New upper bounds for Ramanujan primes

TL;DR

This paper establishes new upper bounds for Ramanujan primes, showing that for large n, these primes are less than certain prime indices defined by a specific asymptotic formula involving logarithms.

Contribution

It provides a novel asymptotic upper bound for Ramanujan primes using prime index comparisons and logarithmic functions.

Findings

For large n, R_n < p_{[ ext{alpha}]} with specified alpha

The bounds depend on epsilon, j(n), and logarithmic growth rates

The results hold for all sufficiently large n

Abstract

For $n\ge 1$, the $n^{\rm th}$ Ramanujan prime is defined as the smallest positive integer $R_n$ such that for all $x\ge R_n$, the interval $(\frac{x}{2}, x]$ has at least $n$ primes. We show that for every $\epsilon>0$, there is a positive integer $N$ such that if $\alpha=2n\left(1+\dfrac{\log 2+\epsilon}{\log n+j(n)}\right)$, then $R_n< p_{[\alpha]}$ for all $n>N$, where $p_i$ is the $i^{\rm th}$ prime and $j(n)>0$ is any function that satisfies $j(n)\to \infty$ and $nj'(n)\to 0$.

Equations

$$R_n<p_{[2n(1+ϵ)]}\text{ for all }n>N.$$

$$s<2n\left(1+\frac{3}{\log n+\log (\log n)-4}\right)$$

$$g(n)=\frac{\log n+j(n)}{\log 2+ϵ}.$$

$$g(n)=\frac{\log n+{\log }_{2}n-\log 2-0.5}{\log 2+0.5}.$$

$$U(k)=k\left(\log k+{\log }_{2}k-1+\frac{{\log }_{2}k-2}{\log k}\right)$$

$$L(k)=k\left(\log k+{\log }_{2}k-1+\frac{{\log }_{2}k-2.1}{\log k}\right).$$

$$F(x,n)=U(x)-2L(x-n)=U(x)-2U(x-n)+2f(x-n)$$

$$G(n)=F(\alpha ,n),$$

$$U^{\prime }(x)=\log x+{\log }_{2}x-\frac{1}{\log x}+\frac{3}{{\log }^{2}x}-\frac{{\log }_{2}x}{{\log }^{2}x}+\frac{{\log }_{2}x}{\log x}$$

$$A=\log \left(\frac{\alpha }{\alpha -n}\right)+\log \left(\frac{\log (\alpha )}{\log (\alpha -n)}\right)+t(n),$$

$$\frac{1}{2}{G}^{\prime }(n)=U^{\prime }(\alpha )\left(1+{\left(\frac{n}{g}\right)}^{\prime }\right)+\left(1+2{\left(\frac{n}{g}\right)}^{\prime }\right)(f^{\prime }(\alpha -n)-U^{\prime }(\alpha -n))$$

$$L^{\prime }(x)=\log x+{\log }_{2}x+\frac{{\log }_{2}x}{\log x}-\frac{{\log }_{2}x}{{\log }^{2}x}-\frac{1.1}{\log x}+\frac{3.1}{{\log }^{2}x}$$

$$A-\log 2+f^{\prime }(\alpha -n)$$

$$=$$

$$T=\log \left(\frac{1+\frac{1}{g(n)}}{1+\frac{2}{g(n)}}\right)-\frac{{\log }_2(\alpha -n)}{\log (\alpha -n)}-\frac{1}{\log \alpha }-\frac{{\log }_2\alpha }{{\log }^2\alpha }+\frac{3}{{\log }^2\alpha }-\frac{3.1}{{\log }^2(\alpha -n)}<0$$

$$g(n)=\frac{\log n+j(n)}{\log 2+ϵ},$$

$$-2ϵ-\delta <G^{\prime }(n)<-2ϵ+\delta ,$$

$$\frac{1+\log n}{\log n(\log n+{\log }_{2}n-\log 2-ϵ)}<\frac{{ϵ}_1}{\log 2+ϵ}.$$

$$\frac{ng(n{)}^{\prime }}{g(n{)}^2}=\frac{(\log 2+ϵ)(1+\log n)}{\log n(\log n+{\log }_{2}n-\log 2-ϵ{)}^2}<\frac{{ϵ}_1}{\log n+{\log }_{2}n-\log 2-ϵ}.$$

