The Riemann Zeta function, $\zeta \left(s\right)$, is defined by the series:

$\zeta \left(s\right)=\frac{1}{1^s}+\frac{1}{2^s}+\frac{1}{3^s}+\cdots =\underset{n=1}{\overset{\infty }{{\displaystyle \sum }}}{n}^{-s},ℝ\left(s\right)>1$ (1)

The importance of the relation between the Riemann hypothesis and primes is highlighted by a one-on-one correspondence between the non-trivial roots of the function and the primes.

The Riemann Hypothesis, proposed by Bernhard Riemann [1] in 1859, conjectures about the distribution of prime numbers and the relation to the Riemann Zeta function, $\zeta \left(s\right)$. It correlates the non-trivial zeros of the ζ-function with the primes if the zeros of the ζ-function have a real part, $\frac{1}{2}$. This hypothesis (RH) is crucial to the understanding of the distribution of primes.

The Robin criterion first specified Guy Robin [2] in 1984 relates the truth of the RH hypothesis to a direct correlation to the statement (Robin’s inequality)

$F_n=\frac{\sigma \left(n\right){\text{e}}^{-\gamma }}{n\log \log \left(n\right)}<1$ (2)

if and only if the set of numbers (I call them the Robin Integers, $R_n$):

$\begin{array}{l}{R}_n\in [3,4,5,6,8,9,10,12,16,18,20,24,30,36,48,60,\\ 72,84,120,180,240,360,720,840,2520,5040]\end{array}$

are the only integers that violate the inequality,

$F_n=\frac{\sigma \left(n\right){\text{e}}^{-\gamma }}{n\log \log \left(n\right)}<1,$ (3)

This paper also studies these functions and their interrelations and hence provides an objective insight into the beauty of these relations to the Riemann Hypothesis. Further, the importance of functions such as the Γ-functions, the Bernoulli functions, $B_m$, and their relationships to the ζ-function is also explored. The Bernoulli function is intimately related to the ζ-function by the relations:

$\zeta \left(2m\right)=\frac{2^{\left(2m-1\right)}{\pi }^{\left(2m\right)}\left|B_{2m}\right|}{\left(2m\right)!},\zeta \left(1-2m\right)=-\frac{B_{2m}}{2m}$ (4)

It is also very clear that the Γ-function, (Γ(z)), is intimately to Riemann’s ζ-function. The Γ-function was pioneered round about 1729, by the Swiss mathematician Leonhard Eular [3] . Eular’s deep insights into Γ-function led to numerous results in probability theory and in statistics. The particular form of the Γ-function used in this paper was developed by Carl Freidman Gauss [4] . Gauss’s work led to the famous reflection formula of the ζ-function, and numerous new results that will be presented in this paper. Certain invariant relations of the Γ-function are developed in this paper to show the connections of the Γ-function to other functions.

With this in mind, the present paper explores these relationships with some profound results. The distribution of primes and the confluence of the Robin criterion is clearly shown in Figures 1-3 , and Figure 4 . Results in prime numbers such as the twin prime conjecture are also discussed.

The ζ-function for primes is given by the prime, p, relation:

$\zeta \left(s\right)=\underset{p}{{\displaystyle \prod }}\frac{1}{1-p^{-s}}ℝ\left(s\right)>1$ (5)

There are key relations for primes that will be highlighted by this article. In general, these relations stem from the Γ-function and in particular, the Gauss’s Γ-function product formula. The intriguing relations of the Sum of Divisors function $\sigma \left(m\right)$, with the $\zeta$-function and the Riemann Hypothesis, were developed by several pioneers including Guy Robin [2] . The Sums of Divisors is defined as

$\sigma \left(m\right)=\underset{d|n}{{\displaystyle \sum }}d$ (6)

The uniformity of the function for primes, p, $\sigma \left(p^x\right)=\left(p^{x+1}-1\right)/\left(p-1\right)$, provides a way to link the $\sigma \left(m\right)$function to the Robin inequality and hence, the Riemann Hypothesis.

Note: If ${\displaystyle {\prod }_{k=1}^K{p}_k^{a_k}}$ is the prime factorization of m, where m, is an integer with K factors, and $a_k$ are particular numbers, then,

$\sigma \left(m\right)=\{\begin{array}{ll}\underset{k=1}{\overset{K}{{\displaystyle \prod }}}\frac{p_k^{{\left(a_k+1\right)}^r}}{p_k^r-1}\hfill & r>0\hfill \\ \underset{k=1}{\overset{K}{{\displaystyle \prod }}}\left(a_k+1\right)\hfill & r=0\hfill \end{array}$ (7)

where, $\gamma =0.5772156649\cdots$ is the Euler-constant.

Several attempts to refine the fundamental relations for $F_n$ as a referendum on the Riemann hypothesis have been made with great success. P. Solé and M. Planat [5] , showed in 1962 that, if $N_n$ is the primordial number of index-n, defined as the product of the first n primes, with $N_n={\displaystyle {\prod }_{k=1}^n{p}_k}$, then,

$\underset{n\to \infty }{\lim }{N}_n=\frac{{\text{e}}^{\gamma }}{\zeta \left(2\right)}~1.08,$ (8)

J. B. Roser ( [6] , Theorem 15], proved that there are inequalities between Euler’s totient function $\phi \left(n\right)$, and ${\text{e}}^{\gamma }n\log \log \left(n\right)$,

$\frac{n}{\phi \left(n\right)}\le {\text{e}}^{\gamma }n\log \log \left(n\right),n\ge 3$ (9)

In 1984 Heath-Brown [7] , showed that there are infinitely many positive integers m, for which the divisor function has $d\left(m\right)=d\left(m+B\right)$, where B is an integer. C.G. Pinner [8] also demonstrated that there are repeated values of the divisor function. Y.J. Chloe, N. Lichiardopol, and P. Moree [9] also studied the Robin criteria as did Sayak Chakrabarty [10] , and they found possible correlations of the repeated divisor function with Highly composite numbers. Jean-Louis Nicolas [11] , proved that there are an infinite number of values of $n\ge 3$, that satisfy the inequality (5).

