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