The Gamma-function, denoted as ( $\Gamma \left(s\right)$), was first introduced by Swiss mathematician Leonhard Euler [1] in 1729. Euler’s deep insights into Γ-function led to numerous results that provide key insights into many fields of mathematics including Probability theory and Statistics. Other major contributions to the development of the Γ-function used in this paper were developed by Carl Freidman Gauss [2] . Gauss’s work led to the famous reflection formula of the $\zeta$-function, and numerous new results that will be presented in this paper. A key insight into the Γ-function is its multiplicative nature. So far, there has been little development in the additive representation of the Γ-function. The form of the Γ-function ( [3] , p. 895):

$\begin{array}{l}\Gamma \left(s\right)~z^{s-\frac{1}{2}}{\text{e}}^{-s}\sqrt{2\pi }\left\{1+\frac{1}{12z}+\frac{1}{288s^2}-\frac{139}{51840s^3}-\frac{571}{2488320s^4}+O\left(s^{-5}\right)\right\},\\ \left[\left|\arg s\right|<\pi \right]\end{array}$ (1)

For $s$ real and positive, the remainder of the series is less than the last term that is retained.

Similar series exists for $\ln \left(\text{Γ}\left(s\right)\right)$. It will be significant if other forms of these series can be found.

The product form of the Γ-function due to Gauss, provides further insights into many relations that will be developed in this paper. The product form is given by, ( [3] , p. 896):

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

Certain invariant relations of the product Γ-function will be developed in this paper to show the connections of the Γ-function to other functions, particularly the Riemann-Zeta function, denoted by $\zeta \left(s\right)$. The Γ-function will not be complete without the mention of the $\zeta$-function. The $\zeta$-function, is defined by the series:

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

The importance of the $\zeta$-function is its relation to the distribution of primes and the Riemann hypothesis. There is a one-on-one correspondence between the non-trivial roots of the function and the primes. The $\zeta$-function also has a product relation for primes $p$, given by ( [3] , p. 1037);

$\zeta \left(s\right)=\underset{p}{{\displaystyle \prod }}\left(\frac{1}{1-p^{-s}}\right),ℝ\left(s\right)>1$ (4)

Both the $\zeta$-function, and the Γ-function are factorable. These two functions are related by the $\zeta$-function reflection formula developed by Gauss given by ( [3] , p. 1038):

$\text{Γ}\left(\frac{s}{2}\right){\pi }^{-\frac{s}{2}}\zeta \left(s\right)=\text{Γ}\left(\frac{1-s}{2}\right){\pi }^{\frac{s-1}{2}}\zeta \left(1-s\right)$ (5)

These relations are well studied, and they provide a wealth of information in Number theory and many disciplines in Mathematics.

The Riemann Hypothesis, proposed by Bernhard Riemann [4] in 1859, conjectures about the distribution of prime numbers and their relation to the Riemann Zeta function, $\zeta \left(s\right)$. It correlates the non-trivial zeros of the $\zeta$-function with the primes if the zeros of the $\zeta$-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 [5] in 1984 relates the truth of the RH hypothesis to a direct correlation to the statement (Robin’s inequality)

$F_n=\frac{\sigma \left(n\right){\text{e}}^{-\gamma }}{n\log \log \left(n\right)}<1$ (6)

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,

$F_n=\frac{\sigma \left(n\right){\text{e}}^{-\gamma }}{n\log \log \left(n\right)}<1,$ (7)

We will deal with this function in its various forms in this paper. This paper also studies these functions and their interrelations and hence provides an objective insight to 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 $\zeta$-function is also explored.

The Bernoulli function is intimately related to the $\zeta$-function by the relations ( [3] , p. 1048):

$\zeta \left(2m\right)=\frac{2^{\left(2m-1\right)}{\pi }^{\left(2m\right)}\left|B_{2m}\right|}{\left(2m\right)!},\zeta \left(1-2m\right)=-\frac{B_{2m}}{2m}$ (8)

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 will be clearly shown. Results in prime numbers such as the twin prime conjecture will also be discussed. There are key relations for primes that will be highlighted in this article. In general, these relations stem from the $\zeta$-function, and 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 was developed by several pioneers including Guy Robin [5] . The Sums of Divisors is defined as

$\sigma \left(m\right)=\underset{d|n}{{\displaystyle \sum }}d$ (9)