$$A+f^{\prime }(\alpha -n)-\log 2<\log \left(\frac{\log (2.4n)}{\log n}\right)+\frac{{\log }_2(2.4n)}{\log (2n)}+\frac{1.1}{\log n}+\frac{{\log }_2(1.4n)}{{\log }^{2}n}.$$

$$\log \left(\frac{\log (2.4n)}{\log n}\right)<\frac{{ϵ}_2}{5}$$

$$\frac{{\log }_2(2.4n)}{\log (2n)}+\frac{1.1}{\log n}+\frac{{\log }_2(1.4n)}{{\log }^{2}n}<\frac{4{ϵ}_2}{5}.$$

$$\frac{A+f^{\prime }(\alpha -n)}{-A+U^{\prime }(\alpha -n)-2f^{\prime }(\alpha -n)}<\frac{\log 2+{ϵ}_2}{\log n+{\log }_{2}n-\log 2-{ϵ}_2}.$$

$$\frac{A+f^{\prime }(\alpha -n)}{-A+U^{\prime }(\alpha -n)-2f^{\prime }(\alpha -n)}+\frac{ng(n{)}^{\prime }}{g(n{)}^2}<\frac{\log 2+{ϵ}_1+{ϵ}_2}{\log n+{\log }_{2}n-\log 2-ϵ}=\frac{1}{g(n)}.$$

Full text

1 Introduction

For $n\geq 1$, the $n^{\rm th}$ Ramanujan prime is defined as the smallest positive integer $R_{n}$, such that for all $x\geq R_{n}$, the interval $(\frac{x}{2},x]$ has at least $n$ primes. Note that by the minimality condition, $R_{n}$ is prime and the interval $(\frac{R_{n}}{2},R_{n}]$ contains exactly $n$ primes. Let $R_{n}=p_{s}$, where $p_{i}$ denotes the $i^{\rm th}$ prime. Sondow [7] showed that $p_{2n}<R_{n}<p_{4n}$ for all $n$, and conjectured that $R_{n}<p_{3n}$ for all $n$. This conjecture was proved by Laishram [4], and the upper bound $p_{3n}$ improved by various authors ([1], [8]). Subsequently, Srinivasan [9] and Axler [1] improved these bounds by showing that for every $\epsilon>0$, there exists an integer $N$ such that

[TABLE]

Using the method in [9] (outlined below), a further improvement was presented by Srinivasan and Nicholson, who proved that

[TABLE]

for all $n>241$. The above result follows from a special case of our main theorem given below. Yang and Togbe [11], also used the method in [9], to give tight upper and lower bounds for $R_{n}$ for large $n$ (greater than $10^{300}$). For some interesting generalizations of Ramanujan primes the reader may refer to [2], [5] and [6].

The main idea in [9] is to define a function $F(x)$ that is decreasing for $x\geq 2n$ and that satisfies $F(s)>0$. Then, an $\alpha>2n$ is found such that $F(\alpha)<0$ for $n>N$, which would imply that $s<\alpha$ for $n>N$ given the decreasing nature of $F$. We employ a variation of this method, where we first show that $F(\alpha)$ is a decreasing function for $n>N$. Then we find an integer greater than $N$ for which $F(\alpha)<0$, which leads us to the desired result. Our main result is the following.

Theorem 1.1**.**

Let $R_{n}=p_{s}$ and $\epsilon>0$. Let $j(n)>0$ be a function such that $j(n)\to\infty$ and $nj^{\prime}(n)\to 0$ as $n\to\infty$ and let

[TABLE]

Then there exists a positive integer $N$ such that for all $n>N$, we have $s<\alpha$, where $\alpha=2n\left(1+\frac{1}{g(n)}\right)$.

In the following corollary we record a bound obtained with $\epsilon=0.5$, where $j(n)$ is chosen so as to minimize the number of calculations. Similar results can be given for smaller values of $\epsilon$ (with different $j(n)$) where the determination of $N$ depends solely on computational power.

Corollary 1.1**.**

Let $R_{n}=p_{s}$. Then for $n>43$ we have $s<2n\left(1+\frac{1}{g(n)}\right)$, where

[TABLE]

2 The basic functions and lemmas

We will use the following bounds for the $k^{\rm th}$ prime given by Dusart.

Lemma 2.1**.**

The following hold for the $k^{\rm th}$ prime $p_{k}$.

$p_{k}>k\left(\log k+\log_{2}k-1+\frac{\log_{2}k-2.1}{\log k}\right)$* for all $k\geq 3$.* 2. 2.