In recent times, I focused on iterations of number theoretic functions. In the paper the Towering Zeta Function [12] , I show that the Zeta function can be iterated T-times, denoted as ${\zeta }^T\left(z\right)$ which tends to a constant when T approaches infinity:

$\begin{array}{c}\underset{T\to \infty }{\lim }{\zeta }^T\left(z\right)=\zeta \left(\underset{T\to \infty }{\lim }{\text{slog}}_n\left(z\right)\right)=\zeta \left(z_w\right)=z_w\\ =-0.29590500557213955\cdots \end{array}$

I also showed that the Zeta function only converges to the real constant when the complex part of conjugate Zeta function roots, T, are expressed in the exponential form:

${\text{e}}^{2i\theta }=\frac{\frac{1}{2}+iT}{\frac{1}{2}-iT},\text{where}\theta ={\tan }^{-1}2T,$

In 2024, I showed [13] that Perron’s formula given by:

$\psi \left(x\right)=x-\underset{\rho }{{\displaystyle \sum }}\frac{x^{\rho }}{\rho }-\frac{1}{2}\ln \left(1-x^{-2}\right)-\ln 2\pi$ (10)

Involves the complex roots $\rho$, of the Zeta function can be iterated to condense to the $k^{th}$ iterant:

$x-{\Psi }^k\left(x\right)=k\left(x-\Psi \left(x\right)\right)$ (11)

The rational functions of the roots of the Riemann zeta function are also expressed as rational Gamma functions. The behavior of the Gamma function for complex arguments restricts the validity of these rational arguments to arguments $\Re \left(s\right)>0$. As will be shown, this restriction also applies to the roots.

The first method that shows that the roots of the function must lie on the half-line is obtained by means of the Gauss-gamma function. It is shown that the gamma function has properties that impose conditions for the functional relations on the half-line.

The second method also uses the complex form of the Γ-function in [14] . In both these methods, an analysis of the relations that involve the Zeta functions and the gamma functions is used to conditionally show that the half-line plays an important role in many functions. In a first demonstration of the relationship of the Γ-function to the ½-line, invariants of the Γ-function are used to demonstrate how the ½ line features in a lot of results. In a second method, the relationship between ζ-function, the Bernoulli numbers, $B_n$ and Γ-function is explored to precipitate unexpected results about the rational relations of complex Γ-function, and its specific influence on Riemann conjecture. The rational functions of the non-trivial roots of the ζ-function happen to be expressible as relations between the Bernoulli numbers, $B_n$ and Γ-function and the complex roots. Both these methods produce indisputable verification that the non-trivial roots of the ζ-function lie on the half-line with no assumptions made. The process is a natural consequence of a special property of these functions.

Any function with a relational product $\left\{n\cdot y\right\}$, can be represented by the Sums of Divisor function, $\sigma \left(m\right)$. Here is a simple example:

$\log \left(n\cdot y\right)=\log n+\log y,$ (12)

Then, if $n\cdot y=m$, we can put $n=\sigma \left(m\right)$, $y=\frac{m}{\sigma \left(m\right)}$, and so,

$\log \left(m\right)=\log \sigma \left(m\right)+\log \frac{m}{\sigma \left(m\right)}$ (13)

Here is another example:

If $n\cdot y=m$, we can put $n=\sigma \left(m\right)$, $y=\frac{m}{\sigma \left(m\right)}$, and so, applied to the formula:

$\cos \left(n\cdot y\right)=\cos \left(y\right)\underset{k=1}{\overset{\frac{n-1}{2}}{{\displaystyle \prod }}}\left(1-\frac{{\sin }^2\left(y\right)}{{\sin }^2\frac{\left(2k-1\right)\pi }{2n}}\right),\left[n\text{is}\text{even}\right]$ (14)

By using the sum of divisor function, we find the relation:

$\cos \left(m\right)=\cos \left(\frac{m}{\sigma \left(m\right)}\right)\underset{k=1}{\overset{\frac{\sigma \left(m\right)-1}{2}}{{\displaystyle \prod }}}\left(1-\frac{{\sin }^2\left(\frac{m}{\sigma \left(m\right)}\right)}{{\sin }^2\frac{\left(2k-1\right)\pi }{2\sigma \left(m\right)}}\right),\left[\sigma \left(m\right)\text{is}\text{odd}\right]$ (15)

Interestingly, (14) and (15) differentiate between odd and even values of $\sigma \left(m\right)$. Since primes have $\sigma \left(p\right)=p+1$, an even number the relation (14) does not apply to primes. Since $p+1$ is always even except for the prime 2! Since $\sigma \left(2\right)=3$,

$\cos \left(2\right)=\cos \left(\frac{2}{3}\right)\underset{k=1}{\overset{1}{{\displaystyle \prod }}}\left(1-\frac{{\sin }^2\left(\frac{2}{3}\right)}{{\sin }^2\frac{\left(2k-1\right)\pi }{6}}\right),\left[\sigma \left(2\right)\text{is}\text{odd}\right]$ (16)

$\begin{array}{c}-0.4161468365\cdots =0.7858872608\cdots \left(1-\frac{0.3823812134\cdots }{0.2500000000}\right)\\ =-0.4161468365\cdots \end{array}$ (17)

The fact that the sum of divisor function $\sigma \left(m\right)$, can be manipulated this way leads to some interesting formulas that can produce significant and unexpected results.