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,

$\sigma \left(m\right)=\{\begin{array}{ll}\underset{k=1}{\overset{K}{{\displaystyle \prod }}}\frac{p_k^{{\left(a_k+1\right)}^r}}{p_k^r-1}\hfill & r>0\hfill \\ \underset{k=1}{\overset{K}{{\displaystyle \prod }}}\left(a_k+1\right)\hfill & r=0\hfill \end{array}$ (10)

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 [6] , 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,

$\underset{n\to \infty }{\lim }{N}_n=\frac{{\text{e}}^{\gamma }}{\zeta \left(2\right)}~1.08,$ (11)

J. B. Roser ( [7] , Theorem 15), proved that there are inequalities between Euler’s totient function $\phi \left(n\right)$, and ${\text{e}}^{\gamma }n\log \log \left(n\right)$,

$\frac{n}{\phi \left(n\right)}\le {\text{e}}^{\gamma }n\log \log \left(n\right),n\ge 3$ (12)

In 1984 Heath-Brown [8] , 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 [9] also demonstrated that there are repeated values of the divisor function. Y. J. Chloe, N. Lichiardopol, and P, Moree [10] also studied the Robin criteria as did Sayak Chakrabarty [11] , and they found possible correlations of the repeated divisor function with Highly composite numbers. Jean-Louis Nicolas [12] proved that there are an infinite number of values of $n\ge 3$, that satisfy the inequality (12).

In recent times, I focused on iterations of number theoretic functions. In the paper the Towering Zeta Function [13] , 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:

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

I also showed that the Zeta function only converge 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 [14] that Perron’s formula given by:

$\psi \left(x\right)=x-\underset{\rho }{{\displaystyle \sum }}\frac{x^{\rho }}{\rho }-\frac{1}{2}\ln \left(1-x^{-2}\right)-\ln 2\pi$ (13)

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

$x-{\Psi }^k\left(x\right)=k\left(x-\Psi \left(x\right)\right)$ (14)

Such insights could give us a significant understanding of some of the most complex problems in mathematics.

I first want to introduce the curious fact that 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:

$\log \left(n\cdot y\right)=\log n+\log y,$ (15)

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

$\log \left(m\right)=\log \sigma \left(m\right)+\log \frac{m}{\sigma \left(m\right)}$ (16)

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 ( [3] , p. 41):

$\cos \left(n\cdot y\right)=\cos \left(y\right)\underset{k=1}{\overset{\frac{n-1}{2}}{{\displaystyle \prod }}}\left(1-\frac{{\sin }^2\left(y\right)}{{\sin }^2\frac{\left(2k-1\right)\pi }{2n}}\right),\left[n\text{is}\text{even}\right]$ (17)

$\sin \left(n\cdot y\right)=n\sin \left(y\right)\cos \left(y\right)\underset{k=1}{\overset{\frac{n-1}{2}}{{\displaystyle \prod }}}\left(1-\frac{{\sin }^2\left(y\right)}{{\sin }^2\frac{k\pi }{n}}\right),\left[n\text{is}\text{even}\right]$ (18)

By using the sum of divisor function, $n=\sigma \left(m\right)$, $y=\frac{m}{\sigma \left(m\right)}$, we find the relation:

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

Interestingly, (17) and (19) for example, 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 (18) does not apply to primes. Since $p+1$ is always even except for the prime 2! Since $\sigma \left(2\right)=3$,

$\cos \left(2\right)=\cos \left(\frac{2}{3}\right)\underset{k=1}{\overset{1}{{\displaystyle \prod }}}\left(1-\frac{{\sin }^2\left(\frac{2}{3}\right)}{{\sin }^2\frac{\left(2k-1\right)\pi }{6}}\right),\left[\sigma \left(2\right)\text{is}\text{odd}\right]$ (20)

$\begin{array}{c}-0.4161468365\cdots =0.7858872608\cdots \left(1-\frac{0.3823812134\cdots }{0.2500000000}\right)\\ =-0.4161468365\cdots \end{array}$ (21)

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:

$n=\sigma \left(m\right),x=\frac{n}{\sigma \left(m\right)},$ (22)

in the following trigonometric relations, we get:

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

(23)

