• (3) If $\frac{\pi }{2}<x<\frac{2\pi }{3}$, then $\frac{\pi }{3}<\pi -x<\frac{ \pi }{2}$. By Lemma 4 and 5 we have

  $$ -\cos x=\cos (\pi -x)>\frac{2}{\pi }\left( \frac{\pi }{2}-(\pi -x)\right) = \frac{2}{\pi }\left( x-\frac{\pi }{2}\right) >0, $$

  and

  $$ -\cos x=\cos (\pi -x)=\cos \left( \frac{\pi }{3}+\left( \frac{2\pi }{3} -x\right) \right) <\frac{1}{2}-\frac{\sqrt[2]{3}}{2}\left( \frac{2\pi }{3} -x\right) <\frac{1}{2}. $$

  Thus

  $$ \left\{ \begin{array}{c} -\frac{1}{2}<-\frac{1}{2}+\frac{\sqrt[2]{3}}{2}\left( \frac{2\pi }{3} -x\right) <\cos x<-\frac{2}{\pi }\left( x-\frac{\pi }{2}\right) <0 \\[5pt] -2<2\cos x-1<-1 \\ \sqrt[2]{3}\left( \frac{2\pi }{3}-x\right) <2\cos x+1<1 \end{array} \right. . $$

  Hence

  $$ \left\{ \begin{array}{c} 1<\frac{1}{\cos x(2\cos x-1)}<\frac{\pi }{2\left( x-\frac{\pi }{2}\right) } \\[5pt] -\frac{\sqrt[2]{3}\pi }{6\left( x-\frac{\pi }{2}\right) \left( \frac{2\pi }{3 }-x\right) }<\frac{1}{\cos x(2\cos x+1)}<-2 \end{array} \right. . $$

• (4) If $\frac{2\pi }{3}<x<\pi$, then $0<\pi -x<\frac{\pi }{3}$. By Lemma 5 we have

  $$ 1>-\cos x=\cos (\pi -x)=\cos \left( \frac{\pi }{3}-\left( x-\frac{2\pi }{3} \right) \right) >\frac{1}{2}+\frac{\sqrt[2]{3}}{4}\left( x-\frac{2\pi }{3} \right) >\frac{1}{2}. $$

  Thus

  $$ \left\{ \begin{array}{c} -1<\cos x<-\frac{1}{2}-\frac{\sqrt[2]{3}}{4}\left( x-\frac{2\pi }{3}\right) <-\frac{1}{2} \\[5pt] -3<2\cos x-1<-2 \\[5pt] -1<2\cos x+1<-\frac{\sqrt[2]{3}}{2}\left( x-\frac{2\pi }{3}\right) \end{array} \right. . $$

  Hence

  $$ \left\{ \begin{array}{c} \frac{1}{3}<\frac{1}{\cos x(2\cos x-1)}<1 \\[5pt] 1<\frac{1}{\cos x(2\cos x+1)}<\frac{4\sqrt[2]{3}}{3\left( x-\frac{2\pi }{3} \right) } \end{array} \right. . $$

□

Lemma 7

Let k,l be any positive integers. If 0 < a < k < l < b, then

$$ \underset{x=k}{\overset{l}{\sum }}\frac{1}{x-a}\leq \frac{1}{k-a}+\ln \frac{ l-a}{k-a}, $$

and

$$ \underset{x=k}{\overset{l}{\sum }}\frac{1}{b-x}\leq \frac{1}{b-l}+\ln \frac{ b-k}{b-l}. $$

Proof

The first result follows from

$$ \underset{x=k}{\overset{l}{\sum }}\frac{1}{x-a}\leq \frac{1}{k-a} +\int\nolimits_{k}^{l}\frac{dx}{x-a}\text{ and }\int \frac{dx}{x-a}=\ln (x-a), $$

and the second result follows from

$$ \underset{x=k}{\overset{l}{\sum }}\frac{1}{b-x}\leq \frac{1}{b-l} +\int\nolimits_{k}^{l}\frac{dx}{b-x}\text{ and }\int \frac{dx}{b-x}=-\ln (b-x). $$