For example, if we put:

$n=\sigma \left(m\right),x=\frac{n}{\sigma \left(m\right)},$ (18)

in the following trigonometric relations, we get:

$\begin{array}{l}\sin \left(m\right)=\sigma \left(m\right)\sin \left(\frac{m}{\sigma \left(m\right)}\right)\cos \left(\frac{m}{\sigma \left(m\right)}\right)\underset{k=1}{\overset{\frac{\sigma \left(m\right)-2}{2}}{{\displaystyle \prod }}}\left(1-\frac{{\sin }^2\left(\frac{m}{\sigma \left(m\right)}\right)}{{\sin }^2\frac{\pi k}{\sigma \left(m\right)}}\right)\\ \cos \left(m\right)=\underset{k=1}{\overset{\sigma \left(m\right)}{{\displaystyle \prod }}}\left(1-\frac{{\sin }^2\left(\frac{m}{2\sigma \left(m\right)}\right)}{{\sin }^2\frac{\left(2k-1\right)\pi }{2\sigma \left(m\right)}}\right)\end{array}\}\left[\sigma \left(m\right)\text{is}\text{even}\right]$ (19)

$\begin{array}{l}\sin \left(m\right)=\sigma \left(m\right)\sin \left(\frac{n}{\sigma \left(m\right)}\right)\underset{k=1}{\overset{\frac{\sigma \left(m\right)-2}{2}}{{\displaystyle \prod }}}\left(1-\frac{{\sin }^2\left(\frac{m}{\sigma \left(m\right)}\right)}{{\sin }^2\frac{k\pi }{\sigma \left(m\right)}}\right)\\ \cos \left(m\right)=\cos \left(\frac{m}{\sigma \left(m\right)}\right)\underset{k=1}{\overset{\frac{\sigma \left(m\right)-1}{2}}{{\displaystyle \prod }}}\left(1-\frac{{\sin }^2\left(\frac{m}{\sigma \left(m\right)}\right)}{{\sin }^2\frac{\left(2k-1\right)\pi }{2\sigma \left(m\right)}}\right)\end{array}\}\left[\sigma \left(m\right)\text{is}\text{odd}\right]$ (20)

Any factorizable integer function can be done this way. Of course, the construct restricts the functions we use to define the integers, m. Clearly, the relationship of integer functions to the Gauss Γ-function is clear. Again, we can represent the relationship of the Gauss gamma function ( [14] , p. 896] as:

$\Gamma \left(y\cdot n\right)={\left(2\pi \right)}^{\frac{1-y}{2}}{y}^{\left(n\cdot y\right)-\frac{1}{2}}\underset{k=0}{\overset{y-1}{{\displaystyle \prod }}}\Gamma \left(n+\frac{k}{y}\right)$ (21)

Let $g\left(z\right)$ be a function with k-factors, then for any given factor, $f_j\left(m\right)$, $j<k$,

$g\left(z\right)=f_j\left(z\right)\left({\displaystyle {\prod }_{n=1}^{j-1}{f}_n\left(z\right)}{\displaystyle {\prod }_{n=j+1}^k{f}_n\left(z\right)}\right)$

and,

$\Gamma \left(g\left(z\right)\right)={\left(2\pi \right)}^{\frac{1-\frac{f\left(z\right)}{f_j\left(m\right)}}{2}}{\left(\frac{f\left(z\right)}{f_j\left(m\right)}\right)}^{f\left(z\right)-\frac{1}{2}}\underset{k=0}{\overset{\frac{f\left(z\right)}{f_j\left(m\right)}-1}{{\displaystyle \prod }}}\Gamma \left(\frac{f\left(z\right)+k}{\frac{f\left(z\right)}{f_j\left(m\right)}}\right)\text{for}\text{any}j.$ (22)

The relation (22) is only true if $f_j\left(m\right)|g\left(z\right)\in \mathbb{Z}$. Observe that while $g\left(z\right)$, is uniquely defined as a real or complex function, $\frac{g\left(z\right)}{f_j\left(m\right)}$ is not uniquely defined as an integer function since many functions $f_j\left(m\right)$, can satisfy the relation (22) for some $g\left(z\right)$. However, there may exist multiple functions for which $\left\{\frac{g\left(z\right)}{f_a\left(m\right)},\frac{g\left(z\right)}{f_b\left(m\right)},\frac{g\left(z\right)}{f_c\left(m\right)},\cdots \right\}\in \mathbb{Z}$. In such a case, $g\left(z\right)$ has multiple integer-factors. It is interesting to note that the product form of two variables of the gamma function is behaving like a non-commutative function (I use that word loosely) since, $\Gamma \left(n\cdot y\right)\ne \Gamma \left(y\cdot n\right)$, when the roles of n, and y, are interchanged. This may provide a means to understand the sum of divisors function $\sigma \left(v\right)$. Suppose, we consider values of $n=\sigma \left(v\right)$, $y=\frac{m}{\sigma \left(v\right)}$ and, let $y\cdot n=m$, then,

$\Gamma \left(m\right)={\left(2\pi \right)}^{\frac{1-\sigma \left(v\right)}{2}}\sigma {\left(v\right)}^{m-\frac{1}{2}}\underset{k=0}{\overset{\sigma \left(v\right)-1}{{\displaystyle \prod }}}\Gamma \left(\frac{m+k}{\sigma \left(v\right)}\right)$ (23)

The relation (23) is invariant to the substation $\sigma \left(v\right)\to \sigma \left(v+\mu \right)$. An Interchange of the variables, gives:

$\Gamma \left(y\cdot n\right)={\left(2\pi \right)}^{\frac{1-y}{2}}{y}^{\left(n\cdot y\right)-\frac{1}{2}}\underset{k=0}{\overset{y-1}{{\displaystyle \prod }}}\Gamma \left(n+\frac{k}{y}\right)$ (24)

then,

$\Gamma \left(m\right)={\left(2\pi \right)}^{\frac{1-\frac{m}{\sigma \left(v\right)}}{2}}{\left(\frac{m}{\sigma \left(v\right)}\right)}^{m-\frac{1}{2}}\underset{k=0}{\overset{\frac{m}{\sigma \left(v\right)}-1}{{\displaystyle \prod }}}\Gamma \left(\sigma \left(v\right)+\frac{k}{\frac{m}{\sigma \left(v\right)}}\right)$ (25)

Note that the multiplicated variables $y\cdot n=m$, has not changed m, and, it is the multiplication operation that generates a distinct choice of $\sigma \left(v\right)|m$. There is a distinct difference between $\sigma \left(v\right)$, and $\sigma \left(v+\mu \right)$, since the expression is not invariant to substitution $\sigma \left(v\right)\to \sigma \left(v+\mu \right)$, unless $\left\{\frac{m}{\sigma \left(v\right)},\frac{m}{\sigma \left(v+\mu \right)}\right\}\in \mathbb{Z}$ for any $\mu$. A function $g\left(z\right)$, that may have factors that are integers and complex numbers can be represented as a product of its factors such that the following theorem applies.

