## A new result on the frequency with which Erdos and Turan inequalities occur

Author : MARTIN MAULHARDT

ABSTRACT.

In our previous two papers we have proved by a different and simpler method than the one used by Erdos and Turan that the inequality $p_{k+1} - p_k \leq p_k - p_{k-1}$ is satisfied infinitely many times in the set of prime numbers. We have also extend this property to any large set. That is, these inequalities are a consequence of the divergence of the series of the reciprocals of any large set. It is not a consequence of the fact that any number is a unique product of primes. In our second paper we continued the work of Erdos and Turan proving that the more general inequalities $p_{k+r} - p_k \leq p_k - p_{k-r}$ of which the case $r=1$ is just an example are satisfied infinitely many times for any large set (in particular the set of prime numbers) for every $r \in N$. Unfortunately Erdos and Turan did not consider the problem of the frequency with which these inequalities occur. We have found an answer and it is on this answer that this paper is all about.

Note : The reader might want to check our first two works on this same webpage if he does not feel familiar with some theorems and/or definitions.

1. Introduction and Basic Definitions

Definition 1. A triple of natural numbers $(a,b,c)$ is said to have the closer property if $c-b \leq b-a$.

For example, the triple $(7,11,13)$ has the closer property. So does the triple $(23,29,31)$

The convergence criteria and the three closer primes theorem of our first work of this webpage prove that :

Theorem 2. The set $P$ of prime numbers has infinitely many triples of consecutive prime numbers with the closer property.

The fact that these triples are infinite is a good result but it demands more precision. For example, if we put in correspondence all triples of consecutive prime numbers with the natural numbers i.e. we labeled : $T_1 = (2,3,5)$ $T_2 = (3,5,7)$ $T_3 = (5,7,11)$

.................................. $T_n = (p_n,p_{n+1},p_{n+2})$

some of these will have the closer property and some will not. It might happen that, starting at some point on, the triples with the closer property are very unusual. For example they might be only the followings : $T_{100} , T_{10,000} , T_{1,000,000} , T_{100,000,000} ........... T_{100^{n}} .........$ .

In this case the indexes of the triples with the closer property are in geometric rate.

Or it might happen that they are the followings : $T_{100} , T_{121} , T_{144} , T_{169} ........... T_{n^2}...........$.

In both cases they are infinite and the reader of this paper realizes at once that, in some sense, in the second case we have more triples with the closer property than in the first.

We now extend the definition of the closer property for some sequences of natural numbers.

Definition 3. Let $a_n$ be an infinite strictly increasing sequence of natural numbers  say $A = (a_1, a_2, \dots a_n \dots )$. We say that A has the closer property if there are infinite triples of consecutive elements of A with the closer property.

For example, the sequence, $a_n=(1,2,4,5,7,8, \dots 3n+1,3n+2, \dots)$ has the closer property. So does the sequence of prime numbers and any large set considered as a sequence.

2. Main result of this paper

We are now ready to state and prove the main result of this paper.

We will prove that in the sequence of prime numbers $P = (2,3,5, \dots ,p_n, \dots )$ the set of indexes $I$ with triples satisfying the closer property ALSO has the closer property. That is, there is no $n_0 \in N$ from which starting at that $n_0$ the indexes of the triples with the closer property separate more and more.

Let us consider the following sequence of increasing natural numbers : $a_n = (1, 3; 4, 6, 9; 10, 12, 15, 19; 20, 22, 25, 29, 34; \dots)$

This sequence if formed by the following law :

The differences between consecutive terms are $d_n = (2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 6,\dots)$

We now prove a Lemma related to this sequence $a_n$

Lemma 4. The series of the reciprocals of $a_n$ is convergent, i.e : $\sum\limits_{n=1}^{\infty}\frac{1}{a_n} < \infty$

Proof. We write the series in the explicit form $1 + \frac{1}{3} + \frac{1}{4} + \frac{1}{6} + \frac{1}{9} + \frac{1}{10} + \frac{1}{12} + \frac{1}{15} + \frac{1}{19} +\dots$.

Let us now associate its terms in the following way : $(1 + \frac{1}{3}) + (\frac{1}{4} + \frac{1}{6} + \frac{1}{9}) + (\frac{1}{10} + \frac{1}{12} + \frac{1}{15} + \frac{1}{19}) +\dots$.

We have 2 terms in the first parenthesis, 3 terms in the second parenthesis, 4 terms in the third parenthesis, etc. Using the comparison criteria we found that this series  sums less than the following series : $(1 + 1) + (\frac{1}{4} + \frac{1}{4} + \frac{1}{4}) + (\frac{1}{10} + \frac{1}{10} + \frac{1}{10} + \frac{1}{10}) +\dots =$. $2 + \frac{3}{4} + \frac{4}{10} + \frac{5}{20} +\dots +\frac{n+1}{\frac{n^3+3n^2+2n}{6}} +\dots =$ $\sum\limits_{n=1}^{\infty}\frac{6(n+1)}{n^3+3n^2+2n} < \infty$

and this concludes the proof of the lemma.

We are now ready to prove the following important theorem :

Theorem 5. Let $A=(a_1, a_2,\dots ,a_n,\dots )$ be such that $\sum\limits_{n=1}^{\infty}\frac{1}{a_n} = \infty$. Let $T_n= (a_n,a_{n+1},a_{n+2})$. Let the indexes of the infinitely many triples $T_n$ with the closer property be $I = (i_o,i_1,i_2, \dots ,i_n, \dots )$. Then $I$ also has the closer property, i.e.  there exists infinitely many triples $(i_n,i_{n+1},i_{n+2})$ such that $i_{n+2} - i_{n+1} \leq i_{n+1} - i_n$.

Proof. Suppose that starting at some $k \in N$ all indexes of triples with the closer property satisfy the inequality $i_{n+2} - i_{n+1} > i_{n+1} - i_n$. Let the triple $T_k$ be $(a_k,a_{k+1},a_{k+2})$. Consider the sequence of the $a_n$ starting at $k$. Let this sequence be : $a_k,a_{k+1},a_{k+2}, \dots$. (We assume $k>3$). Now $a_k \geq 1,a_{k+1}\geq 3,a_{k+2}\geq 4, a_{k+3}\geq 6,a_{k+4}\geq 9,a_{k+5}\geq 10, \dots$.

The reciprocals of this $a_k$ form a convergent series by comparing this series with the one of lemma 4. But the series of the reciprocals of the $a_n$ starting at $n=1$ is divergent. Absurd! Thus the set of the indexes of triples with the closer property ALSO has the closer property.

Corollary 6. Let $P$ be the set of prime numbers. Let the infinite triples of primes with the closer property be $T_{i_0},T_{i_1},T{i_2},\dots T_{i_n},\dots$. Then there exists infinitely many $i_n$ such that $i_{n+2}-i_{n+1}\leq i_{n+1}-i_n$.