$\begin{array}{l}\sin \left(m\right)=\sigma \left(m\right)\sin \left(\frac{n}{\sigma \left(m\right)}\right)\underset{k=1}{\overset{\frac{\sigma \left(m\right)-2}{2}}{{\displaystyle \prod }}}\left(1-\frac{{\sin }^2\left(\frac{m}{\sigma \left(m\right)}\right)}{{\sin }^2\frac{k\pi }{\sigma \left(m\right)}}\right)\\ \cos \left(m\right)=\cos \left(\frac{m}{\sigma \left(m\right)}\right)\underset{k=1}{\overset{\frac{\sigma \left(m\right)-1}{2}}{{\displaystyle \prod }}}\left(1-\frac{{\sin }^2\left(\frac{m}{\sigma \left(m\right)}\right)}{{\sin }^2\frac{\left(2k-1\right)\pi }{2\sigma \left(m\right)}}\right)\end{array}\}\left[\sigma \left(m\right)\text{is}\text{odd}\right]$ (24)

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 ( [3] , p.896) as:

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

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

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

The relation (26) is only true if $f_j\left(m\right)\in \mathbb{Z}\left(integers\right).$ Observe that there may exist multiple integer factors of $g\left(z\right)$ that satisfy (26). 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, $\text{Γ}\left(n\cdot y\right)\ne \text{Γ}\left(y\cdot n\right)$, 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,

$\text{Γ}\left(m\right)={\left(2\pi \right)}^{\frac{1-\sigma \left(v\right)}{2}}\sigma {\left(v\right)}^{m-\frac{1}{2}}\underset{k=0}{\overset{\sigma \left(v\right)-1}{{\displaystyle \prod }}}\text{Γ}\left(\frac{m+k}{\sigma \left(v\right)}\right)$ (27)

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

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

then,

$\begin{array}{c}\text{Γ}\left(m\right)={\left(2\pi \right)}^{\frac{1-\sigma \left(v+\mu \right)}{2}}{\left(\frac{m}{\sigma \left(v+\mu \right)}\right)}^{m-\frac{1}{2}}\underset{k=0}{\overset{\sigma \left(v+\mu \right)-1}{{\displaystyle \prod }}}\text{Γ}\left(1+\frac{k}{\sigma \left(v+\mu \right)}\right)\\ \end{array}$ (29)

A function $g\left(z\right)$, that has integers and complex factors can be represented as a product of its factors such that the following Theorem 1 applies.

Theorem 1. Let $f_j\left(z\right)>0$, represent one integer factor of $g\left(z\right)$, then,

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

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

The significance of the Theorem 1 is its consequences for prime numbers, and their relations to functions like the $\zeta$-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,

$g\left(z\right)={\displaystyle {\prod }_{n=1}^k{f}_n\left(z\right)}=f_j\left(z\right)\left({\displaystyle {\prod }_{n=1}^{j-1}{f}_n\left(z\right)}{\displaystyle {\prod }_{n=j+1}^k{f}_n\left(z\right)}\right)$ (31)

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

$\text{Γ}\left(n\cdot y\right)={\left(2\pi \right)}^{\frac{1-n}{2}}{n}^{\left(n\cdot y\right)-\frac{1}{2}}\underset{k=0}{\overset{n-1}{{\displaystyle \prod }}}\text{Γ}\left(y+\frac{k}{n}\right)$ (32)

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

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

$\text{Γ}\left(g\left(z\right)\right)={\left(2\pi \right)}^{\frac{1-f_j\left(z\right)}{2}}{\left(f_j\left(z\right)\right)}^{g\left(z\right)-\frac{1}{2}}\underset{k=0}{\overset{f_j\left(z\right)-1}{{\displaystyle \prod }}}\text{Γ}\left(\frac{g\left(z\right)+k}{f_j\left(z\right)}\right)$ (33)

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 $\text{Γ}\left(g\left(z\right)\right)$ invariant. However, this is also true for any integer factors, $m$ of $f_v\left(z\right)$, then, for any $m$,

$\text{Γ}\left(g\left(z\right)\right)={\left(2\pi \right)}^{\frac{1-m}{2}}{m}^{g\left(z\right)-\frac{1}{2}}\underset{k=0}{\overset{m-1}{{\displaystyle \prod }}}\text{Γ}\left(\frac{g\left(z\right)+k}{m}\right)$ (34)

remains invariant to the substitutions $f_j\left(z\right)\to f_v\left(z\right)\to m$.