THEOREM 1: Let $f_j\left(z\right)>0$, represent an integer factor of $g\left(z\right)$, then,

$\Gamma \left(g\left(z\right)\right)={\left(2\pi \right)}^{\frac{1-f_j\left(z\right)}{2}}{\left(f_j\left(z\right)\right)}^{g\left(z\right)-\frac{1}{2}}\underset{k=0}{\overset{f_j\left(z\right)-1}{{\displaystyle \prod }}}\Gamma \left(\frac{g\left(z\right)+k}{f_j\left(z\right)}\right)$ (26)

is invariant with respect to choices of any of the factors $f_v\left(z\right)$, of $g\left(z\right)$.

The significance of the THEOREM 1, is its consequences for prime numbers, and their relations to functions like the ζ-function and the sum of divisors function, $\sigma \left(m\right)$ and primes.

PROOF:

Let $f_j\left(z\right)$, be some j^th integer factor of k-factors a real or complex function $g\left(z\right)$. Then,

$g\left(z\right)={\displaystyle {\prod }_{n=1}^k{f}_n\left(z\right)}=f_j\left(z\right)\left({\displaystyle {\prod }_{n=1}^{j-1}{f}_n\left(z\right)}{\displaystyle {\prod }_{n=j+1}^k{f}_n\left(z\right)}\right)$ (27)

The Gauss gamma product formula is a simple relation given by:

$\Gamma \left(n\cdot y\right)={\left(2\pi \right)}^{\frac{1-n}{2}}{n}^{\left(n\cdot y\right)-\frac{1}{2}}\underset{k=0}{\overset{n-1}{{\displaystyle \prod }}}\Gamma \left(y+\frac{k}{n}\right)$ (28)

Then, since $f_j\left(z\right)$, is an integer-factor of $g\left(z\right)$, we have, putting $n=f_j\left(z\right)$,

$\Gamma \left(f_j\left(z\right)\left(\underset{n=1}{\overset{j-1}{{\displaystyle \prod }}}{f}_n\left(z\right)\underset{n=j+1}{\overset{k}{{\displaystyle \prod }}}{f}_n\left(z\right)\right)\right)={\left(2\pi \right)}^{\frac{1-f_j\left(z\right)}{2}}{\left(f_j\left(z\right)\right)}^{g\left(z\right)-\frac{1}{2}}\underset{k=0}{\overset{f_j\left(z\right)-1}{{\displaystyle \prod }}}\Gamma \left(\frac{g\left(z\right)+k}{f_j\left(z\right)}\right)$ (29)

If there any other integer factor labelled here $f_v\left(z\right),\in Z$, then, the substitution $f_j\left(z\right)\to f_v\left(z\right)$ leaves $\Gamma \left(g\left(z\right)\right)$ invariant.

$\Gamma \left(g\left(z\right)\right)={\left(2\pi \right)}^{\frac{1-f_v\left(z\right)}{2}}{\left(f_v\left(z\right)\right)}^{g\left(z\right)-\frac{1}{2}}\underset{k=0}{\overset{f_v\left(z\right)-1}{{\displaystyle \prod }}}\Gamma \left(\frac{g\left(z\right)+k}{f_v\left(z\right)}\right)$ (30)

THEOREM 2: If $\rho$ is a non-trivial root of the ζ-function, then,

$\underset{k=2}{\overset{\infty }{{\displaystyle \prod }}}\left(\frac{\zeta \left(1-k^{-\rho }\right)}{2\Gamma \left(k^{-\rho }\right)\zeta \left(k^{-\rho }\right)\cos \left(\frac{\pi \left(k^{-\rho }\right)}{2}\right)}\right)=2\pi$ (31)

PROOF:

From the reflection formula,

$2^{1-z}\Gamma \left(z\right)\zeta \left(z\right)\cos \left(\frac{\pi z}{2}\right)={\pi }^z\zeta \left(1-z\right)$ (32)

$\Gamma \left(g\left(z\right)\right)=\frac{{\left(2\pi \right)}^{g\left(z\right)}\zeta \left(1-g\left(z\right)\right)}{2\zeta \left(g\left(z\right)\right)\cos \left(\frac{\pi g\left(z\right)}{2}\right)}$ (33)

DEFINITION 1: Define the transformation

${\left(2\pi \right)}^{-g\left(z\right)}=\frac{\zeta \left(1-g\left(z\right)\right)}{2\Gamma \left(g\left(z\right)\right)\zeta \left(g\left(z\right)\right)\cos \left(\frac{\pi \left(g\left(z\right)\right)}{2}\right)}$ (34)

as the Pi-transformation.

Then, the left-hand-side is separable into real and complex parts.

