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I am a physicist looking for a PhD position in theoretical physics in the realm of quantum information or related topics in quantum mechanics concerned with fundamental properties like entanglement or superposition and validity of theoretical assumptions and predictions.

To live in an asymmetric world, you should better be symmetric. However if you live in a symmetric world, asymmetry suffices.

I have also a strong interest in number theory because the universe of (natural) numbers is similar to an enigma. It is like the prototyp of all theories of maximal stability and we learn from it fundamental rules how to encode information with minimal relative overhead.

Some results of my personal research (currently most of them is a collection of identities of number theoretic functions I found by myself unless otherwise mentioned which seem important for me, if you find errors feel free to send me a hint - i will apply the correction and a note):

## The illusion of the natural number

When we are dealing with a natural number ${\displaystyle n\in \mathbb {N} +1}$ by the fundamental theorem of arithmetic we are dealing obviously with prime numbers. We may decompose ${\displaystyle n=\prod _{i=1}^{\infty }p_{i}^{m_{i}}}$. In the linear representation we have ${\displaystyle n=\dots p_{i-1}\underbrace {p_{i}\dots p_{i}} _{m_{i}}p_{i+1}\dots }$ which means that we write down only prime numbers. However to quantify the multiplicity ${\displaystyle m_{i}}$ of each prime number ${\displaystyle p_{i}}$ we resort to natural numbers again. We may overcome this border by recursively decompose each multiplicity ${\displaystyle m_{i}}$ and switch to an exponential representation. Again we see only prime numbers, e.g. ${\displaystyle 256=2^{2^{3}}}$, ${\displaystyle 455625=3^{2*3}5^{2^{2}}}$ but if we want to quantify the exponential levels where the primes appear we use again natural numbers. We might say in the last example that the multiplicity for ${\displaystyle p_{2}=3}$ is ${\displaystyle m_{2}=6}$ and the level is ${\displaystyle l_{2}=2}$. We can now represent every level ${\displaystyle l_{i}}$ again in an exponential representation and also every other quantification of the number ${\displaystyle n}$ like the number of distinct prime numbers in ${\displaystyle n}$ and the number of levels of levels and so on. We say that a representation of a natural number is in prime number language if all of its quantifications are closed under prime numbers meaning that they are expressable only with prime numbers (or - including the ${\displaystyle 1}$ - the set of smallest coprime numbers). The general question is now if there is a system that quantifies every natural number in prime number language. If it exists then a natural number would be like an illusion. It seems natural for this question to be answered that square-free numbers play a critical role in this game and also the number of quantifiable properties of a natural number ${\displaystyle n}$. We now try to make the statement precise which serves us also as a check of the problem if it is well posed: Let ${\displaystyle {\text{P}}}$ be the set of all quantifiable properties of a natural number so that ${\displaystyle \forall P\in {\text{P}}\colon \forall n\in \mathbb {N} +1\colon P\left(n\right)\in \mathbb {N} +1}$ is a quantification of ${\displaystyle n}$. It is clear that if ${\displaystyle \forall P\in {\text{P}}\colon \forall n\in \mathbb {N} +1\colon P\left(n\right)\in \mathbb {P} \cup \left\{1\right\}}$ holds we would be done but this is in general not the case. If ${\displaystyle P\left(n\right)\not \in \mathbb {P} \cup \left\{1\right\}}$ we look at ${\displaystyle \forall Q\in {\text{P}}\colon \left(Q\circ P\right)\left(n\right)}$. A solution to the illusion problem would be to find ${\displaystyle {\text{P}}}$ such that there is always a finite chain of composition of elements from ${\displaystyle {\text{P}}}$ which evaluates to an element of ${\displaystyle \mathbb {P} \cup \left\{1\right\}}$.

Definition (Basis of a semi-group): Let ${\displaystyle \left({\text{G}},\circ \right)}$ be a countable semi-group and ${\displaystyle {\text{B}}\subset {\text{G}}}$ with ${\displaystyle \left|{\text{B}}\right|\leq \aleph _{0}}$. Then ${\displaystyle {\text{B}}}$ is called a basis for ${\displaystyle {\text{G}}}$ iff

${\displaystyle {\text{G}}=<{\text{B}}>_{\circ }}$.

Example: ${\displaystyle <\mathbb {P} \cup \left\{1\right\}>_{*}=\mathbb {N} +1}$

${\displaystyle <\left\{\emptyset \right\}\cup \cup _{p\in \mathbb {P} }\left\{\left\{p\right\}\right\}>_{\cup }\cong \left\{n\in \mathbb {N} +1|\mu \left(n\right)^{2}=1\right\}}$

Remark: Both examples are commutative and include a neutral element which makes them a commutative monoid. Please note that in the last example the square-free numbers form a semi-group if the multiplicity ${\displaystyle m_{i}}$ saturates to ${\displaystyle 1}$ under multiplication. Thus we find a morphism between the lhs and the rhs.

Definition (Minimal basis):

${\displaystyle \forall b\in {\text{B}}_{\min }\colon <{\text{B}}_{\min }\setminus \left\{b\right\}>_{\circ }\neq <{\text{B}}_{\min }>_{\circ }}$

Lemma (Semi-group): Let ${\displaystyle \left({\text{G}},\circ \right)}$ be a countable semi-group such that ${\displaystyle {\text{G}}=<{\text{G}}>_{\circ }}$. Then ${\displaystyle \exists {\text{B}}\subset {\text{G}}\colon {\text{G}}=<{\text{B}}>_{\circ }}$.

Proof: Let ${\displaystyle \forall {\text{B}}\subset {\text{G}}\colon {\text{G}}\neq <{\text{B}}>_{\circ }}$. Then ${\displaystyle \exists g_{0},g_{1},g\in {\text{G}}\colon g=g_{0}\circ g_{1}}$. But then we can decompose ${\displaystyle g_{0},g_{1}}$ recursively and we get a binary tree. If the tree would have a finite level then we could take the leafes as elements of a set ${\displaystyle {\text{B}}}$. So it follows that there must be at least one path of length ${\displaystyle \aleph _{0}}$ but then ${\displaystyle g\not \in {\text{G}}}$ by assumption. ${\displaystyle \square }$

## Symbolic calculus

Definition (Free abelian monoid)

Let ${\displaystyle {\text{S}}}$ be any set. Then ${\displaystyle {\text{S}}^{*}}$ with the following properties

• ${\displaystyle {\text{S}}\cup \left\{\epsilon \right\}\subset {\text{S}}^{*}}$
• ${\displaystyle \forall u,v\in {\text{S}}^{*}\colon u\circ v\in \mathbb {{\text{S}}^{*}} }$
• ${\displaystyle \forall u,v\in {\text{S}}^{*}\colon u\circ v=v\circ u}$
• ${\displaystyle \forall u\in {\text{S}}^{*}\colon u\circ \epsilon =u=\epsilon \circ u}$

is called the free abelian monoid with ${\displaystyle \circ }$ beeing the concatenation.