□

Lemma 8

Let k,l be any positive integers, and b > a > 0 be any real numbers. If a < k < l < (a + b)/2, then

$$ \underset{x=k}{\overset{l}{\sum }}\frac{1}{(x-a)(b-x)}\leq \frac{1}{ (k-a)(b-k)}+\frac{1}{b-a}\cdot \ln \frac{(l-a)(b-k)}{(b-l)(k-a)}, $$

and if (a + b)/2 < k < l < b, then

$$ \underset{x=k}{\overset{l}{\sum }}\frac{1}{(x-a)(b-x)}\leq \frac{1}{ (l-a)(b-l)}+\frac{1}{b-a}\cdot \ln \frac{(l-a)(b-k)}{(b-l)(k-a)}. $$

Proof

Since $\frac{1}{(x-a)(b-x)}$ is monotonically decrease when a < x < (a + b)/2, and monotonically increase when (a + b)/2 < x < b, the result follows from

$$ \int \frac{dx}{(x-a)(b-x)}=\frac{1}{b-a}\cdot \ln \frac{(x-a)}{(b-x)} $$

and

$$ \left\{ \begin{array}{c} \underset{x=k}{\overset{l}{\sum }}\frac{1}{(x-a)(b-x)}\leq \frac{1}{ (k-a)(b-k)}+\int\nolimits_{k}^{l}\frac{dx}{(x-a)(b-x)},\text{ }\ \ \text{if \ \ }a<k<l<(a+b)/2, \\ \underset{x=k}{\overset{l}{\sum }}\frac{1}{(x-a)(b-x)}\leq \frac{1}{ (l-a)(b-l)}+\int\nolimits_{k}^{l}\frac{dx}{(x-a)(b-x)},\text{ \ \ if \ \ } (a+b)/2<k<l<b. \end{array} \right. $$

□

Proof

(Proof of Lemma 3) It’s clear that S(p − γ) = − S(γ), i.e. |S(p − γ)| = |S(γ)|. So we need only to show the case γ = (p + 3)/2. In this case, we have

$$ \begin{array}{rcl} S\left( \frac{p+3}{2}\right) &=&\!\!\overset{p-1}{\underset{c=1}{\sum }}\tan \left( \frac{\pi c(p+3)}{2p}\right) \tan \left( \frac{\pi c}{p}\right) = \overset{p-1}{\underset{c=1}{\sum }}\tan \left( \frac{\pi c}{2}+\frac{3\pi c }{2p}\right) \tan \left( \frac{\pi c}{p}\right) \\ &=&\!\!\underset{ \begin{array}{c} 1\!\leq\! c\!\leq\! p\!-\!1 \\ 2|c \end{array} }{\sum }\tan \left( \frac{3\pi c}{2p}\right) \tan \left( \frac{\pi c}{p} \right) -\underset{ \begin{array}{c} 1\!\leq\! c\!\leq\! p\!-\!1\!\! \\ 2\nmid c \end{array} }{\sum }\cot \left( \frac{3\pi c}{2p}\right) \tan \left( \frac{\pi c}{p} \right) \end{array} $$

On the other hand, it can be shown that $\tan \left( \frac{3\pi c}{2p} \right) \tan \left( \frac{\pi c}{p}\right) =-1+\frac{1}{\cos \left( \frac{ \pi c}{p}\right) \left( 2\cos \left( \frac{\pi c}{p}\right) -1\right) }$, and $\cot \left( \frac{3\pi c}{2p}\right) \tan \left( \frac{\pi c}{p}\right) =1+\frac{1}{\cos \left( \frac{\pi c}{p}\right) \left( 2\cos \left( \frac{\pi c}{p}\right) +1\right) }$, thus we have

$$\label{eqn11} \begin{array}{rcl} S\left( \frac{p+3}{2}\right) &=&-(p-1)+\underset{ \begin{array}{c} 1\leq c\leq p-1 \\ 2|c \end{array} }{\sum }\frac{1}{\cos \left( \frac{\pi c}{p}\right) \left( 2\cos \left( \frac{\pi c}{p}\right) -1\right) } \notag \\ &&-\underset{ \begin{array}{c} 1\leq c\leq p-1 \\ 2\nmid c \end{array} }{\sum }\frac{1}{\cos \left( \frac{\pi c}{p}\right) \left( 2\cos \left( \frac{\pi c}{p}\right) +1\right) } \notag \\ &=&-(p-1)+I_{e}-I_{o}, \end{array} $$