If $g\left(z\right)={\displaystyle {\sum }_k{z}_k}$, i.e., the sum of k-terms, and functions, then, the Pi-transformation gives:

${\left(2\pi \right)}^{-g\left(z_k\right)}={\left(2\pi \right)}^{-{\displaystyle {\sum }_k{z}_k}}=\underset{k}{{\displaystyle \prod }}\left(\frac{\zeta \left(1-g\left(z_k\right)\right)}{2\Gamma \left(g\left(z_k\right)\right)\zeta \left(g\left(z_k\right)\right)\cos \left(\frac{\pi \left(g\left(z_k\right)\right)}{2}\right)}\right)$ (35)

For example, let $g\left(z\right)=\sigma +i\tau$, then, from (35),

${\left(2\pi \right)}^{-\left(\sigma +i\tau \right)}=\frac{\zeta \left(1-\sigma -i\tau \right)}{2\Gamma \left(\sigma +i\tau \right)\zeta \left(\sigma +i\tau \right)\cos \left(\frac{\pi \left(\sigma +i\tau \right)}{2}\right)}$ (36)

And let $g\left(z\right)=-\left(\sigma +i\tau \right)y$, where y is any function, then, from (33),

${\left(2\pi \right)}^{\left(\sigma +i\tau \right)y}=\frac{\zeta \left(1+\left(\sigma +i\tau \right)y\right)}{2\zeta \left(-\left(\sigma +i\tau \right)y\right)\cos \left(\frac{-\pi \left(\sigma +i\tau \right)y}{2}\right)\Gamma \left(-\left(\sigma +i\tau \right)y\right)}$ (37)

${\left(2\pi \right)}^{-\left(\rho +i\tau \right)+\left(\rho +i\tau \right)y}=\frac{\frac{\zeta \left(1-\sigma -i\tau \right)}{2\Gamma \left(\sigma +i\tau \right)\zeta \left(\sigma +i\tau \right)\cos \left(\frac{\pi \left(\sigma +i\tau \right)}{2}\right)}}{\frac{\zeta \left(1+\left(\sigma +i\tau \right)y\right)}{2\zeta \left(-\left(\sigma +i\tau \right)y\right)\cos \left(\frac{-\pi \left(\sigma +i\tau \right)y}{2}\right)\Gamma \left(-\left(\sigma +i\tau \right)y\right)}}$ (38)

As an example, put $y=\rho -i\tau$

${\left(2\pi \right)}^{{\sigma }^2+{\tau }^2-\sigma -i\tau }=\frac{\zeta \left(-{\sigma }^2-{\tau }^2\right)\zeta \left(1-\sigma -i\tau \right)\Gamma \left(-{\sigma }^2-{\tau }^2\right)\cos \left(\frac{\pi \left(-{\sigma }^2-{\tau }^2\right)}{2}\right)}{\zeta \left(1+{\sigma }^2+{\tau }^2\right)\zeta \left(1+\sigma +i\tau \right)\Gamma \left(\sigma +i\tau \right)\cos \left(\frac{\pi \left(\sigma +i\tau \right)}{2}\right)}$ (39)

The relation (39) is again separable into real and complex parts for products of n-functions since the left-hand side is separable into products, ${\displaystyle {\prod }_{k=1}^{n}F\left(\arg z_k\right)}$. It is obvious that the powers of $\left(2\pi \right)$ on the right-hand-side (LHS) determine completely the arguments on the left-hand-side (RHS), and in general, the form ${\left(2\pi \right)}^{g\left(z\right)}$ determines the RHS as having arguments

$\underset{k=1}{\overset{n}{{\displaystyle \prod }}}F\left(z_k\right)={\left(2\pi \right)}^{{\displaystyle {\sum }_{n=1}^n{z}_k}}$ (40)

As another example, taking the product relation for the arguments, $g\left(z\right)=\frac{1}{2}+iT$, $f\left(z\right)=\frac{1}{2}-iT$.

Then,

$\begin{array}{l}{\left(2\pi \right)}^{-\frac{1}{2}-iT}=2\left[\frac{\zeta \left(\frac{1}{2}-iT\right)}{\Gamma \left(\frac{1}{2}+iT\right)\zeta \left(\frac{1}{2}+iT\right)\cos \left(\frac{\pi \left(\frac{1}{2}+iT\right)}{2}\right)}\right],\\ {\left(2\pi \right)}^{-\frac{1}{2}+iT}=2\left[\frac{\zeta \left(\frac{1}{2}+iT\right)}{\Gamma \left(\frac{1}{2}-It\right)\zeta \left(\frac{1}{2}-iT\right)\cos \left(\frac{\pi \left(\frac{1}{2}-iT\right)}{2}\right)}\right]\end{array}$ (41)

$2\pi =4\left[\frac{\zeta \left(\frac{1}{2}-iT\right)}{\Gamma \left(\frac{1}{2}+iT\right)\zeta \left(\frac{1}{2}+iT\right)\cos \left(\frac{\pi \left(\frac{1}{2}+iT\right)}{2}\right)}\right]\left[\frac{\zeta \left(\frac{1}{2}+iT\right)}{\Gamma \left(\frac{1}{2}-It\right)\zeta \left(\frac{1}{2}-iT\right)\cos \left(\frac{\pi \left(\frac{1}{2}-iT\right)}{2}\right)}\right]$ (42)

The product of the relation (42) is independent of the ζ-function, and relies only on the relation:

$4\left\{\Gamma \left(\frac{1}{2}+iT\right)\Gamma \left(\frac{1}{2}-iT\right)\right\}\left(\cos \left(\frac{\pi \left(\frac{1}{2}+iT\right)}{2}\right)\cos \left(\frac{\pi \left(\frac{1}{2}-iT\right)}{2}\right)\right)=2\pi$ (43)

Now,

${\left(2\pi \right)}^{-g\left(z\right)}=\underset{k-1}{\overset{\infty }{{\displaystyle \prod }}}\left(\frac{\zeta \left(1-g\left(z\right)\right)}{2\Gamma \left(g\left(z\right)\right)\zeta \left(g\left(z\right)\right)\cos \left(\frac{\pi \left(g\left(z\right)\right)}{2}\right)}\right)$ (44)

Put $g\left(z\right)=\zeta \left(z\right)$, then,

${\left(2\pi \right)}^{-1-{\displaystyle {\sum }_{k=2}^{\infty }{k}^{-z}}}=\frac{1}{2\pi }\underset{k=2}{\overset{\infty }{{\displaystyle \prod }}}\left(\frac{\zeta \left(1-k^{-z}\right)}{2\Gamma \left(k^{-z}\right)\zeta \left(k^{-z}\right)\cos \left(\frac{\pi \left(k^{-z}\right)}{2}\right)}\right)={\left(2\pi \right)}^{-\zeta \left(z\right)}$ (45)

