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Analogues of Cramer’s conjecture

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Abstract: In this post, I present my approach to Cramer’s conjecture on the gap between consecutive primes. However since the sequence of primes p_n are just a special of the sequence of numbers asymptotically equidistributed modulo one s_n, I have considered the general problem of estimating the gap between consecutive terms of s_n. Applying this general theory on prime numbers show why Cramer’s conjecture should possibly be true.

1. Introduction

One of the most famous open problems in the theory of primes is the gap between consecutive primes. Given two consecutive prime numbers p_{n+1} and p_n, how large can the difference g(n)=p_{n+1}-p_n be? The answer to this question has baffled mathematicians for almost a century; however continuous progress has been since 1920 when Harald Cramer’s proved on the assumption of the Riemann Hypothesis, that

p_{n+1} - p_n = O(p_n^{0.5}\ln p_n).

Cramer’s Conjecture: Cramer applied probabilistic methods to estimate an upper bound on g(n) and in 1936, he suggested that

\displaystyle{\limsup_{n \rightarrow \infty}\frac{p_{n+1} - p_n}{\ln ^2 p_n} = 1}.

This statement is known as the Cramer’s conjecture. Intuitively, this means the gaps between consecutive primes are always small, and it quantifies asymptotically just how small they can be. This conjecture has neither been proved nor disproved. In Cramer’s model one assumes that the probability that a natural number x is prime is 1/\ln x. This model was consistent with empirical and  in 1966, Gallagher showed that the Hardy-Littlewood conjecture on k-tuples was consistent with Cramer’s model.

Cramer-Granville’s Conjecture: Despite so much evidence in support of Cramer’s probabilistic model, in 1985, Helmut Maier proved a result that actually contradicts the predictions of Cramer’s. The problem with Cramer’s model is that it fails to take into account divisibility. Thus, for the primes p > 2, the probability that p+1 is prime is not 1/\ln(p + 1) as suggested by the Cramer model but rather 0 since p+1 is even. Further p+2 is always odd; therefore it is twice as likely to be prime as a random natural number. Thus n and n+2 being primes are not independent events. Based on Maier’s result, in 1995, Andrew Granville refined Cramer’s model and conjectured that

\displaystyle{\limsup_{n \rightarrow \infty}\frac{p_{n+1} - p_n}{\ln ^2 p_n} \ge 2e^{-\gamma}}

where \gamma is the Euler-Mascheroni constant. The modified statement

p_{n+1} - p_n < M\ln ^2 p_n

with M>1 is known as the Cramer-Granville’s Conjecture.

Firoozbakht’s conjecture:  In 1982, Iranian mathematician Farideh Firoozbakht the following conjecture which is lesser known but appears to have far reaching consequences in the study of the gap between consecutive primes that the sequence p_n^{1/n} decreases for all n \ge 1. This conjecture has been computationally verified for all primes up to 10^{12}. From the prime number theorem, it can be shown that Firoozbakht’s conjecture implies

p_{n+1} - p_n < \ln ^2 p_n - \ln p_n + 1.

This upper bound is not only stronger that the Cramer’s conjecture but it also contradicts Granville’s limit 2e^{-\gamma}. Thus either Cramer-Granville’s Conjecture is false or the Firoozbakht’s conjecture is false. But which one if false? We don’t know for sure. Currently the best unconditional result is far from either; g(n)=O(p_n^{0.535}) which was proved by R. Baker and G. Harman.

2. Sufficient conditions for Cramer’s conjecture

Lemma 1. Let c_n be the largest real such that

\displaystyle{p_n ^{\frac{1}{n}} > \Big(1-\frac{c_n \ln p_n}{n^2}\Big)p_{n+1} ^{\frac{1}{n+1}}}.

If there exists a natural number n_0 such that for all n>n_0,

  1. c_n \le 0 then Firoozbakht’s conjecture or,
  2. c_n is a positive constant then the Cramer-Granville conjecture is true.

This can easily be proved by finding the upper bound on p_{n+1} - p_n from the inequality mentioned in the lemma.

Let p(n) = p_n ^{1/n}. If p(n) > p(n+1), we define a_n as

\displaystyle{a_n = -\frac{1}{\ln n}\ln\Big(\frac{p(n)}{p(n+1)} - 1\Big)}.

Lemma 2If there exists a natural number n_0 such that for all n>n_0, a_n \ge 0 then Firoozbakht’s conjecture is true and this would imply that

p_{n+1} - p_n < \ln^2 p_n - \ln p_n + 1.

This lemma is also proved by assuming the hypothesis in the statement of the lemma and finding the upper bound on p_{n+1} - p_n. Thus we see that we can gain new insights in to the difference between consecutive primes by studying the quantity p(n) = p_n ^{1/n} and more specifically, we need a good estimation of the ratio p(n)/p(n+1).

3. Gap between consecutive terms of sequences asymptotically equidistributed mod 1

The following idea is drawn form Firoozbakht’s conjecture. Let s_n be asymptotically equidistributed modulo one and let h(n)=s_n^{1/n}. If we can find a lower bound on h(n)/h(n+1) than by algebraic manipulations, we can obtain an upper bound on the gap s_{n+1} - s_n. The first task is to find a good estimate of h(n)=s_n^{1/n}. Since s_n is asymptotically equidistributed mod 1, we have

\displaystyle{\lim_{n \rightarrow \infty}\frac{1}{n}\sum_{r \le n}f\Big(\frac{s_r}{s_n}\Big) = \int_{0}^{1}f(x)dx}.

Taking f(x) =x^{1/n}, and simplifying we have:

Lemma 3If s_n is asymptotically equidistributed mod 1 and \lim_{n \rightarrow \infty}s_n^{1/n} = 1, then

\displaystyle{s_n^{1/n} = \frac{n+1}{n^2}\sum_{r \le n}s_r^{\frac{1}{n}} + o(1)}.

Let j(n) = \frac{n+1}{n^2}\sum_{r \le n}s_r^{\frac{1}{n}} and let \epsilon _n = o(1) be the error term so that h(n) = j(n) + \epsilon _n.

Lemma 4. Assuming the hypothesis of Lemma 3, there exists a positive intger n_0 and a positive constant c such that for all n > n_0,

\displaystyle{\frac{j(n)}{j(n+1)} > 1 - \frac{c\ln s_n}{n^2}}.

Although h(n) = j(n) + o(1), Lemma 4 does not directly imply that

\displaystyle {\frac{h(n)}{h(n+1)} > 1 - \frac{c\ln s_n}{n^2}}.

should hold for all sufficiently large n. However if this relation were true than we would have been able to show that

\displaystyle{s_{n+1}-s_n < \frac{(2+\epsilon) s_n \ln s_n}{n} + 1}.

To be continued.


[1] http://arxiv.org/abs/1010.1399



1 Comment

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