Corollary (Free abelian monoid)

Let ${\displaystyle {\mathcal {S}}}$ be a family of monoids and ${\displaystyle \cap _{{\text{S}}\in {\mathcal {S}}}{\text{S}}\neq \emptyset }$. Then

• ${\displaystyle \left(\cap _{{\text{S}}\in {\mathcal {S}}}{\text{S}}\right)^{*}\subseteq \cap _{{\text{S}}\in {\mathcal {S}}}{\text{S}}^{*}}$.

Let ${\displaystyle {\text{S}}}$ be any countable set and ${\displaystyle a\in {\text{S}}^{\mathbb {N} }}$ a bijection. Then

• ${\displaystyle \forall u\in {\text{S}}^{*}\colon u=\circ _{i\in \mathbb {N} }a_{i}^{e_{a_{i}}\left(u\right)}}$ where ${\displaystyle e_{a_{i}}\left(u\right)}$ is the occurence number of ${\displaystyle a_{i}}$ in ${\displaystyle u}$.

Proof: First statement: Let ${\displaystyle u_{0}\in \left(\cap _{{\text{S}}\in {\mathcal {S}}}{\text{S}}\right)^{*}}$. Then ${\displaystyle \exists v_{1}\in \cap _{{\text{S}}\in {\mathcal {S}}}{\text{S}}\colon \exists u_{1}\in \left(\cap _{{\text{S}}\in {\mathcal {S}}}{\text{S}}\right)^{*}\colon u_{0}=v_{1}\circ u_{1}}$. Since ${\displaystyle u_{1}}$ is from a monoid we can repeat this step a finite times for ${\displaystyle u_{i}}$ and we get a sequence ${\displaystyle v_{1},\cdots ,v_{n}\in \cap _{{\text{S}}\in {\mathcal {S}}}{\text{S}}}$ with ${\displaystyle u_{0}=\circ _{i=1}^{n}v_{i}}$. Since ${\displaystyle {\text{S}}\subset {\text{S}}^{*}}$ this implies ${\displaystyle u_{0}\in \cap _{{\text{S}}\in {\mathcal {S}}}{\text{S}}^{*}}$.

The second statement follows directly from the abelian property of the monoid.

Definition (Primorial)

Let ${\displaystyle {\text{S}}}$ be any countable set and ${\displaystyle a\in {\text{S}}^{\mathbb {N} }}$ a bijection. Then ${\displaystyle \circ _{i=0}^{a_{u}^{-1}}a_{i}\leftarrow u\colon {\text{S}}^{*}\leftarrow {\text{S}}\colon \#}$ and ${\displaystyle \epsilon \#=\epsilon }$.

Definition (Primorial iterates)

Let ${\displaystyle {\text{S}}}$ be any infinite countable set and ${\displaystyle a\in {\text{S}}^{\mathbb {N} }}$ a bijection. Then we define:

• ${\displaystyle \#\colon {\text{S}}^{*}\rightarrow \left({\text{S}}\#\right)^{*}\colon u\rightarrow \circ _{i=0}^{\infty }\left(a_{i}\#\right)^{e_{a_{i}}\left(u\right)}}$
• ${\displaystyle \forall n\in \mathbb {N} \colon \#_{n}\colon \#^{n}{\text{S}}^{*}\rightarrow \#^{n+1}{\text{S}}^{*}\colon u\rightarrow \#u}$

Lemma (Properties of primorial iterates)

• ${\displaystyle \#_{0}=\#}$
• ${\displaystyle \forall n\in \mathbb {N} \colon \#^{n}{\text{S}}^{*}=\left(\#^{n}{\text{S}}\right)^{*}}$
• ${\displaystyle \forall n\in \mathbb {N} \colon \#^{n+1}{\text{S}}^{*}\subset \#^{n}{\text{S}}^{*}}$
• ${\displaystyle \forall n\in \mathbb {N} \colon \#_{n}}$ is an isomorphism.
• ${\displaystyle \cap _{n\in \mathbb {N} }\#^{n}{\text{S}}=\{\epsilon ,a_{0}\}}$

Proof The first property is directly by definition since ${\displaystyle \left(\#^{n}{\text{S}}\right)^{*}={\text{S}}^{*}}$: ${\displaystyle \#_{0}\colon {\text{S}}^{*}\rightarrow \left(\#{\text{S}}\right)^{*}\colon u\rightarrow \#u}$.

Now the second one goes by induction. Because of ${\displaystyle \forall n\in \mathbb {N} \colon \#a_{n}=a_{n}\#}$, ${\displaystyle \#}$ is a homomorphism. Since it is also invertible on ${\displaystyle a_{n}}$, ${\displaystyle \#}$ is bijective because the image is precisely ${\displaystyle \left({\text{S}}\#\right)^{*}}$. Suppose now that ${\displaystyle \#_{n}}$ is an isomorphism. Now ${\displaystyle \#_{n+1}u=\#_{n}\#u}$ but ${\displaystyle \#}$ and ${\displaystyle \#_{n}}$ are isomorphisms so has to be ${\displaystyle \#_{n+1}}$.

The third property follows again by induction. We have ${\displaystyle \left(\#{\text{S}}\right)^{*}\subset {\text{S}}^{*}}$. Now suppose ${\displaystyle \left(\#^{n}{\text{S}}\right)^{*}\subset \left(\#^{n-1}{\text{S}}\right)^{*}}$. Then ${\displaystyle \left(\#^{n+1}{\text{S}}\right)^{*}=\left(\#\#^{n}{\text{S}}\right)^{*}=\#\left(\#^{n}{\text{S}}\right)^{*}\subset \#\left(\#^{n-1}{\text{S}}\right)^{*}=\left(\#^{n}{\text{S}}\right)^{*}}$.

The last property can be proven by considering the fixpoints of ${\displaystyle \#}$. Since ${\displaystyle \#\epsilon =\epsilon }$ and ${\displaystyle \#a_{0}=a_{0}\#=a_{0}}$ we know already two fixpoints. Now consider ${\displaystyle a_{1}}$. Since ${\displaystyle \#a_{1}=a_{1}\#=a_{0}a_{1}}$ we will increase the sequence length under every iteration of ${\displaystyle \#}$.