If $z=\rho$, is a root, then, $\zeta \left(\rho \right)=0$, the non-trivial roots follow:

$\underset{k=2}{\overset{\infty }{{\displaystyle \prod }}}\left(\frac{\zeta \left(1-k^{-\rho }\right)}{2\Gamma \left(k^{-\rho }\right)\zeta \left(k^{-\rho }\right)\cos \left(\frac{\pi \left(k^{-\rho }\right)}{2}\right)}\right)=2\pi$ (46)

Note that the conjugate relations on ½ line give,

$4\left[\frac{\zeta \left(\frac{1}{2}-iT\right)}{\Gamma \left(\frac{1}{2}+iT\right)\zeta \left(\frac{1}{2}+iT\right)\cos \left(\frac{\pi \left(\frac{1}{2}+iT\right)}{2}\right)}\right]\left[\frac{\zeta \left(\frac{1}{2}+iT\right)}{\Gamma \left(\frac{1}{2}-It\right)\zeta \left(\frac{1}{2}-iT\right)\cos \left(\frac{\pi \left(\frac{1}{2}-iT\right)}{2}\right)}\right]=2\pi$ (47)

Then, using the fact that

${\left(2\pi \right)}^{-g\left(z\right)}=\frac{\zeta \left(1-g\left(z\right)\right)}{2\zeta \left(g\left(z\right)\right)\cos \left(\frac{\pi g\left(z\right)}{2}\right)\left[\Gamma \left(g\left(z\right)\right)\right]}$ (48)

Put $g\left(z\right)=\zeta \left(z\right)-\frac{1}{z-1}$, and separate the terms for $\frac{1}{z-1}$, then,

$\begin{array}{l}{\left(2\pi \right)}^{-1-{\displaystyle {\sum }_{k=2}^{\infty }{k}^{-z}}+\frac{1}{z-1}}\\ =\frac{1}{2\pi }\left[\frac{\zeta \left(1+\frac{1}{z-1}\right)}{2\Gamma \left(-\frac{1}{z-1}\right)\zeta \left(-\frac{1}{z-1}\right)\cos \left(-\frac{\pi \left(\frac{1}{z-1}\right)}{2}\right)}\right]\underset{k=2}{\overset{\infty }{{\displaystyle \prod }}}\left(\frac{\zeta \left(1-k^{-z}\right)}{2\Gamma \left(k^{-z}\right)\zeta \left(k^{-z}\right)\cos \left(\frac{\pi \left(k^{-z}\right)}{2}\right)}\right)\\ =\frac{\zeta \left(1-\zeta \left(z\right)+\frac{1}{z-1}\right)}{2\Gamma \left(\zeta \left(z\right)-\frac{1}{z-1}\right)\zeta \left(\zeta \left(z\right)-\frac{1}{z-1}\right)\cos \left(\frac{\pi \left(\zeta \left(z\right)-\frac{1}{z-1}\right)}{2}\right)}\end{array}$ (49)

Then when $z=\rho$, a root of the ζ-function,

$\begin{array}{l}{\left(2\pi \right)}^{-1-{\displaystyle {\sum }_{k=2}^{\infty }{k}^{-z}}+\frac{1}{z-1}}\\ =\frac{1}{2\pi }\left[\frac{\zeta \left(\frac{\rho }{\rho -1}\right)}{\Gamma \left(\frac{1}{1-\rho }\right)\zeta \left(\frac{1}{1-\rho }\right)\cos \left(\frac{\pi }{2-2\rho }\right)}\right]\underset{k=2}{\overset{\infty }{{\displaystyle \prod }}}\left(\frac{\zeta \left(1-k^{-z}\right)}{2\Gamma \left(k^{-z}\right)\zeta \left(k^{-z}\right)\cos \left(\frac{\pi \left(k^{-z}\right)}{2}\right)}\right)\\ =\frac{\zeta \left(\frac{\rho }{\rho -1}\right)}{\Gamma \left(\frac{1}{1-\rho }\right)\zeta \left(\frac{1}{1-\rho }\right)\cos \left(\frac{\pi }{2-2\rho }\right)}\end{array}$ (50)

$\underset{k=2}{\overset{\infty }{{\displaystyle \prod }}}\left(\frac{\zeta \left(1-k^{-\rho }\right)}{2\Gamma \left(k^{-\rho }\right)\zeta \left(k^{-\rho }\right)\cos \left(\frac{\pi \left(k^{-\rho }\right)}{2}\right)}\right)=2\pi$ (51)

Hence the relation is uniquely satisfied by the roots $\rho$, of the ζ-function when $g\left(\rho \right)=\frac{1}{1-\rho }$.

${\left(2\pi \right)}^{\frac{1}{\rho -1}}=\frac{\zeta \left(\frac{\rho }{\rho -1}\right)}{2\Gamma \left(\frac{1}{\rho -1}\right)\zeta \left(\frac{1}{\rho -1}\right)\cos \left(\frac{\pi }{2}\left(\frac{1}{\rho -1}\right)\right)}$ (52)

Also note that $\underset{z\to \infty }{\lim }g\left(z\right)=\underset{z\to \infty }{\lim }\left(\zeta \left(z\right)-\frac{1}{z-1}\right)=\lambda \approx 0.5772156649\cdots$, where $\lambda$ is the Euler-Mascheroni constant [14] . Let us not separate the terms, then, the Cauchy principal value exists

$\underset{z\to \infty }{\lim }\left({\left(2\pi \right)}^{-\zeta \left(z\right)+\frac{1}{z-1}}\right)={\left(2\pi \right)}^{-\lambda }$ (53)