$p_{k}<k\left(\log k+\log_{2}k-1+\frac{\log_{2}k-2}{\log k}\right)$* for all $k\geq 688383$.*

Proof.

See [3] ∎

Let

[TABLE]

and

[TABLE]

Note that $U(x)=L(x)+f(x)$ where $f(x)=\dfrac{0.1x}{\log x}$. We define

[TABLE]

and

[TABLE]

where $\alpha=2n\left(1+\frac{1}{g(n)}\right)$ and $g(n)$ is a function that satisfies $g(n)\geq 1$ and $g(n)\to\infty$ as $n\to\infty$.

Lemma 2.2**.**

Let $R_{n}=p_{s}$. Then the following hold.

$p_{s-n}<\frac{1}{2}p_{s}.$** 2. 2.

$2n<\alpha<2.4n$* for all $n>43$.* 3. 3.

$F(x,n)$* is a decreasing function for all $x\geq 2n$ and $F(s,n)>0$ for $n\geq 688383$.*

Proof.

For parts 1 and 2 see [9, Lemma 2.1] and [9, Remark 2.1] respectively. For part 3 see [11]. ∎ ∎

The following lemma contains useful results that include an expression for the derivative $G^{\prime}(n)$ in terms of the function $U(x)$.

Lemma 2.3**.**

Let $A=U^{\prime}(\alpha)-U^{\prime}(\alpha-n)$. Then the following hold.

$A=A(n)\to\log 2$* as $n\to\infty$.* 2. 2.

$\frac{1}{2}G^{\prime}(n)=A+f^{\prime}(\alpha-n)+\left(\frac{n}{g(n)}\right)^{\prime}(A-U^{\prime}(\alpha-n)+2f^{\prime}(\alpha-n)).$** 3. 3.

$L^{\prime}(x)>\log x+\log_{2}x$* for $x>20$.* 4. 4.

$A+f^{\prime}(\alpha-n)-\log 2<\log\left(\frac{\log\alpha}{\log(\alpha-n)}\right)+\frac{\log_{2}\alpha}{\log\alpha}+\frac{1.1}{\log(\alpha-n)}+\frac{\log_{2}(\alpha-n)}{\log^{2}(\alpha-n)}$.

Proof.

We have

[TABLE]

and hence

[TABLE]

where $t(n)\to 0$ as $n\to\infty$. As $\alpha=2n\left(1+\frac{1}{g(n)}\right)$ and $g(n)\to\infty$, we have $A\to\log 2$.

For the second part of the lemma, $G(n)=U(\alpha)-2U(\alpha-n)+2f(\alpha-n)$, which gives $G^{\prime}(n)=U^{\prime}(\alpha)\alpha^{\prime}-2U^{\prime}(\alpha-n)(\alpha^{\prime}-1)+2f^{\prime}(\alpha-n)(\alpha^{\prime}-1)$. As $\alpha^{\prime}=2+2\left(\frac{n}{g(n)}\right)^{\prime}$, we have

[TABLE]

and the result follows by the definition of $A$.

For part 3 we have

[TABLE]

from which the claim follows as for $n>20$ we have $\frac{\log_{2}x}{\log x}-\frac{\log_{2}x}{\log^{2}x}-\frac{1.1}{\log x}>0$.

For the last part, we have

[TABLE]

where

[TABLE]

as $\frac{3}{\log^{2}\alpha}-\frac{3.1}{\log^{2}(\alpha-n)}<0$. ∎ ∎

3 Proofs of main results

The following lemma shows that $G^{\prime}(n)$ is a decreasing function for large $n$, which is crucial in the proof of Theorem 1.1.

Lemma 3.1**.**

Let $\epsilon>0$ and

[TABLE]

where $j(n)>0$ is a function that satisfies $j(n)\to\infty$ and $nj^{\prime}(n)\to 0$ as $n\to\infty$. Then $G^{\prime}(n)\to-2\epsilon$.

Proof.