## Simple residue rules

• Let ${\displaystyle n=\sum _{i=0}^{k}a_{i}\left(mb+r\right)^{i}}$. Then ${\displaystyle n\equiv _{b}\sum _{i=0}^{k}a_{i}r^{i}}$.
• Let ${\displaystyle n=\sum _{i=0}^{k}a_{i}\left(2b\right)^{i}}$. Then ${\displaystyle n\equiv _{2}a_{0}}$.
• Let ${\displaystyle n=\sum _{i=0}^{k}a_{i}\left(2b+1\right)^{i}}$. Then ${\displaystyle n\equiv _{2}\sum _{i=0}^{k}a_{i}}$.
• Let ${\displaystyle n=\sum _{i=0}^{k}a_{i}b^{i}}$. Then ${\displaystyle n\equiv _{b\pm 1}\sum _{i=0}^{k}\left(\mp 1\right)^{i}a_{i}}$.
• Let ${\displaystyle n=\sum _{i=0}^{kl-1}a_{i}b^{i}}$ and ${\displaystyle b^{l}\equiv _{d}r}$. Then ${\displaystyle n\equiv _{d}\sum _{j=0}^{k-1}r^{j}\sum _{i=0}^{l-1}a_{lj+i}b^{i}}$

## Generating function of x mod y

• ${\displaystyle \forall y\in \mathbb {N} +1\colon \sum _{x=1}^{\infty }\left(x{\bmod {y}}\right)z^{x}=\sum _{i=0}^{\infty }z^{yi}\sum _{x=0}^{y-1}xz^{x}={\frac {z-z^{y}\left(y+z-yz\right)}{\left(1-z^{y}\right)\left(1-z\right)^{2}}}}$
• ${\displaystyle \sum _{y=1}^{\infty }\left(1-z\right)^{2}\sum _{x=0}^{\infty }\left(\left(x+1\right){\bmod {y}}\right)z^{x}-{\frac {1}{1-z^{y}}}=-1+\sum _{y=1}^{\infty }\left(\sigma _{1}\left(y\right)-\sigma _{0}\left(y\right)-\sigma _{1}\left(y+1\right)\right)z^{y}}$ (The rhs follows from the special Lambert series.)
• ${\displaystyle \left(1-z^{2}\right)^{2}\sum _{y=1}^{\infty }\left(1-z^{y}\right)\sum _{x=1}^{\infty }\left(x{\bmod {y}}\right)z^{x+y}={\frac {z^{3}}{1-z}}}$

## Moments of the harmonic summatory function

${\displaystyle \forall i,m\in \mathbb {N} \colon i^{m}=-(-1)^{m}\sum _{j=0}^{m-1}(-1)^{j}{\binom {m}{j}}H_{-j}\left(i\right)}$

Proof:

{\displaystyle {\begin{aligned}\forall i,m\in \mathbb {N} \colon &-(-1)^{m}\sum _{j=0}^{m-1}(-1)^{j}{\binom {m}{j}}H_{-j}\left(i\right)\\=&-(-1)^{m}\sum _{j=0}^{m-1}(-1)^{j}{\binom {m}{j}}{\frac {1}{j+1}}\sum _{k=0}^{j}\left(-1\right)^{k}{\binom {j+1}{k}}B_{k}i^{j-k+1}\\=&-(-1)^{m}\sum _{k=0}^{m-1}\left(-1\right)^{k}\sum _{j=0}^{m-k-1}{\binom {m}{j+k}}{\frac {1}{j+k+1}}{\binom {j+k+1}{j}}B_{j}i^{k+1}\\=&-(-1)^{m}\sum _{k=0}^{m-1}\left(-1\right)^{k}{\frac {1}{k+1}}{\binom {m}{k}}i^{k+1}\sum _{j=0}^{m-k-1}{\binom {m-k}{j}}B_{j}\\=&i^{m}\end{aligned}}}

We used for the first step Faulhaber's formula and for the last step the identity: ${\displaystyle \sum _{j=0}^{m-1}{\binom {m}{j}}B_{j}=\delta _{1m}}$.

• Let ${\displaystyle a_{n}\in \mathbb {N} ^{\mathbb {N} }}$ and ${\displaystyle A_{x}\left(n\right)\equiv \sum _{i=1}^{n}H_{-x}\left(a_{n}\left(i\right)\right)}$. Then ${\displaystyle \forall m>0\in \mathbb {N} \colon \sum _{i=1}^{n}a_{n}\left(i\right)^{m}=-(-1)^{m}\sum _{i=0}^{m-1}(-1)^{i}{\binom {m}{i}}A_{i}\left(n\right)}$
• Let ${\displaystyle a_{n}=id}$. Then ${\displaystyle \forall m>0\in \mathbb {N} \colon H_{-m}\left(n\right)=-(-1)^{m}\sum _{j=1}^{n}\sum _{i=0}^{m-1}(-1)^{i}{\binom {m}{i}}H_{-i}\left(j\right)}$