From the Gauss Γ-product formula,

$\text{Γ}\left(m\right)={\left(2\pi \right)}^{\frac{1-\sigma \left(m\right)}{2}}\sigma {\left(m\right)}^{m-\frac{1}{2}}\underset{k=0}{\overset{\sigma \left(m\right)-1}{{\displaystyle \prod }}}\text{Γ}\left(\frac{m+k}{\sigma \left(m\right)}\right)$ (35)

It is clear that the following relations are equal,

${\left(2\pi \right)}^{\frac{1-m}{2}}{m}^{m-\frac{1}{2}}\underset{k=0}{\overset{m-1}{{\displaystyle \prod }}}\text{Γ}\left(\frac{m+k}{m}\right)={\left(2\pi \right)}^{\frac{1-\sigma \left(m\right)}{2}}\sigma {\left(m\right)}^{m-\frac{1}{2}}\underset{k=0}{\overset{\sigma \left(m\right)-1}{{\displaystyle \prod }}}\text{Γ}\left(\frac{m+k}{\sigma \left(m\right)}\right)$ (36)

Then, for all real numbers $m$,

${\left(2\pi \right)}^{\frac{\sigma \left(m\right)-m}{2}}{\left(\frac{m}{\sigma \left(m\right)}\right)}^{m-\frac{1}{2}}\frac{{{\displaystyle \prod }}_{k=0}^{m-1}\text{Γ}\left(\frac{m+k}{m}\right)}{{{\displaystyle \prod }}_{k=0}^{\sigma \left(m\right)-1}\text{Γ}\left(\frac{m+k}{\sigma \left(m\right)}\right)}=1$ (37)

Since, for all primes, $m=p$, $\sigma \left(p\right)=p+1$, from (37), we get for all primes, $p$:

$\sqrt{2\pi }\frac{p^{p-\frac{1}{2}}{{\displaystyle \prod }}_{k=0}^{p-1}\text{Γ}\left(\frac{p+k}{p}\right)}{\sigma {\left(p\right)}^{p-\frac{1}{2}}{{\displaystyle \prod }}_{k=0}^{\sigma \left(p\right)-1}\text{Γ}\left(\frac{p+k}{\sigma \left(p\right)}\right)}=1$ (38)

and so, (38) is not invariant to the substitutions, $\sigma \left(p\right)\to \sigma \left(p+j\right)$, unless $\left\{p,p+j\right\}\in \text{primes}$.

The curious properties of the Γ-function can be made analogous to the divisor function when we study Heath Brown [15] . I will just highlight some relations that are of interest in mathematics. To further show how the Γ-function is related to primes, we consider a theorem by D.R. Heath Brown [15] on repeated values of the sums of the divisors, $\sigma \left(x\right)$.

Theorem 2. (D. R. Heath-Brown [15] )

For any positive integer N there exist N distinct natural numbers $a_n$ with the following properties. For $m\ne n$ write $\left|a_m-a_n\right|=d\left(x\right)$. Then $d\left(x\right)$ divides $a_m$ and $a_n$. Moreover,

$d\left(a_m\right)d\left(\frac{a_n}{d\left(x\right)}\right)=d\left(a_n\right)d\left(\frac{a_m}{d\left(x\right)}\right)$

Theorem 2 is the first clue that there is a finite number of integers N in the Robin set. Let us explore this further. We can generalize Heath Brown to the Γ-function as follows using the sums of divisors, $\sigma \left(x\right)$ in the following lemma 1.

Lemma 1: For $m\ne n$ write $\left|a_m-a_n\right|=\sigma \left(x\right)$. For any positive integer $K_{\sigma \left(x\right),a_m}$, there exist distinct natural numbers $\sigma \left(x\right),a_m$ with the following properties, $\sigma \left(x\right)|\left\{a_m,a_n\right\}$, moreover, the integers $K_{\sigma \left(x\right),a_m}$ are unique for values of $\sigma \left(x\right)$, and for the case $\sigma \left(x\right)=1$, the integers $a_n\in \text{primes}$, $p$, and $a_m=\sigma \left(p\right)$.