(10)

where

$$ \begin{array}{rcl} I_{e} &=&\underset{ \begin{array}{c} 1\leq c\leq p-1 \\ 2|c \end{array} }{\sum }\frac{1}{\cos \left( \frac{\pi c}{p}\right) \left( 2\cos \left( \frac{\pi c}{p}\right) -1\right) } \\ &=&\overset{(p-1)/2}{\underset{c=1}{\sum }}\frac{1}{\cos \left( \frac{2\pi c }{p}\right) \left( 2\cos \left( \frac{2\pi c}{p}\right) -1\right) }, \end{array} $$

and

$$ \begin{array}{rcl} I_{o} &=&\underset{ \begin{array}{c} 1\leq c\leq p-1 \\ 2\nmid c \end{array} }{\sum }\frac{1}{\cos \left( \frac{\pi c}{p}\right) \left( 2\cos \left( \frac{\pi c}{p}\right) +1\right) } \\ &=&\overset{(p-1)/2}{\underset{c=1}{\sum }}\frac{1}{\cos \left( \frac{\pi (2c-1)}{p}\right) \left( 2\cos \left( \frac{\pi (2c-1)}{p}\right) +1\right) } . \end{array} $$

Next we estimate I _e and I _o respectively.

Let $x=\frac{2\pi c}{p},$ then from Lemma 6 we know that

$$ \left\{ \begin{array}{cc} 1<\frac{1}{\cos x\left( 2\cos x-1\right) }<\frac{4\sqrt[2]{3}}{3\left( \frac{ \pi }{3}-\frac{2\pi c}{p}\right) }, & \text{if }1\leq c\leq \left\lfloor \frac{p}{6}\right\rfloor \text{,} \\[10pt] -\frac{\sqrt[2]{3}\pi }{6\left( \frac{2\pi c}{p}-\frac{\pi }{3}\right) \left( \frac{\pi }{2}-\frac{2\pi c}{p}\right) }<\frac{1}{\cos x\left( 2\cos x-1\right) }<-2,\text{ } & \text{if }\left\lceil \frac{p}{6}\right\rceil \leq c\leq \left\lfloor \frac{p}{4}\right\rfloor \text{,} \\[10pt] 1<\frac{1}{\cos x\left( 2\cos x-1\right) }<\frac{\pi }{2\left( \frac{2\pi c}{ p}-\frac{\pi }{2}\right) }, & \text{\ if }\left\lceil \frac{p}{4} \right\rceil \leq c\leq \left\lfloor \frac{p}{3}\right\rfloor \text{,} \\[10pt] \frac{1}{3}<\frac{1}{\cos x\left( 2\cos x-1\right) }<1, & \text{if } \left\lceil \frac{p}{3}\right\rceil \leq c\leq \frac{p-1}{2}\text{.} \end{array} \right. $$

That is,

$$ \left\{ \begin{array}{cc} 1<\frac{1}{\cos x\left( 2\cos x-1\right) }<\frac{2\sqrt[2]{3}p}{3\pi \left( \frac{p}{6}-c\right) }, & \text{if }1\leq c\leq \left\lfloor \frac{p}{6} \right\rfloor \text{,} \\[10pt] -\frac{\sqrt[2]{3}p^{2}}{24\pi \left( c-\frac{p}{6}\right) \left( \frac{p}{4} -c\right) }<\frac{1}{\cos x\left( 2\cos x-1\right) }<-2,\text{ } & \text{if } \left\lceil \frac{p}{6}\right\rceil \leq c\leq \left\lfloor \frac{p}{4} \right\rfloor \text{,} \\[10pt] 1<\frac{1}{\cos x\left( 2\cos x-1\right) }<\frac{p}{4\left( c-\frac{p}{4} \right) }, & \text{\ if }\left\lceil \frac{p}{4}\right\rceil \leq c\leq \left\lfloor \frac{p}{3}\right\rfloor \text{,} \\[10pt] \frac{1}{3}<\frac{1}{\cos x\left( 2\cos x-1\right) }<1, & \text{if } \left\lceil \frac{p}{3}\right\rceil \leq c\leq \frac{p-1}{2}\text{.} \end{array} \right. $$