## The sum of the sum of moments of divisors up to a natural number ${\displaystyle n\in \mathbb {N} }$

• ${\displaystyle \Sigma _{x}\left(n\right)\equiv \sum _{i=1}^{n}\sigma _{x}\left(i\right)=\sum _{i=1}^{n}i^{x}\sigma _{-x}\left(i\right)=\sum _{i=1}^{n}\left\lfloor {\frac {n}{i}}\right\rfloor i^{x}=\sum _{i=1}^{n}H_{-x}\left(\left\lfloor {\frac {n}{i}}\right\rfloor \right)}$
• ${\displaystyle \forall n\in \mathbb {N} \colon H_{z}\left(n\right)=\sum _{i=1}^{n}\mu \left(i\right)\Sigma _{-z}\left(\left\lfloor {\frac {n}{i}}\right\rfloor \right)}$
• ${\displaystyle \forall n\in \mathbb {N} \colon \forall z\in \mathbb {C} \colon n=\sum _{i=1}^{n}{\frac {\mu \left(i\right)}{i^{-z}}}\Sigma _{z}\left(\left\lfloor {\frac {n}{i}}\right\rfloor \right)}$
• ${\displaystyle \forall m\in \mathbb {N} +1\colon \sum _{i=1}^{n}\left\lfloor {\frac {n}{i}}\right\rfloor ^{m}=-(-1)^{m}\sum _{i=0}^{m-1}(-1)^{i}{\binom {m}{i}}\Sigma _{i}\left(n\right)}$
• ${\displaystyle \Sigma _{0}\left(n\right)=\sum _{i=1}^{n}\left\lfloor {\frac {n}{i}}\right\rfloor }$
• ${\displaystyle \Sigma _{1}\left(n\right)={\frac {1}{2}}\sum _{i=1}^{n}\left\lfloor {\frac {n}{i}}\right\rfloor ^{2}+\left\lfloor {\frac {n}{i}}\right\rfloor ={\frac {1}{2}}\Sigma _{0}\left(n\right)+{\frac {1}{2}}\sum _{i=1}^{n}\left\lfloor {\frac {n}{i}}\right\rfloor ^{2}}$
• ${\displaystyle \Sigma _{2}\left(n\right)={\frac {1}{6}}\sum _{i=1}^{n}2\left\lfloor {\frac {n}{i}}\right\rfloor ^{3}+3\left\lfloor {\frac {n}{i}}\right\rfloor ^{2}+\left\lfloor {\frac {n}{i}}\right\rfloor =\Sigma _{1}\left(n\right)-{\frac {1}{3}}\Sigma _{0}\left(n\right)+{\frac {1}{3}}\sum _{i=1}^{n}\left\lfloor {\frac {n}{i}}\right\rfloor ^{3}}$
• ${\displaystyle \Sigma _{3}\left(n\right)={\frac {1}{4}}\sum _{i=1}^{n}\left\lfloor {\frac {n}{i}}\right\rfloor ^{4}+2\left\lfloor {\frac {n}{i}}\right\rfloor ^{3}+\left\lfloor {\frac {n}{i}}\right\rfloor ^{2}={\frac {3}{2}}\Sigma _{2}\left(n\right)-\Sigma _{1}\left(n\right)+{\frac {1}{4}}\Sigma _{0}\left(n\right)+{\frac {1}{4}}\sum _{i=1}^{n}\left\lfloor {\frac {n}{i}}\right\rfloor ^{4}}$
• ${\displaystyle \sum _{i=1}^{n}\left\lfloor {\frac {n}{i}}\right\rfloor ^{m}-\sum _{i=1}^{n-1}\left\lfloor {\frac {n-1}{i}}\right\rfloor ^{m}=-(-1)^{m}\sum _{i=0}^{m-1}(-1)^{i}{\binom {m}{i}}\sigma _{i}\left(n\right)}$
• ${\displaystyle \forall x\in \mathbb {R} _{+}\colon \sum _{i=1}^{n}\left\lfloor {\frac {n}{i}}\right\rfloor ^{x}\geq \sum _{i=1}^{n}\left({\frac {n+1}{i}}-1\right)^{x}=\sum _{i=1}^{n}\left({\frac {n+1}{i}}-1\right)^{-x}}$
• ${\displaystyle \sum _{i=1}^{n}\left\lfloor {\frac {n}{i}}\right\rfloor ^{x}=\sum _{i=0}^{n-1}\left\lfloor {\frac {1}{1-{\frac {i}{n}}}}\right\rfloor ^{x}=\sum _{i=0}^{n-1}\left\lfloor \sum _{j=0}^{\infty }{\frac {i^{j}}{n^{j}}}\right\rfloor ^{x}=\sum _{i=0}^{n-1}\left\lfloor \lim _{q\to \infty }{\frac {n-i{\frac {i^{q}}{n^{q}}}}{n-i}}\right\rfloor ^{x}}$
• Let ${\displaystyle \mathbb {N} ^{\mathbb {N} _{n}}={\text{N}}_{>}\cup {\text{N}}_{=}\cup {\text{N}}_{<}}$ where ${\displaystyle \forall q\in {\text{N}}_{\circ }\colon \sum _{i=0}^{n-1}\left({\frac {n-i{\frac {i^{q_{i}}}{n^{q_{i}}}}}{n-i}}\right)^{x}\circ \sum _{i=1}^{n}\left\lfloor {\frac {n}{i}}\right\rfloor ^{x}}$. Then ${\displaystyle \sum _{i=1}^{n}\left\lfloor {\frac {n}{i}}\right\rfloor ^{x}-\sum _{i=1}^{n-1}\left\lfloor {\frac {n-1}{i}}\right\rfloor ^{x}\leq \inf _{q_{1}\in {\text{N}}_{>},q_{2}\in {\text{N}}_{<}}\sum _{i=0}^{n-1}\left({\frac {n-i{\frac {i^{q_{1i}}}{n^{q_{1i}}}}}{n-i}}\right)^{x}-\sum _{i=0}^{n-2}\left({\frac {n-1-i{\frac {i^{q_{2i}}}{\left(n-1\right)^{q_{2i}}}}}{n-1-i}}\right)^{x}}$.

## A bound on the sum of divisors of a natural number ${\displaystyle n\in \mathbb {N} }$

We have in general the identity: ${\displaystyle \forall z\in \mathbb {C} \colon \sigma _{z}\left(n\right)=n^{z}\sigma _{-z}\left(n\right)}$ or ${\displaystyle \forall z\in \mathbb {C} \colon n^{-{\frac {z}{2}}}\sigma _{z}\left(n\right)=n^{\frac {z}{2}}\sigma _{-z}\left(n\right)}$ which is invariant under reflection ${\displaystyle z\rightarrow -z}$. First we proof the general property ${\displaystyle \forall x>0\colon \forall d<{\sqrt {n}}\colon \forall k<{\sqrt {n}}-d\colon {\frac {n^{x}}{d^{x}}}+{d^{x}}\geq {\frac {n^{x}}{\left(d+k\right)^{x}}}+{\left(d+k\right)^{x}}}$

Proof: ${\displaystyle {\frac {n^{x}}{d^{x}}}+{d^{x}}-{\frac {n^{x}}{\left(d+k\right)^{x}}}-{\left(d+k\right)^{x}}=\left(\left(d+k\right)^{x}-d^{x}\right)\left(\left({\frac {n}{d\left(d+k\right)}}\right)^{x}-1\right)>0}$

A relative tight upper bound can then be found as ${\displaystyle \sigma _{x}\left(n\right)\leq n^{x}H_{x}\left(\sigma _{0}\left(n\right)\right)}$ where the harmonic number can be coarse bounded as ${\displaystyle H_{x}\left(\sigma _{0}\left(n\right)\right)\leq \left({\frac {1-\sigma _{0}\left(n\right)^{1-x}}{x-1}}+1\right)}$. Using the symmetry of the divisors we get an even better upper bound: ${\displaystyle \sigma _{x}\left(n\right)\leq n^{x}H_{x}\left(\left\lfloor {\frac {\sigma _{0}\left(n\right)}{2}}\right\rfloor \right)+H_{-x}\left(\left\lfloor {\frac {\sigma _{0}\left(n\right)}{2}}\right\rfloor \right)+\delta _{\left\lfloor {\sqrt {n}}\right\rfloor {\sqrt {n}}}n^{\frac {x}{2}}}$.