$\frac{\Gamma \left(a_m\right)\Gamma \left(\frac{a_n}{\sigma \left(x\right)}\right)}{\Gamma \left(a_n\right)\Gamma \left(\frac{a_m}{\sigma \left(x\right)}\right)}=K_{a_n,a_m},K_{a_n,a_m}\in Z$

Proof: Put $\left|a_m-a_n\right|=\sigma \left(x\right)$,

$\frac{\Gamma \left(a_m\right)\Gamma \left(\frac{a_n}{\sigma \left(x\right)}\right)}{\Gamma \left(a_n\right)\Gamma \left(\frac{a_m}{\sigma \left(x\right)}\right)}=\frac{\Gamma \left(\sigma \left(x\right)+a_n\right)\Gamma \left(\frac{a_n}{\sigma \left(x\right)}\right)}{\Gamma \left(a_n\right)\Gamma \left(\frac{\sigma \left(x\right)+a_n}{\sigma \left(x\right)}\right)}$ (39)

A study of (39) generates integer series for every $a_n$, and $\sigma \left(x\right)$, examples of which are given in Table 1 below.

All these integer sequences are well known. For example, the series for $a_n=3$, is the total number of occurrences of the consecutive pattern 1324 in all permutations of $\left[a_n+3\right]$. By computer analysis, (using Maple 2024), there are exactly N = 7 distinct natural numbers $a_n=\left[11,14,15,15,18,19,21\right]$, $a_m=\left[10,12,13,14,16,17,19\right]$ that satisfy the relation $a_m-a_n=\text{Γ}\left(x\right)$, with the greatest common divisor being $\text{Γ}\left(x\right)=1$, $x=\left\{1,2\right\}$. It is clear that the Heath-Brown’s Lemma 1 in [15] extends to a variety of relations including,

$\frac{\Gamma \left(a_m\right)\Gamma \left(\frac{a_n}{\sigma \left(x\right)}\right)}{\Gamma \left(a_n\right)\Gamma \left(\frac{a_m}{\sigma \left(x\right)}\right)}=K_{a_n,a_m},K_{a_n,a_m}\in Z$ (40)

and, the case of unity used by Heath Brown is just one example of Lemma 1. The trivial case is given by putting $a_m=\sigma \left(p\right)$, $a_n=p$, $\sigma \left(x\right)|a_m,a_n$, where $p$ is a prime. Then, $\sigma \left(x\right)=a_m-a_n=1$.

$\frac{\text{Γ}\left(a_m\right)\text{Γ}\left(\frac{a_n}{\sigma \left(x\right)}\right)}{\text{Γ}\left(a_n\right)\text{Γ}\left(\frac{a_m}{\sigma \left(x\right)}\right)}=1$ (41)

However, one must note that $\sigma \left(x\right)$, is not a cover for all integer values, since there are values $\sigma \left(x\right)\ne m,\left\{m\in \forall \mathbb{Z}\right\}$. To understand this, we can now use the general relation:

$\text{Γ}\left(a_m\right)={\left(2\pi \right)}^{\frac{1-\sigma \left(x\right)}{2}}{\left(\sigma \left(x\right)\right)}^{a_m-\frac{1}{2}}\underset{k=0}{\overset{\sigma \left(x\right)-1}{{\displaystyle \prod }}}\text{Γ}\left(\frac{a_m}{\sigma \left(x\right)}+\frac{k}{\sigma \left(x\right)}\right)$ (42)

$\text{Γ}\left(\frac{a_m}{\sigma \left(x\right)}\right)={\left(2\pi \right)}^{\frac{1-a_m}{2}}{\left(a_m\right)}^{\frac{a_m}{\sigma \left(x\right)}-\frac{1}{2}}\underset{k=0}{\overset{a_m-1}{{\displaystyle \prod }}}\text{Γ}\left(\frac{1}{\sigma \left(x\right)}+\frac{k}{a_m}\right)$ (43)

And dividing (42) by (43), we get:

$\frac{\text{Γ}\left(a_m\right)}{\text{Γ}\left(\frac{a_m}{\sigma \left(x\right)}\right)}=\frac{{\left(2\pi \right)}^{\frac{1-\sigma \left(x\right)}{2}}{\left(\sigma \left(x\right)\right)}^{a_m-\frac{1}{2}}{{\displaystyle \prod }}_{k=0}^{\sigma \left(p\right)-1}\text{Γ}\left(\frac{a_m}{\sigma \left(x\right)}+\frac{k}{\sigma \left(x\right)}\right)}{{\left(2\pi \right)}^{\frac{1-a_m}{2}}{\left(a_m\right)}^{\frac{a_m}{\sigma \left(x\right)}-\frac{1}{2}}{{\displaystyle \prod }}_{k=0}^{a_m-1}\text{Γ}\left(\frac{1}{\sigma \left(x\right)}+\frac{k}{a_m}\right)}$ (44)