We have $\left(\frac{n}{g(n)}\right)^{\prime}=\frac{(\log 2+\epsilon)(\log n+j(n)-1-nj^{\prime}(n))}{(\log n+j(n))^{2}}$ and therefore $\left(\frac{n}{g(n)}\right)^{\prime}\to 0$ as $n\to\infty$. By our assumption on $j(n)$ it follows that $\frac{j(n)}{\log n}\to 0$ which gives $\left(\frac{n}{g(n)}\right)^{\prime}\log(\alpha-n)\to\log 2+\epsilon$ (as $\frac{\log(\alpha-n)}{\log n}\to 1$). It is easy to see that $\left(\frac{n}{g(n)}\right)^{\prime}\log_{2}(\alpha-n)\to 0$. It follows that $\left(\frac{n}{g(n)}\right)^{\prime}U^{\prime}(\alpha-n)\to\log 2+\epsilon$ (see equation (1)). Lastly note that $f^{\prime}(x)\to 0$ as $x\to\infty$. The result follows now on using all the above and the fact that $A\to\log 2$ (Lemma 2.3 part 1) in part 2 of Lemma 2.3. ∎

Proof of Theorem 1.1 We will first show that there exists a positive integer $N$, such that $G(n)<0$ for $n>N$. We have $G^{\prime}(n)\to-2\epsilon$ by the lemma above, which means that if $0<\delta<2\epsilon$, then there exists an integer $M$, such that for all $n>M$ we have $|G^{\prime}(n)+2\epsilon|<\delta$, that is

[TABLE]

for all $n>M$. Let $a$ and $b$ be two integers such that $M<a<b$. Then $G(b)-G(a)=\int_{a}^{b}G^{\prime}(n)dn<(b-a)(-2\epsilon+\delta)<0$. If $a$ is fixed, it follows that $G(b)<G(a)+(b-a)(-2\epsilon+\delta)<0$ for large $b$. Therefore there exists a positive integer $N>M$, such that for all $n>N$, we have $G(n)=F(\alpha,n)<0$.

We may assume that $N>688383$ so that from Lemma 2.2, part 3 we have $F(s,n)>0$. Moreover, from the same lemma we have $F(x,n)$ is decreasing for $x\geq 2n$. As $s$ and $\alpha$ are both bigger than $2n$, we have $s<\alpha$ for $n>N$ and the result follows. ∎ ∎

Proof of Corollary

Let $\epsilon=\epsilon_{1}+\epsilon_{2}=0.5$. We will first show that for $n>688383$ we have $G^{\prime}(n)<0$.

Let $\epsilon_{1}=0.1$. It is easy to verify that for $n>688383$ we have

[TABLE]

It follows that for all $n>688383$

[TABLE]

Next, we will show that $A+f^{\prime}(\alpha-n)-\log 2<\epsilon_{2}$.

Using Lemma 2.3, part 4 and Lemma 2.2 part 2, we have

[TABLE]

Observe that for $n>36734$

[TABLE]

as $\log\left(\frac{\log(2.4n)}{\log n}\right)<\frac{\epsilon_{2}}{5}$ holds if $\frac{\log(2.4n)}{\log n}<e^{\frac{\epsilon_{2}}{5}}$, that is if $2.4n<n^{e^{\frac{\epsilon_{2}}{5}}}$. The above holds if $2.4<n^{e^{\frac{\epsilon_{2}}{5}-1}}$ or $n>36734$.

Computation yields that for $n>688383$

[TABLE]

From equations (3)-(5) we have $A+f^{\prime}(\alpha-n)-\log 2<\epsilon_{2}$. From Lemma 2.3 part 3, $L^{\prime}(\alpha-n)=U^{\prime}(\alpha-n)-f^{\prime}(\alpha-n)>\log(\alpha-n)+\log_{2}(\alpha-n)>\log n+\log_{2}n$ and hence for $n>688383$ we have

[TABLE]

As $\epsilon_{1}+\epsilon_{2}=\epsilon$, equations (2) and (6) give

[TABLE]

From Lemma 2.3, part 2, noting that $\left(\frac{n}{g(n)}\right)^{\prime}=\frac{1}{g(n)}-\frac{ng(n)^{\prime}}{g(n)^{2}}$, we have $G^{\prime}(n)<0$ for all $n>688383$. Also, $G(688383)<0$ and hence we conclude that $G(n)<0$ for $n>688383$.

From Lemma 2.2, part 3 we have $F(s,n)>0$ and $F(x,n)$ is decreasing for $x\geq 2n$. As $s$ and $\alpha$ are both bigger than $2n$, it follows that $s<\alpha$ for $n>688383$. That the result holds for $43<n\leq 688383$ is a simple calculation. ∎

Remark 3.1**.**

Similar results for lower bounds for $R_{n}$ can be given using $G(x,n)=L(x)-2U(x-n+1)$ instead of $F(x,n)$.