• If we restrict ourself only to odd natural numbers ${\displaystyle 2\mathbb {N} +1=\left\{2\right\}^{\perp }}$ we may get an other bound because the spacing of possible divisors is at least ${\displaystyle 2}$. We can express this as:
{\displaystyle {\begin{aligned}\forall n\in \mathbb {N} \colon \sigma _{x}\left(2n+1\right)\leq &\left(2n+1\right)^{x}\left(H_{x}\left(2\left\lfloor {\frac {\sigma _{0}\left(2n+1\right)}{2}}\right\rfloor \right)-2^{-x}H_{x}\left(\left\lfloor {\frac {\sigma _{0}\left(2n+1\right)}{2}}\right\rfloor \right)\right)\\&+H_{-x}\left(2\left\lfloor {\frac {\sigma _{0}\left(2n+1\right)}{2}}\right\rfloor \right)-2^{x}H_{-x}\left(\left\lfloor {\frac {\sigma _{0}\left(2n+1\right)}{2}}\right\rfloor \right)+\delta _{\left\lfloor {\sqrt {2n+1}}\right\rfloor {\sqrt {2n+1}}}\left(2n+1\right)^{\frac {x}{2}}\end{aligned}}}
• I we consider instead the set ${\displaystyle 6\mathbb {N} +1\cup 6\mathbb {N} +5=\left\{2,3\right\}^{\perp }}$ we have a minimal periodic spacing of ${\displaystyle 4,2}$ which we can express algebraically as ${\displaystyle 3-\left(-1\right)^{i}}$ (we have for every periodic sequence ${\displaystyle a,b}$ algebraically ${\displaystyle {\frac {a+b}{2}}-\left(-1\right)^{i}{\frac {a-b}{2}}}$ which shows that every such sequence is isomorphic to an alternating sequence) and therefore we get
{\displaystyle {\begin{aligned}\forall n\in 6\mathbb {N} \pm 1\colon \sigma _{x}\left(n\right)&\leq \delta _{\left\lfloor {\sqrt {n}}\right\rfloor ,{\sqrt {n}}}n^{\frac {x}{2}}+\sum _{i=0}^{{\frac {\sigma _{0}\left(n\right)}{2}}-1}{\frac {n^{x}}{\left(3i+{\frac {3}{2}}-{\frac {\left(-1\right)^{i}}{2}}\right)^{x}}}+\left(3i+{\frac {3}{2}}-{\frac {\left(-1\right)^{i}}{2}}\right)^{x}\end{aligned}}}

We might also find always a better upper bound for any multiplicative function ${\displaystyle f}$ under the following assumption: Let ${\displaystyle n_{1},n_{2}\in \mathbb {N} \wedge n_{1}\perp n_{2}}$ and ${\displaystyle f\left(n_{1}n_{2}\right)=f\left(n_{1}\right)f\left(n_{2}\right)\leq g\left(n_{1}n_{2}\right)\leq g\left(n_{1}\right)f\left(n_{2}\right)}$. Then ${\displaystyle f\left(n_{1}\right)\leq {\frac {g\left(n_{1}n_{2}\right)}{f\left(n_{2}\right)}}\leq g\left(n_{1}\right)}$.

### On Robin's criterion for RH

• ${\displaystyle \sigma _{-1}\left(n^{k}\right)<\sigma _{-1}\left(n^{k+1}\right)}$
• ${\displaystyle \sigma _{-1}\left(n\right)={\frac {n}{\varphi \left(n\right)}}\prod _{p|n}1-p^{-e_{n}\left(p\right)-1}}$
• ${\displaystyle \forall n\in \mathbb {N} \colon \sup \left\{\sigma _{-1}\left(m\right)|{\text{rad}}\left(m\right)|n\right\}={\frac {n}{\varphi \left(n\right)}}}$
• ${\displaystyle \lim _{k\to \infty }\sigma _{-1}\left(n^{k}\right)={\frac {n}{\varphi \left(n\right)}}}$
• ${\displaystyle \forall n\in \mathbb {N} \colon \max \left\{{\frac {m}{\varphi \left(m\right)}}|m\leq n\right\}={\frac {n'}{\varphi \left(n'\right)}}}$ with ${\displaystyle n'=\max \left\{\prod _{i=1}^{m}p_{i}\leq n|m\in \mathbb {N} \right\}}$
• ${\displaystyle \forall n\in \mathbb {N} \colon \max \left\{{\frac {m}{\varphi \left(m\right)}}|m<\#n\right\}=\prod _{i=1}^{n-1}{\frac {p_{i}}{p_{i}-1}}}$

Lemma

Let ${\displaystyle p_{i}\in \mathbb {P} }$ the strictly increasing sequence of prime numbers and ${\displaystyle i,j\in \mathbb {N} }$. Then

${\displaystyle \sigma _{-1}\left(p_{i}\right)>{\frac {p_{i+j+1}}{\varphi \left(p_{i+j+1}\right)}}}$.

Proof: This follows from the fact that ${\displaystyle \forall r>1\colon {\frac {r+{\frac {1}{r}}}{r-1}}-{\frac {r+1}{r}}={\frac {2}{r\left(r-1\right)}}>0}$. ${\displaystyle \square }$

Lemma (Exchange of exponents)

Let ${\displaystyle p_{i}\in \mathbb {P} }$ the strictly increasing sequence of prime numbers and ${\displaystyle i,j,a,b\in \mathbb {N} }$ with ${\displaystyle j>i}$ and ${\displaystyle b>a}$. Then

${\displaystyle \sigma _{-1}\left(p_{i}^{a}p_{j}^{b}\right)<\sigma _{-1}\left(p_{i}^{b}p_{j}^{a}\right)}$.

Proof: It is enough to consider the derivative of ${\displaystyle f\left(r\right)={\frac {1-r^{-a-1}}{1-r^{-b-1}}}}$ for ${\displaystyle r>1}$. We have ${\displaystyle f'\left(r\right)={\frac {r^{-a+b-1}g\left(r\right)}{\left(r^{b+1}-1\right)^{2}}}>0}$ with ${\displaystyle g\left(r\right)=-(b+1)r^{a+1}+a\left(r^{b+1}-1\right)+r^{b+1}+b}$ because ${\displaystyle g'\left(r\right)=(a+1)(b+1)\left(r^{b}-r^{a}\right)>0}$ and ${\displaystyle g\left(1\right)=0}$. Therefore ${\displaystyle \forall \epsilon >0\colon f\left(r\right). ${\displaystyle \square }$

Definition (Asymptotic equivalence)

${\displaystyle \left[m\right]\equiv \left\{n\in \mathbb {N} |{\frac {n}{\varphi \left(n\right)}}=m\right\}}$

Corollary (Representation of a natural number)

Let ${\displaystyle n\in \mathbb {N} }$. Then there exists ${\displaystyle m\in \mathbb {N} ^{\times \infty }}$ with ${\displaystyle \left(m_{i},m_{j}\right)=1+\delta _{i,j}\left(m_{i}-1\right)}$ so that ${\displaystyle n=\prod _{i=1}^{\infty }m_{i}^{i}}$. In particular we make the identification ${\displaystyle n\cong m}$.