${\left(2\pi \right)}^{-\lambda }=\underset{z\to \infty }{\lim }\left(\frac{\zeta \left(1-\zeta \left(z\right)+\frac{1}{z-1}\right)}{2\Gamma \left(\zeta \left(z\right)-\frac{1}{z-1}\right)\zeta \left(\zeta \left(z\right)-\frac{1}{z-1}\right)\cos \left(\frac{\pi \left(\zeta \left(z\right)-\frac{1}{z-1}\right)}{2}\right)}\right)$ (54)

${\left(2\pi \right)}^{-\lambda }=\frac{\zeta \left(1-\lambda \right)}{2\Gamma \left(\lambda \right)\zeta \left(\lambda \right)\cos \left(\frac{\pi \left(\lambda \right)}{2}\right)}=0.34616095200041890772=C_{\infty }$ (55)

It must be noted that to a large number of decimal places, the constant $C_{\infty }\approx \pi -e-\lambda +\frac{1}{2}$:

Then, we can conclude that for all the Stieltjes numbers ${\lambda }_k$, follow:

$\boxed{{\left(2\pi \right)}^{-{\lambda }_k}=\frac{\zeta \left(1-{\lambda }_k\right)}{2\Gamma \left({\lambda }_k\right)\zeta \left({\lambda }_k\right)\cos \left(\frac{\pi \left({\lambda }_k\right)}{2}\right)}}$ (56)

From the Gauss Γ-product formula [14] , (21),

$\Gamma \left(m\right)={\left(2\pi \right)}^{\frac{1-\sigma \left(m\right)}{2}}\sigma {\left(m\right)}^{m-\frac{1}{2}}\underset{k=0}{\overset{\sigma \left(m\right)-1}{{\displaystyle \prod }}}\Gamma \left(\frac{m+k}{\sigma \left(m\right)}\right)$ (57)

It is clear that the following relations are equal,

${\left(2\pi \right)}^{\frac{1-m}{2}}{m}^{m-\frac{1}{2}}\underset{k=0}{\overset{m-1}{{\displaystyle \prod }}}\Gamma \left(\frac{m+k}{m}\right)={\left(2\pi \right)}^{\frac{1-\sigma \left(m\right)}{2}}\sigma {\left(m\right)}^{m-\frac{1}{2}}\underset{k=0}{\overset{\sigma \left(m\right)-1}{{\displaystyle \prod }}}\Gamma \left(\frac{m+k}{\sigma \left(m\right)}\right)$ (58)

Then, for all real numbers m,

${\left(2\pi \right)}^{\frac{\sigma \left(m\right)-m}{2}}{\left(\frac{m}{\sigma \left(m\right)}\right)}^{m-\frac{1}{2}}\frac{{{\displaystyle \prod }}_{k=0}^{m-1}\Gamma \left(\frac{m+k}{m}\right)}{{{\displaystyle \prod }}_{k=0}^{\sigma \left(m\right)-1}\Gamma \left(\frac{m+k}{\sigma \left(m\right)}\right)}=1$ (59)

Since, for all primes, $m=p$, $\sigma \left(p\right)=p+1$, from (59), we get for all primes, p:

$\sqrt{2\pi }\frac{p^{p-\frac{1}{2}}{{\displaystyle \prod }}_{k=0}^{p-1}\Gamma \left(\frac{p+k}{p}\right)}{\sigma {\left(p\right)}^{p-\frac{1}{2}}{{\displaystyle \prod }}_{k=0}^{\sigma \left(p\right)-1}\Gamma \left(\frac{p+k}{\sigma \left(p\right)}\right)}=1$ (60)

and so, (60) is not invariant to the substitutions, $\sigma \left(p\right)\to \sigma \left(p+j\right)$, unless $\left\{p,p+j\right\}\in \text{Primes}$.

In 1999, Pierre Dusart [15] showed that every k^th prime $p_k$ is greater than

$p_k>k\left(\ln p+\ln \ln p-1\right)\text{for}k\ge 2.$

Contributions by G. H. Hardy, Ramanujan [ [16] , §9, p. 141] also demonstrated bounds for primes. Leonhard Euler [17] presented to the St. Petersburg Academy on April 6, 1752. The paper “Observatio de summis divisorum, Novi Commentarii”. In that work, he presented a relation between the Triangular numbers and the Sums of Divisor function $\sigma \left(p\right)$. In this paper, I will show that the triangular numbers are indeed related to periods generated between integers as a result of Robin’s inequality.

THEOREM 3: There are an infinite number twin primes $\left\{p,p+2\right\}$ that have the restriction,

$\frac{{\left(\frac{\sigma \left(p\right)}{\sigma \left(p+2\right)}\right)}^2}{{\left(2\pi \right)}^{p+2-\frac{\sigma \left(p\right)}{2}-\frac{\sigma \left(p+2\right)}{2}}}\le 1,\forall \left\{p,p+2\right\}\in \text{twin}\text{primes}$ (61)

PROOF:

From the expression (61), put the prime relation for $p,p+2$,

$\left[\sqrt{2\pi }\frac{p^{p-\frac{1}{2}}{{\displaystyle \prod }}_{k=0}^{p-1}\Gamma \left(\frac{p+k}{p}\right)}{\sigma {\left(p\right)}^{p-\frac{1}{2}}{{\displaystyle \prod }}_{k=0}^{\sigma \left(p\right)-1}\Gamma \left(\frac{p+k}{\sigma \left(p\right)}\right)}\right]\left[\sqrt{2\pi }\frac{p^{p+2-\frac{1}{2}}{{\displaystyle \prod }}_{k=0}^{p+2-1}\Gamma \left(\frac{p+2+k}{p}\right)}{\sigma {\left(p+2\right)}^{p+2-\frac{1}{2}}{{\displaystyle \prod }}_{k=0}^{\sigma \left(p+2\right)-1}\Gamma \left(\frac{p+2+k}{\sigma \left(p+2\right)}\right)}\right]=1$ (62)