Expanding,

$\begin{array}{c}\frac{\Gamma \left(a_m\right)\Gamma \left(\frac{a_n}{\sigma \left(x\right)}\right)}{\Gamma \left(a_n\right)\Gamma \left(\frac{a_m}{\sigma \left(x\right)}\right)}={\left(2\pi \right)}^{\frac{\sigma \left(x\right)-a_n}{2}}{\left(\sigma \left(x\right)\right)}^{\frac{\left(1+2a_m-2a_n\right)\sigma \left(x\right)-2a_m}{2\sigma \left(x\right)}}{\left(a_n\right)}^{\frac{2a_n-\sigma \left(x\right)}{2\sigma \left(x\right)}}\\ \times \frac{{{\displaystyle \prod }}_{k=0}^{\sigma \left(x\right)-1}\Gamma \left(\frac{a_m+k}{\sigma \left(x\right)}\right){{\displaystyle \prod }}_{k=0}^{a_n-1}\Gamma \left(\frac{a_n+k\sigma \left(x\right)}{a_n\sigma \left(x\right)}\right)}{{{\displaystyle \prod }}_{k=0}^{\sigma \left(x\right)-1}\Gamma \left(\frac{a_n+k}{\sigma \left(x\right)}\right){{\displaystyle \prod }}_{k=0}^{\sigma \left(x\right)-1}\Gamma \left(\frac{a_m+k\sigma \left(x\right)}{{\left(\sigma \left(x\right)\right)}^2}\right)}\end{array}$ (45)

Then, since $a_m-a_n=\sigma \left(x\right)$, and if $a_n=p$, $a_m=\sigma \left(p\right)$, $\sigma \left(x\right)=1$, then for primes $p$,

$1={\left(2\pi \right)}^{\frac{1}{2}-\frac{p}{2}}{p}^{p-\frac{1}{2}}\frac{1}{\text{Γ}\left(p\right)}\underset{k=0}{\overset{p-1}{{\displaystyle \prod }}}\text{Γ}\left(\frac{p+k}{p}\right)$ (46)

which can be obtained from the original Γ-product formula for primes.

$\sqrt{2\pi }\frac{p^{p-\frac{1}{2}}{{\displaystyle \prod }}_{k=0}^{p-1}\text{Γ}\left(\frac{p+k}{p}\right)}{\sigma {\left(p\right)}^{p-\frac{1}{2}}{{\displaystyle \prod }}_{k=0}^{\sigma \left(p\right)-1}\text{Γ}\left(\frac{p+k}{\sigma \left(p\right)}\right)}=1\left\{p\text{is}\text{prime}\right\}$ (47)

Combining the two relations (46) and (47), we get:

${\left(2\pi \right)}^{\frac{1}{2}-\frac{p}{2}}\frac{p^{p-\frac{1}{2}}}{\text{Γ}\left(p\right)}\underset{k=0}{\overset{p-1}{{\displaystyle \prod }}}\text{Γ}\left(\frac{p+k}{p}\right)=\sqrt{2\pi }\frac{p^{p-\frac{1}{2}}{{\displaystyle \prod }}_{k=0}^{p-1}\text{Γ}\left(\frac{p+k}{p}\right)}{\sigma {\left(p\right)}^{p-\frac{1}{2}}{{\displaystyle \prod }}_{k=0}^{\sigma \left(p\right)-1}\text{Γ}\left(\frac{p+k}{\sigma \left(p\right)}\right)}$ (48)

$\frac{1}{{\left(2\pi \right)}^{\frac{p}{2}}\text{Γ}\left(p\right)}=\frac{1}{\sigma {\left(p\right)}^{p-\frac{1}{2}}{{\displaystyle \prod }}_{k=0}^{\sigma \left(p\right)-1}\text{Γ}\left(\frac{p+k}{\sigma \left(p\right)}\right)}$ (49)

$\sigma {\left(p\right)}^{p-\frac{1}{2}}=\frac{{\left(2\pi \right)}^{\frac{p}{2}}\text{Γ}\left(p\right)}{{{\displaystyle \prod }}_{k=0}^{\sigma \left(p\right)-1}\text{Γ}\left(\frac{p+k}{\sigma \left(p\right)}\right)}$ (50)