Proof: Just consider the prime factorization of ${\displaystyle n}$ and regroup the factors according to it's prime multiples in increasing order. The co-prime condition follows immediately.

Remark: Clearly there will some ${\displaystyle j}$ so that ${\displaystyle \forall i\geq j\colon m_{i}=1}$.

Definition (Dedekind ${\displaystyle \Psi _{t}}$) ${\displaystyle \forall t\in \mathbb {C} \colon \Psi _{t}\left(n\right)\equiv n\prod _{p|n}{\frac {p-p^{1-t}}{p-1}}}$

${\displaystyle \forall t\in \mathbb {C} \colon \psi _{t}\left(n\right)\equiv {\frac {\Psi _{t}\left(n\right)}{n}}}$

This representation has the advantage that we have ${\displaystyle \sigma _{1}\left(n\right)=\prod _{i=1}^{\infty }\Psi _{i+1}\left(m_{i}\right)}$.

Corollary (Dedekind ${\displaystyle \psi _{t}}$) Let ${\displaystyle n\in \mathbb {N} }$. Then ${\displaystyle \exists t\in \mathbb {R} _{+}\colon \sigma _{-1}\left(n\right)=\psi _{t}\left({\text{rad}}\left(n\right)\right)}$.

Lemma (Dedekind ${\displaystyle \psi _{t}}$) Let ${\displaystyle n\in \mathbb {N} ,p\in \mathbb {P} }$ and ${\displaystyle \sigma _{-1}\left(n\right)=\psi _{t}\left({\text{rad}}\left(n\right)\right)}$.

• ${\displaystyle p|n\Rightarrow \exists t
• ${\displaystyle p\not |n\Rightarrow \exists t>t'\in \mathbb {R} _{+}\colon \sigma _{-1}\left(np\right)=\psi _{t'}\left({\text{rad}}\left(np\right)\right)}$

Definition (Extremal primes)

Let ${\displaystyle n\in \mathbb {N} }$ given as above so that we have ${\displaystyle n=\prod _{i=1}^{\infty }m_{i}^{i}}$. We define:

• ${\displaystyle {\underline {p}}_{i}\equiv \min _{p|m_{i}}p}$
• ${\displaystyle {\overline {p}}_{i}\equiv \max _{p|m_{i}}p}$

Definition (Primorial)

${\displaystyle x\#\equiv \prod _{i=1}^{\pi \left(x\right)}p_{i}}$

Corollary (Primorial) Let ${\displaystyle n\in \mathbb {N} }$. Then ${\displaystyle n\#={\text{rad}}\left(n!\right)}$.

Lemma (Product of primorials)

Let ${\displaystyle n}$ be a product of primorials. Then ${\displaystyle n=\prod _{i=1}^{\infty }p_{q_{i}}\#^{k_{i}}}$ for some sequences ${\displaystyle q,k\in \mathbb {N} ^{\times \infty }}$.

Remark: If we understand ${\displaystyle p_{q}^{k}}$ pointwise then we find a particular short notation of products of primorials: ${\displaystyle n\cong p_{q}^{k}=\left(p_{q_{1}}^{k_{1}},p_{q_{2}}^{k_{2}},\cdots \right)}$. As an example pick as ${\displaystyle n}$ the ${\displaystyle 2386^{\text{th}}}$ colossally abundant number. It is given in this notation as: ${\displaystyle n\cong \left(p_{1}^{7},p_{2}^{3},p_{3},p_{4},p_{6},p_{12},p_{44},p_{2301}\right)}$. However we can get even more structure:

Definition (Primorial products) Let ${\displaystyle <\mathbb {P} \#>}$ be the set of all finite primorial products. Then we define a completely multiplicative mapping ${\displaystyle \#\colon \mathbb {N} \rightarrow <\mathbb {P} \#>}$ with ${\displaystyle \forall p\in \mathbb {P} \colon \#p=p\#}$.

Lemma (Primorial isomorphism) Let ${\displaystyle <\mathbb {P} \#>}$ be the set of all finite primorial products. Then under the multiplication of ${\displaystyle \mathbb {N} }$ we have ${\displaystyle \mathbb {N} \cong <\mathbb {P} \#>}$.

Proof This is because ${\displaystyle \#}$ is a bijection and a homomorphism.

Corollary (Properties of primorial isomorphism)

• ${\displaystyle \forall n\in \mathbb {N} \colon \#_{n}\colon \#^{n}<\mathbb {P} \#>\rightarrow \#^{n+1}<\mathbb {P} \#>\colon m\rightarrow \#m}$ is an isomorphism.
• ${\displaystyle \forall n,k\in \mathbb {N} \colon \#^{n}2^{k}=2^{k}}$
• ${\displaystyle \cap _{n\in \mathbb {N} }\#^{n}<\mathbb {P} \#>=2^{\mathbb {N} }}$

Corollary (Product of primorials)

Let ${\displaystyle n}$ be a product of primorials. Then ${\displaystyle {\underline {p}}_{i}>1\Rightarrow {\underline {p}}_{i}>{\overline {p}}_{i+1}}$.

Definition (Monotonic bound)

${\displaystyle \forall n\in \mathbb {N} \colon f\left(n\right) and ${\displaystyle f\left(n\right)\rightarrow \infty }$.

