Abstract

# Abstract

An adapted version of the zeta function is defined as $$\theta_m(s)$$, this is used to explore the sequence of prime numbers.

# Introduction

The Riemann zeta function [1] is $\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s} = \prod_{primes}\frac{1}{1-p^{-s}}$

which is also expressed as Euler’s result relating to primes $$p$$. An adaptation of this function can be made $\eta_{m}(s) = \sum_{n=1}^{m} \frac{1}{n^s}$

which for $$s=1$$ becomes the summation for harmonic numbers $$H_m$$. Or for any $$s$$ the generalised harmonic numbers $$H_{m}^{(s)}$$. The generalised harmonic numbers are equivalent to a difference of zeta functions $H_{m}^{(s)} = \zeta(s) - \zeta(s,m+1)$

where $$\zeta(s)$$ is the Riemann zeta function as defined above and $$\zeta(s,m+1)$$ is the Hurwitz zeta function [2] defined as $\zeta(s,a) = \sum_{n=0}^{\infty} \frac{1}{(n+a)^s}$

It is important to note that $$\zeta(1)$$ is divergent and has value $$\infty$$. However, $$H_{m}^{(1)}$$ will tend to $$\zeta(1)$$ as $$m \to \infty$$.

Our function of interest will be defined as $\theta_m(s) = \sum_{n=1}^{m} \bigg\lfloor{\frac{m}{n^s}}\bigg\rfloor$

With the brackets denoting the floor of each term independantly, then summed together. For these functions $$m$$ controls the number in question. The relevence to prime numbers should appear as this function is a measure of the divisibility of a number by all numbers up to that number.

Unlike the $$zeta$$ and $$H$$ functions, as $$m \to \infty$$, $$\AddPointHere$$ ...

The function use could be, for any number $$m \in \mathbb{Z}$$ $\theta_m(1) = \bigg\lfloor\frac{m}{1}\bigg\rfloor + \bigg\lfloor\frac{m}{2}\bigg\rfloor + ... + \bigg\lfloor\frac{m}{m}\bigg\rfloor$

The limit of the sum terminating at $$m$$ could have been $$\infty$$, the result of $$\theta$$ would be the same as the floor of any integer divided by a greater integer will be zero.

Analysing the sequence for different $$m$$, $\theta(1)_1 = \bigg\lfloor\frac{1}{1}\bigg\rfloor = 1\\ \theta(1)_2 = \bigg\lfloor\frac{2}{1}\bigg\rfloor + \bigg\lfloor\frac{2}{2}\bigg\rfloor = 3\\ \theta(1)_3 = \bigg\lfloor\frac{3}{1}\bigg\rfloor + \bigg\lfloor\frac{3}{2}\bigg\rfloor + \bigg\lfloor\frac{3}{3}\bigg\rfloor = 5\\ \theta(1)_4 = \bigg\lfloor\frac{4}{1}\bigg\rfloor + \bigg\lfloor\frac{4}{2}\bigg\rfloor + \bigg\lfloor\frac{4}{3}\bigg\rfloor + \bigg\lfloor\frac{4}{4}\bigg\rfloor = 8 \\ etc.$

Define, $$\theta(1)_0=0$$. Introduce a measure $\Delta\theta(1)_m = \theta(1)_m - \theta(1)_{m-1}$

Analysing the values as $$m$$ progresses $\begin{array}{| c | c | c |} \hline m & \theta(1)_m & \Delta\theta(1)_m \\ \hline 1 & 1 & 1 \\ 2 & 3 & 2 \\ 3 & 5 & 2 \\ 4 & 8 & 3 \\ 5 & 10 & 2 \\ 6 & 14 & 4 \\ 7 & 16 & 2 \\ 8 & 20 & 4 \\ 9 & 23 & 3 \\ 10 & 27 & 4 \\ 11 & 29 & 2 \\ 12 & 35 & 6 \\ 13 & 37 & 2 \\ 14 & 41 & 4 \\ 15 & 45 & 4 \\ 16 & 50 & 5 \\ 17 & 52 & 2 \\ 18 & 58 & 6 \\ 19 & 60 & 2 \\ 20 & 66 & 6 \\ \hline \end{array}$

From analysis of this table, all prime numbers have a $$\Delta\theta$$ value of 2! This could be used as a method to check the primality of a number. Moreover, $$\Delta\theta$$ appears to indicate the number of divisors a number has, relating this property to not only that number but also the number preceding it.

So from this function is is possible to say, for some number $$N$$, if that number is prime or if its not prime, how many divisors it has.

According to [3], the number of divisors a number of the form $$N = p_{1}^{a_1}p_{2}^{a_2}p_{3}^{a_3}...$$, that is, a product of it’s prime factors $$p_i$$ to some power $$a_i$$, then the number of divisors, d is $d(N) = (a_1+1)(a_2+1)(a_3+1)...$

In the notation of [4], the product of primes is over all primes, $N=\prod_{i=1}^{m} p_i^{c(N)_i},$

with $$m$$ the $$m^{th}$$ prime such that $$p_m \le N$$. Now the number of divisors is $d(N)=\prod_{i=1}^{m} (c(N)_i +1)$

Making $$d(N) \equiv \Delta\theta(1)_N$$ we have $\prod_{i=1}^{m} (c(N)_i +1) \equiv \theta_N(1) - \theta_{N-1}(1) \\ \prod_{i=1}^{m} (c(N)_i +1) \equiv \sum_{n=1}^{N} \bigg\lfloor{\frac{N}{n}}\bigg\rfloor - \sum_{n=1}^{N-1} \bigg\lfloor{\frac{N-1}{n}}\bigg\rfloor$

Also from [4] is a definition of the coeficcients $c(N)_i= \sum_{j=1}^{q} \delta_{N mod (p_i)^j}^{0}$

where $$q$$ is the largest $$c(N)$$, which is maximally bound by $$q = \lfloor log_2(N) \rfloor$$

We then have $\prod_{i=1}^{m} \bigg(\sum_{j=1}^{\lfloor log_2(N) \rfloor} \delta_{N mod (p_i)^j}^{0} +1 \bigg) = \sum_{n=1}^{N} \bigg\lfloor{\frac{N}{n}}\bigg\rfloor - \sum_{n=1}^{N-1} \bigg\lfloor{\frac{N-1}{n}}\bigg\rfloor$

# Prime Generation Algorithm

This result links to the patterns show in [5], an algorithm to find primes from a pixel map. The number of divisors for a number on the map is the number of lines intersected traversing across it. This means the function $$\Delta\theta(1)_N$$ is the sum across thhe diagonals of the map, compressing the algorithm from 2 dimensions to 1.

Because of this, it appears much quicker to generate primes from the $$\Delta\theta(1)_N$$ terms in fixed memory. Primes up to around $$50,000$$ were generated.