$\frac{\sigma {\left(p\right)}^{p-\frac{1}{2}}}{{\left(2\pi \right)}^{\frac{p}{2}}\text{Γ}\left(p\right)}\underset{k=0}{\overset{\sigma \left(p\right)-1}{{\displaystyle \prod }}}\text{Γ}\left(\frac{p+k}{\sigma \left(p\right)}\right)=\{\begin{array}{ll}1,\hfill & p\text{is}\text{a}\text{prime}\hfill \\ K_p,\hfill & \text{otherwise}\hfill \end{array}$ (51)

Relation (51) follows a new relation for the Γ-product for primes:

$\boxed{\text{Γ}\left(p\right)={\left(2\pi \right)}^{\frac{-p}{2}}{\left(\sigma \left(p\right)\right)}^{p-\frac{1}{2}}\underset{k=0}{\overset{\sigma \left(p\right)-1}{{\displaystyle \prod }}}\text{Γ}\left(\frac{p+k}{\sigma \left(p\right)}\right),p\in \text{prime}}$ (52)

(52) only holds for primes. On the other hand, the general relation:

$\text{Γ}\left(n\right)={\left(2\pi \right)}^{-\frac{\left(1-\sigma \left(n\right)\right)}{2}}{\left(\sigma \left(n\right)\right)}^{n-\frac{1}{2}}\underset{k=0}{\overset{\sigma \left(n\right)-1}{{\displaystyle \prod }}}\text{Γ}\left(\frac{n+k}{\sigma \left(n\right)}\right)$ (53)

holds for all integer despite the fact that $\sigma \left(n\right)-1\ne n$, when $n$ is not a prime.

The significance of the Heath Brown [15] relation can be understood when one realizes that not all integers are sums of divisors. For example, the following integers are not sums of divisors, hence the series in Table 1 , should not include these integers, as values of $\sigma \left(x\right)$.

$\begin{array}{l}\sigma \left(x\right)\notin [2,5,9,10,11,16,17,19,21,22,23,25,26,27,29,33,34,35,37,\\ 41,43,45,46,47,49,50,51,52,53,55,58,59,61,64,\cdots ],\end{array}$

Hence, Table 1 cannot be valid for all integers, and must be replaced by Table 2 .

The restriction to numbers that are sums of divisors is important. Consider the relation:

$\text{Γ}\left(\sigma \left(v\right)\right)={\left(2\pi \right)}^{\frac{1-\sigma \left(v\right)}{2}}{\left(\sigma \left(v\right)\right)}^{\sigma \left(v\right)-\frac{1}{2}}\underset{k=0}{\overset{\sigma \left(v\right)-1}{{\displaystyle \prod }}}\text{Γ}\left(1+\frac{k}{\sigma \left(v\right)}\right)$ (54)

Surely, (54) does not cover the spectrum, $m\in \forall \mathbb{Z}$. Then, the solutions for which

$\frac{\text{Γ}\left(\sigma \left(a_m\right)\right)\text{Γ}\left(\frac{\sigma \left(a_n\right)}{\sigma \left(x\right)}\right)}{\text{Γ}\left(\sigma \left(a_n\right)\right)\text{Γ}\left(\frac{\sigma \left(a_m\right)}{\sigma \left(x\right)}\right)}=1$ (55)

holds true, belong to the set of numbers that are sums of divisors, i.e., all integers excluding the set:

$\begin{array}{c}{\sigma }^{\nexists }=[2,5,9,10,11,16,17,19,21,22,23,25,26,27,29,33,34,35,\\ 37,41,43,45,46,47,49,50,51,52,53,55,58,59,61,64,\cdots ]\end{array}$

The significance of this exercise is to show that the integers are somehow classified into different groups that contain special properties. For example, the integers that relate to those that comply with the Robin inequality, and those that do not. The same applies in the case of the $\sigma$-function.