Definition (RH bound)

${\displaystyle g_{+}\left(n\right)=\ln \ln n}$

• ${\displaystyle g\left(n\right)=e^{\gamma }g_{+}\left(n\right)}$

${\displaystyle g_{\times }\left(n\right)=\ln n}$

• ${\displaystyle \forall x>1\colon g_{\times }^{2}\left(x\right)=g_{+}\left(x\right)}$
• ${\displaystyle \forall x>1\colon \forall y>0\colon g_{+}\left(x\right)+g_{\times }\left(y\right)=g_{+}\left(x^{y}\right)}$
• ${\displaystyle \forall x,y>1\colon g_{+}\left(x\right)+g_{+}\left(y\right)=g_{+}\left(x^{\ln y}\right)=g_{+}\left(y^{\ln x}\right)}$
• ${\displaystyle \forall x,y>0\colon g_{\times }\left(x\right)g_{\times }\left(y\right)=g_{\times }\left(x^{\ln y}\right)=g_{\times }\left(y^{\ln x}\right)}$
• ${\displaystyle \forall x,y>0\colon g_{\times }\left(x\right)+g_{\times }\left(y\right)=g_{\times }\left(xy\right)}$
• ${\displaystyle \forall x>0,y>1\colon g_{\times }\left(x\right)g_{+}\left(y\right)=g_{\times }\left(\ln x^{\ln y}\right)=g_{\times }\left(x^{\ln \ln y}\right)}$
• ${\displaystyle \forall x,y>1\colon g_{+}\left(xy\right)=g_{+}\left(x\right)+g_{+}\left(\left(xy\right)^{\frac {1}{\ln x}}\right)}$

Lemma (Bound violation)

Let ${\displaystyle n\in \left[m\right]\colon m\geq f\left(n\right)}$ and ${\displaystyle \forall n\in \mathbb {N} \colon f\left(n\right). Then ${\displaystyle m\geq f\left(\min \left[m\right]\right)}$.

Theorem (Violation of montonic bound) Let ${\displaystyle m\in \mathbb {N} }$ and ${\displaystyle f}$ a monotonic bound. Then ${\displaystyle \#\left\{n\in \left[m\right]|m\geq f\left(n\right)\right\}<\infty }$.

Proof: Let ${\displaystyle V=\left\{n\in \left[m\right]|m\geq f\left(n\right)\right\}}$ and ${\displaystyle \#V>0}$. Then by our lemma we may assume that ${\displaystyle m\geq f\left(\min \left[m\right]\right)}$. However ${\displaystyle \forall k\in \mathbb {N} +2\colon \min \left[m\right]^{k}\in \left[m\right]}$ and ${\displaystyle f\left(\min \left[m\right]^{k}\right)>f\left(\min \left[m\right]\right)}$. Because ${\displaystyle \lim _{k\to \infty }f\left(\min \left[m\right]^{k}\right)=\infty }$ ${\displaystyle \exists K\in \mathbb {N} \colon \forall k\geq K\colon f\left(\min \left[m\right]^{k}\right)>m}$. ${\displaystyle \square }$

Lemma: (Violation of bound) Let ${\displaystyle m\in \mathbb {N} }$ and ${\displaystyle f}$ a monotonic bound. Then ${\displaystyle \#\left\{n\in \left[m\right]|\sigma _{-1}\left(n\right)\geq f\left(n\right)\right\}<\infty }$.

Proof: This follows directly by the Theorem and the fact that ${\displaystyle \forall n\in \left[m\right]\colon \sigma _{-1}\left(n\right).

Corollary: (Violation of bound) Let ${\displaystyle m\in \mathbb {N} }$. Then ${\displaystyle \#\left\{n\in \left[m\right]|\sigma _{-1}\left(n\right)\geq g\left(n\right)\right\}<\infty }$.

Lemma: (On the RH bound for even positive integers)

Let ${\displaystyle n=2^{a}\left(2m+1\right)}$ and ${\displaystyle a,m\in \mathbb {N} }$. Then ${\displaystyle \forall m>7\colon \forall a\geq {\frac {\left(\ln(2m+1)\right)^{2}-\ln(2m+1)}{\ln 2}}\colon \sigma _{-1}\left(n\right).

Proof:

We have ${\displaystyle g\left(n\right)=g\left(2^{a}\left(2m+1\right)\right)=g\left(2m+1\right)+g\left(\left(2^{a}(2m+1)\right)^{\frac {1}{\ln(2m+1)}}\right)}$. Now ${\displaystyle \left(2^{a}(2m+1)\right)^{\frac {1}{\ln(2m+1)}}\geq 2m+1\Leftrightarrow a\geq -{\frac {\ln \left((2m+1)^{1-\ln(2m+1)}\right)}{\ln 2}}={\frac {\left(\ln(2m+1)\right)^{2}-\ln(2m+1)}{\ln 2}}}$. Then ${\displaystyle g\left(n\right)\geq 2g\left(2m+1\right)>2{\frac {2m+1}{\varphi \left(2m+1\right)}}={\frac {n}{\varphi \left(n\right)}}>\sigma _{-1}\left(n\right)}$. ${\displaystyle \square }$

Lemma: (On the RH bound for arbitrary powers)

Let ${\displaystyle s>2}$ and ${\displaystyle k\in \mathbb {N} +1}$ and ${\displaystyle a\in \mathbb {R} }$ such that ${\displaystyle k-a\geq 1}$. Then

${\displaystyle \sigma _{-1}\left(s^{k}\right)\geq g\left(s^{k-a}\right)\Rightarrow k.

Proof: Suppose that ${\displaystyle \sigma _{-1}\left(s^{k}\right)\geq g\left(s^{k-a}\right)}$. Now ${\displaystyle g\left(s^{k-a}\right)=g\left(s\right)+e^{\gamma }\ln \left(k-a\right)}$. Therefore ${\displaystyle e^{\gamma }\ln \left(k-a\right)\leq \sigma _{-1}\left(s^{k}\right)-g\left(s\right)<{\frac {s}{\varphi \left(s\right)}}-g\left(s\right)<{\frac {3}{\ln \ln s}}}$. It follows that ${\displaystyle k. ${\displaystyle \square }$

Application (Square-full numbers):

All proper powers of ${\displaystyle n\in \mathbb {N} }$ with ${\displaystyle n\geq e^{e^{\frac {3e^{-\gamma }}{\ln 2}}}}$ fulfill the RH bound. In particular for all such ${\displaystyle n}$ we have: ${\displaystyle \forall k\in \mathbb {N} +2\colon \sigma _{-1}\left(n^{k}\right). Because ${\displaystyle k\geq 2}$ we deduce that all square-full numbers ${\displaystyle n\geq e^{2e^{\frac {3e^{-\gamma }}{\ln 2}}}}$ fulfill the RH bound.

Corollary: (On the RH bound for positive integer powers)

${\displaystyle \forall \epsilon >0\colon \forall n>e^{e^{\frac {3e^{-\gamma }}{\ln \left(1+\epsilon \right)}}}\colon \forall k\in \mathbb {N} +1\colon \sigma _{-1}\left(n^{k}\right)

Lemma: (On the RH bound)

Let ${\displaystyle \delta =\left\lfloor n-{\sqrt {n}}\right\rfloor }$ and ${\displaystyle n=\left(n-\delta \right)^{2\beta }}$ and ${\displaystyle \sigma _{-1}\left(n\right)\leq \sigma _{-1}\left(n^{\beta ^{-1}}\right)}$. Then ${\displaystyle \forall \epsilon >0\colon \forall n-\delta >e^{e^{\frac {3e^{-\gamma }}{\ln \left(1+\epsilon \right)}}}\colon \sigma _{-1}\left(n\right)\geq g\left(n\right)\Rightarrow \beta \leq {\frac {1+\epsilon }{2}}}$.