Thus

$$ \left\{ \begin{array}{c} \left\lfloor \frac{p}{6}\right\rfloor <\overset{\left\lfloor \frac{p}{6} \right\rfloor }{\underset{c=1}{\sum }}\frac{1}{\cos \left( \frac{2\pi c}{p} \right) \left( 2\cos \left( \frac{2\pi c}{p}\right) -1\right) }<\frac{2\sqrt[ 2]{3}p}{3\pi }\cdot \left( \underset{c=1}{\overset{\left\lfloor \frac{p}{6} \right\rfloor }{\sum }}\frac{1}{\frac{p}{6}-c}\right) , \\[15pt] -\frac{\sqrt[2]{3}p^{2}}{24\pi }\cdot \left( \overset{\left\lfloor \frac{p}{4 }\right\rfloor }{\underset{c=\left\lceil \frac{p}{6}\right\rceil }{\sum }} \frac{1}{\left( c-\frac{p}{6}\right) \left( \frac{p}{4}-c\right) }\right) < \overset{\left\lfloor \frac{p}{4}\right\rfloor }{\underset{c=\left\lceil \frac{p}{6}\right\rceil }{\sum }}\frac{1}{\cos \left( \frac{2\pi c}{p} \right) \left( 2\cos \left( \frac{2\pi c}{p}\right) -1\right) }<-2\left( \left\lfloor \frac{p}{4}\right\rfloor -\left\lceil \frac{p}{6}\right\rceil +1\right) , \\[15pt] \left\lfloor \frac{p}{3}\right\rfloor -\left\lceil \frac{p}{4}\right\rceil +1<\overset{\left\lfloor \frac{p}{3}\right\rfloor }{\underset{c=\left\lceil \frac{p}{4}\right\rceil }{\sum }}\frac{1}{\cos \left( \frac{2\pi c}{p} \right) \left( 2\cos \left( \frac{2\pi c}{p}\right) -1\right) }<\frac{p}{4} \cdot \left( \overset{\left\lfloor \frac{p}{3}\right\rfloor }{\underset{ c=\left\lceil \frac{p}{4}\right\rceil }{\sum }}\frac{1}{c-\frac{p}{4}} \right) ,\text{ } \\[15pt] \frac{1}{3}\left( \frac{p-1}{2}-\left\lceil \frac{p}{3}\right\rceil +1\right) <\overset{\frac{p-1}{2}}{\underset{c=\left\lceil \frac{p}{3} \right\rceil }{\sum }}\frac{1}{\cos \left( \frac{2\pi c}{p}\right) \left( 2\cos \left( \frac{2\pi c}{p}\right) -1\right) }<\left( \frac{p-1}{2} -\left\lceil \frac{p}{3}\right\rceil +1\right) . \end{array} \right. $$

By Lemma 7 and 8, asymptotically we have

$$ -\frac{\sqrt[2]{3}}{\pi }p\ln p-2p-7<I_{e}<\left( \frac{2\sqrt[2]{3}}{3\pi }+ \frac{1}{4}\right) p\ln p+3p+1, \label{eqn12} $$

(11)

Let $y=\frac{\pi (2c-1)}{p},$ then from Lemma 6 we know that

$$ \left\{ \begin{array}{cc} \frac{1}{3}<\frac{1}{\cos y\left( 2\cos y+1\right) }<1, & \text{if }1\leq c\leq \left\lfloor \frac{p+3}{6}\right\rfloor \text{,} \\[10pt] 1<\frac{1}{\cos y\left( 2\cos y+1\right) }<\frac{\pi }{2\left( \frac{\pi }{2} -\frac{\pi (2c-1)}{p}\right) },\text{ } & \text{if }\left\lceil \frac{p+3}{6} \right\rceil \leq c\leq \left\lfloor \frac{p+2}{4}\right\rfloor \text{,} \\[10pt] -\frac{\sqrt[2]{3}\pi }{6\left( \frac{\pi (2c-1)}{p}-\frac{\pi }{2}\right) \left( \frac{2\pi }{3}-\frac{\pi (2c-1)}{p}\right) }<\frac{1}{\cos y\left( 2\cos y+1\right) }<-2, & \text{\ if }\left\lceil \frac{p+2}{4}\right\rceil \leq c\leq \left\lfloor \frac{2p+3}{6}\right\rfloor \text{,} \\[10pt] 1<\frac{1}{\cos y\left( 2\cos y+1\right) }<\frac{4\sqrt[2]{3}}{3\left( \frac{ \pi (2c-1)}{p}-\frac{2\pi }{3}\right) }, & \text{if }\left\lceil \frac{2p+3}{ 6}\right\rceil \leq c\leq \frac{p-1}{2}\text{.} \end{array} \right. $$