From the reflection formula,

$2^{1-z}\text{Γ}\left(z\right)\zeta \left(z\right)\cos \left(\frac{\pi z}{2}\right)={\pi }^z\zeta \left(1-z\right)$ (56)

$\text{Γ}\left(g\left(z\right)\right)=\frac{{\left(2\pi \right)}^{g\left(z\right)}\zeta \left(1-g\left(z\right)\right)}{2\zeta \left(g\left(z\right)\right)\cos \left(\frac{\pi g\left(z\right)}{2}\right)}$ (57)

Definition 1: Define the $\pi$-transformation as:

${\left(2\pi \right)}^{-g\left(z\right)}=\frac{\zeta \left(1-g\left(z\right)\right)}{2\text{Γ}\left(g\left(z\right)\right)\zeta \left(g\left(z\right)\right)\cos \left(\frac{\pi \left(g\left(z\right)\right)}{2}\right)}$ (58)

Then, the left-hand-side is separable into real and complex parts. It is important to note that G.H. Hardy [16] , found this function related to prime counting functions. If $g\left(z\right)={\displaystyle {\sum }_k\left(z_k\right)}$, i.e., the sum of $k$-terms, and functions, then, the $\pi$-transformation gives:

${\left(2\pi \right)}^{-g\left(z_k\right)}={\left(2\pi \right)}^{-{\displaystyle {\sum }_k\left(z_k\right)}}=\underset{k}{{\displaystyle \prod }}\left(\frac{\zeta \left(1-g\left(z_k\right)\right)}{2\text{Γ}\left(g\left(z_k\right)\right)\zeta \left(g\left(z_k\right)\right)\cos \left(\frac{\pi \left(g\left(z_k\right)\right)}{2}\right)}\right)$ (59)

Example 1: Let $g\left(z\right)=\sigma +i\tau ,$ then, from (58),

${\left(2\pi \right)}^{-\left(\sigma +i\tau \right)}=\frac{\zeta \left(1-\sigma -i\tau \right)}{2\text{Γ}\left(\sigma +i\tau \right)\zeta \left(\sigma +i\tau \right)\cos \left(\frac{\pi \left(\sigma +i\tau \right)}{2}\right)}$ (60)

And let $g\left(z\right)=-\left(\sigma +i\tau \right)y,$ where $y$ is any function, then, from (58),

${\left(2\pi \right)}^{\left(\sigma +i\tau \right)y}=\frac{\zeta \left(1+\left(\sigma +i\tau \right)y\right)}{2\zeta \left(-\left(\sigma +i\tau \right)y\right)\cos \left(\frac{-\pi \left(\sigma +i\tau \right)y}{2}\right)\text{Γ}\left(-\left(\sigma +i\tau \right)y\right)}$ (61)

${\left(2\pi \right)}^{-\left(\rho +i\tau \right)+\left(\rho +i\tau \right)y}=\frac{\left[\frac{\zeta \left(1-\sigma -i\tau \right)}{2\text{Γ}\left(\sigma +i\tau \right)\zeta \left(\sigma +i\tau \right)\cos \left(\frac{\pi \left(\sigma +i\tau \right)}{2}\right)}\right]}{\left[\frac{\zeta \left(1+\left(\sigma +i\tau \right)y\right)}{2\zeta \left(-\left(\sigma +i\tau \right)y\right)\cos \left(\frac{-\pi \left(\sigma +i\tau \right)y}{2}\right)\text{Γ}\left(-\left(\sigma +i\tau \right)y\right)}\right]}$ (62)

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

${\left(2\pi \right)}^{{\sigma }^2+{\tau }^2-\sigma -i\tau }=\left[\frac{\zeta \left(-{\sigma }^2-{\tau }^2\right)\zeta \left(1-\sigma -i\tau \right)\text{Γ}\left(-{\sigma }^2-{\tau }^2\right)\cos \left(\frac{\pi \left(-{\sigma }^2-{\tau }^2\right)}{2}\right)}{\zeta \left(1+{\sigma }^2+{\tau }^2\right)\zeta \left(1+\sigma +i\tau \right)\text{Γ}\left(\sigma +i\tau \right)\cos \left(\frac{\pi \left(\sigma +i\tau \right)}{2}\right)}\right]$(63)

The relation (63) 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

$\underset{k=1}{\overset{n}{{\displaystyle \prod }}}F\left(z_k\right)={\left(2\pi \right)}^{{\displaystyle {\sum }_{n=1}^n\left(z_k\right)}}$ (64)

Example 2: Let $g\left(z\right)=\frac{1}{2}+iT,f\left(z\right)=\frac{1}{2}-iT$;