Then,

$\frac{{\left(\frac{\sigma \left(p\right)}{\sigma \left(p+2\right)}\right)}^2}{{\left(2\pi \right)}^{p+2-\frac{\sigma \left(p\right)}{2}-\frac{\sigma \left(p+2\right)}{2}}}\le 1,\forall \left\{p,p+2\right\}\in \text{twin}\text{primes}$ (63)

It is clear that if $p<\sigma \left(p\right),p+2<\sigma \left(p+2\right)$, so if p is not a twin prime,

$\begin{array}{l}\left(2\pi \right){\left(\frac{p}{\sigma \left(p\right)}\right)}^{p-\frac{1}{2}}{\left(\frac{p+2}{\sigma \left(p+2\right)}\right)}^{p+\frac{3}{2}}\frac{{{\displaystyle \prod }}_{k=0}^{p-1}\text{Γ}\left(\frac{p+k}{p}\right)}{{{\displaystyle \prod }}_{k=0}^{\sigma \left(p\right)-1}\text{Γ}\left(\frac{p+k}{\sigma \left(p\right)}\right)}\frac{{{\displaystyle \prod }}_{k=0}^{p+1}\text{Γ}\left(\frac{p+2+k}{p+2}\right)}{{{\displaystyle \prod }}_{k=0}^{\sigma \left(p+2\right)-1}\text{Γ}\left(\frac{p+2+k}{\sigma \left(p+2\right)}\right)}\\ \ll 1,\text{if}\left\{p,p+2\right\}\notin \text{twin}\text{primes}\end{array}$(64)

Hence for twin primes, $p,p+2,\frac{\sigma \left(p\right)}{\sigma \left(p+2\right)}=\frac{p+1}{p+3}<1$. Then, it follows that since (64) $\ll 1$, then,

$\begin{array}{l}\frac{\frac{\sigma \left(p\right)}{\sigma \left(p+2\right)}}{\left(2\pi \right){\left(\frac{p}{\sigma \left(p\right)}\right)}^{p-\frac{1}{2}}{\left(\frac{p+2}{\sigma \left(p+2\right)}\right)}^{p+\frac{3}{2}}\frac{{{\displaystyle \prod }}_{k=0}^{p-1}\Gamma \left(\frac{p+k}{p}\right)}{{{\displaystyle \prod }}_{k=0}^{\sigma \left(p\right)-1}\Gamma \left(\frac{p+k}{\sigma \left(p\right)}\right)}\frac{{{\displaystyle \prod }}_{k=0}^{p+1}\Gamma \left(\frac{p+2+k}{p+2}\right)}{{{\displaystyle \prod }}_{k=0}^{\sigma \left(p+2\right)-1}\Gamma \left(\frac{p+2+k}{\sigma \left(p+2\right)}\right)}}\\ \gg 1,\left\{p,p+2\right\}\notin \text{twin}\text{primes}\\ \frac{\frac{\sigma \left(p\right)}{\sigma \left(p+2\right)}}{\left(2\pi \right){\left(\frac{p}{\sigma \left(p\right)}\right)}^{p-\frac{1}{2}}{\left(\frac{p+2}{\sigma \left(p+2\right)}\right)}^{p+\frac{3}{2}}\frac{{{\displaystyle \prod }}_{k=0}^{p-1}\Gamma \left(\frac{p+k}{p}\right)}{{{\displaystyle \prod }}_{k=0}^{\sigma \left(p\right)-1}\Gamma \left(\frac{p+k}{\sigma \left(p\right)}\right)}\frac{{{\displaystyle \prod }}_{k=0}^{p+1}\Gamma \left(\frac{p+2+k}{p+2}\right)}{{{\displaystyle \prod }}_{k=0}^{\sigma \left(p+2\right)-1}\Gamma \left(\frac{p+2+k}{\sigma \left(p+2\right)}\right)}}\\ <1,\left\{p,p+2\right\}\in \text{twin}\text{primes}\end{array}$(65)

The relation (65) determines if any pair of integers, $\left\{p,p+2\right\}\in \text{twin}\text{primes}$exclusively.

Hence,

$\frac{\frac{\sigma \left(p\right)}{\sigma \left(p+2\right)}}{\left(2\pi \right){\left(\frac{p}{\sigma \left(p\right)}\right)}^{p-\frac{1}{2}}{\left(\frac{p+2}{\sigma \left(p+2\right)}\right)}^{p+\frac{3}{2}}\frac{{{\displaystyle \prod }}_{k=0}^{p-1}\Gamma \left(\frac{p+k}{p}\right)}{{{\displaystyle \prod }}_{k=0}^{\sigma \left(p\right)-1}\Gamma \left(\frac{p+k}{\sigma \left(p\right)}\right)}\frac{{{\displaystyle \prod }}_{k=0}^{p+1}\Gamma \left(\frac{p+2+k}{p+2}\right)}{{{\displaystyle \prod }}_{k=0}^{\sigma \left(p+2\right)-1}\Gamma \left(\frac{p+2+k}{\sigma \left(p+2\right)}\right)}}<1$ (66)

$\left(\frac{\sigma \left(p\right)}{\sigma \left(p+2\right)}\right)\frac{{\left(\sigma \left(p\right)\right)}^{p-\frac{1}{2}}{{\displaystyle \prod }}_{k=0}^{\sigma \left(p+2\right)-1}\Gamma \left(\frac{p+2+k}{\sigma \left(p+2\right)}\right){{\displaystyle \prod }}_{k=0}^{\sigma \left(p\right)-1}\Gamma \left(\frac{p+k}{\sigma \left(p\right)}\right)}{\left(2\pi \right){\left(p\right)}^{p-\frac{1}{2}}{{\displaystyle \prod }}_{k=0}^{p-1}\Gamma \left(\frac{p+k}{p}\right){\left(\frac{p+2}{\sigma \left(p+2\right)}\right)}^{p+\frac{3}{2}}{{\displaystyle \prod }}_{k=0}^{p+1}\Gamma \left(\frac{p+2+k}{p+2}\right)}<1$ (67)