Thus

$$ \left\{ \begin{array}{c} \frac{1}{3}\cdot \left\lfloor \frac{p+3}{6}\right\rfloor <\overset{ \left\lfloor \frac{p+3}{6}\right\rfloor }{\underset{c=1}{\sum }}\frac{1}{ \cos \left( \frac{\pi (2c-1)}{p}\right) \left( 2\cos \left( \frac{\pi (2c-1) }{p}\right) +1\right) }<\left\lfloor \frac{p+3}{6}\right\rfloor , \\\\ \left\lfloor \frac{p+2}{4}\right\rfloor -\left\lceil \frac{p+3}{6} \right\rceil +1<\overset{\left\lfloor \frac{p+2}{4}\right\rfloor }{\underset{ c=\left\lceil \frac{p+3}{6}\right\rceil }{\sum }}\frac{1}{\cos \left( \frac{ \pi (2c-1)}{p}\right) \left( 2\cos \left( \frac{\pi (2c-1)}{p}\right) +1\right) }<\frac{p}{4}\cdot \left( \overset{\left\lfloor \frac{p+2}{4} \right\rfloor }{\underset{c=\left\lceil \frac{p+3}{6}\right\rceil }{\sum }} \frac{1}{\frac{p+2}{4}-c}\right) ,\text{ } \\\\ -\frac{\sqrt[2]{3}p^{2}}{24\pi }\cdot \left( \overset{\left\lfloor \frac{2p+3 }{6}\right\rfloor }{\underset{c=\left\lceil \frac{p+2}{4}\right\rceil }{ \sum }}\frac{1}{\left( c-\frac{p+2}{4}\right) \left( \frac{2p+3}{6} -c\right) }\right) <\overset{\left\lfloor \frac{2p+3}{6}\right\rfloor }{ \underset{c=\left\lceil \frac{p+2}{4}\right\rceil }{\sum }}\frac{1}{\cos \left( \frac{\pi (2c-1)}{p}\right) \left( 2\cos \left( \frac{\pi (2c-1)}{p} \right) +1\right) } \\\\ <-2\left( \left\lfloor \frac{2p+3}{6}\right\rfloor -\text{ }\left\lceil \frac{ p+2}{4}\right\rceil +1\right) , \\\\ \frac{p-1}{2}-\left\lceil \frac{2p+3}{6}\right\rceil +1<\overset{\frac{p-1}{2 }}{\underset{c=\left\lceil \frac{2p+3}{6}\right\rceil }{\sum }}\frac{1}{ \cos \left( \frac{\pi (2c-1)}{p}\right) \left( 2\cos \left( \frac{\pi (2c-1) }{p}\right) +1\right) }<\frac{2\sqrt[2]{3}p}{3\pi }\cdot \left( \overset{ \frac{p-1}{2}}{\underset{c=\left\lceil \frac{2p+3}{6}\right\rceil }{\sum }} \frac{1}{c-\frac{2p+3}{6}}\right) . \end{array} \right. $$

Similarly as above, by Lemma 7 and 8, we can also obtain

$$ -\frac{\sqrt[2]{3}}{\pi }p\ln p-2p-7<I_{o}<\left( \frac{2\sqrt[2]{3}}{3\pi }+ \frac{1}{4}\right) p\ln p+3p+1, \label{eqn13} $$

(12)

Combining (10), (11) and (12), we can get

$$ -\left( \frac{5\sqrt[2]{3}}{3\pi }+\frac{1}{4}\right) p\ln p-6p-7<S\left( \frac{p+3}{2}\right) <\left( \frac{5\sqrt[2]{3}}{3\pi }+\frac{1}{4}\right) p\ln p+4p+9. $$

Hence

$$ \left\vert S\left( \frac{p+3}{2}\right) \right\vert <\left( \frac{5\sqrt[2]{3 }}{3\pi }+\frac{1}{4}\right) p\ln p+6p+7. $$

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