Proof: ${\displaystyle \sigma _{-1}\left(n\right)\geq g\left(n\right)\Leftrightarrow \sigma _{-1}\left(\left(n-\delta \right)^{2\beta }\right)\geq g\left(\left(n-\delta \right)^{2\beta }\right)\Leftrightarrow e^{\gamma }\ln \left(2\beta \right)\leq \sigma _{-1}\left(\left(n-\delta \right)^{2\beta }\right)-g\left(n-\delta \right)}$. Now ${\displaystyle \sigma _{-1}\left(\left(n-\delta \right)^{2\beta }\right)\leq \sigma _{-1}\left(\left(n-\delta \right)^{2}\right) and therefore ${\displaystyle e^{\gamma }\ln \left(2\beta \right)\leq e^{\gamma }\ln \left(1+\epsilon \right)\Leftrightarrow \beta \leq {\frac {1+\epsilon }{2}}}$. ${\displaystyle \square }$

Lemma: Let ${\displaystyle s,q\in \mathbb {N} }$ with ${\displaystyle \mu \left(s\right)^{2}=1}$ such that ${\displaystyle \exists 0<\epsilon <1\colon sq>e^{e^{\frac {3e^{-\gamma }}{\ln \left(1+\epsilon \right)}}}}$. Further let ${\displaystyle d\in \mathbb {N} }$ with ${\displaystyle d|s}$. Then ${\displaystyle \forall {\frac {\left(sq\right)^{\epsilon }}{q}}\leq d\leq s\colon \sigma _{-1}\left(sq^{2}d\right).

Proof: ${\displaystyle \sigma _{-1}\left(sq^{2}d\right)\leq \sigma _{-1}\left(s^{2}q^{2}\right) ${\displaystyle \square }$

Remark: This shows essentially that we are blind with this technique to numbers with a long square-free tail. Since ${\displaystyle {\frac {\left(sq\right)^{\epsilon }}{q}}\rightarrow 0}$ for ${\displaystyle q\rightarrow \infty }$ we have control over the quadratic component but not over ${\displaystyle s}$.

Lemma: Let ${\displaystyle n\in \mathbb {N} }$ and ${\displaystyle p\in \mathbb {P} }$ such that ${\displaystyle p|n}$. Then ${\displaystyle \sigma _{-1}\left({\frac {n}{p}}\right).

Remark: The second assumption is equivalent to: ${\displaystyle \exists p\in \mathbb {P} \colon p|n\colon e_{p}\left(n\right)\geq {\frac {\log \left({\frac {(p-1)\log \log n}{\log \log n-\log \log {\frac {n}{p}}}}+1\right)}{\log p}}-1\equiv h_{p}\left(n\right)}$. But this condition is not always fulfilled. A counter example is ${\displaystyle n=160626866400}$ which is a colossally abundant number. The existence of a counter example leads us to the conclusion that we have to investigate further the numbers ${\displaystyle n}$ which have ${\displaystyle \forall p|n\colon e_{p}\left(n\right).

Lemma (Bound on the largest prime factor) Let ${\displaystyle n\in \mathbb {N} }$ and ${\displaystyle \sigma _{-1}\left(n\right) may be eventually false. Then ${\displaystyle \forall p|n\colon p<{\overline {p}}}$ with ${\displaystyle h_{\overline {p}}\left(n\right)=1\Leftrightarrow \left(\log {\frac {n}{\overline {p}}}\right)^{{\overline {p}}+1}=\left(\log n\right)^{\overline {p}}}$.

Lemma (Necessary condition for the failure of Robin's inequality) Let ${\displaystyle n\in \mathbb {N} }$, ${\displaystyle m\in \mathbb {N} \setminus {\mathcal {A}}}$ where ${\displaystyle {\mathcal {A}}}$ is the exceptionel set where Robin's inequality is false. Then ${\displaystyle \forall mg\left(n\right)-g\left(m\right)}$.

Proof: By failure of Robin's inequality we have ${\displaystyle \forall mg\left(m\right)}$. But ${\displaystyle m\in \mathbb {N} \setminus {\mathcal {A}}}$ and therefore ${\displaystyle g\left(m\right)>\sigma _{-1}\left(m\right)}$.

Lemma (Sufficient condition for the truth of Robin's inequality) Let ${\displaystyle n,o\in \mathbb {N} ,m\in \mathbb {N} \setminus {\mathcal {A}}}$ and ${\displaystyle o\leq n^{\frac {1}{\log m}}}$ sucht that ${\displaystyle g\left(o\right)>\sigma _{-1}\left(n\right)-\sigma _{-1}\left(m\right)}$. Then ${\displaystyle \sigma _{-1}\left(n\right).

Proof: By assumption we have ${\displaystyle \sigma _{-1}\left(n\right)-\sigma _{-1}\left(m\right). This implies the stronger condition: ${\displaystyle \sigma _{-1}\left(n\right).

## The sum of the count of unordered bipartite factorizations up to a natural number ${\displaystyle n\in \mathbb {N} }$

${\displaystyle \Sigma _{0,2}\left(n\right)=\sum _{i=0}^{n-1}\left\lfloor {\frac {1}{2}}\left({\sqrt {\left(i^{2}+4n\right)}}-i\right)\right\rfloor =\sum _{i=1}^{n}\left\lfloor {\frac {\sigma _{0}\left(i\right)+1}{2}}\right\rfloor ={\frac {1}{2}}\Sigma _{0}\left(n\right)+{\frac {1}{2}}\left\lfloor {\sqrt {n}}\right\rfloor }$

## The modular multiplication table of the finite cyclic group of natural numbers

We may assume here that we have a Hilbert space ${\displaystyle {\mathcal {H}}}$ with dimension ${\displaystyle d_{\mathcal {H}}\in \mathbb {N} }$ and an orthonormal base ${\displaystyle {\mathcal {B}}_{\mathcal {H}}}$. The space of ${\displaystyle \left(1,1\right)}$-tensors is given by ${\displaystyle {\text{M}}_{\mathcal {H}}}$ and ${\displaystyle \forall M\in {\text{M}}_{\mathcal {H}}\colon \forall 1\leq i,j\leq d_{\mathcal {H}}\colon M_{ij}}$ are the basis coefficients.

Now for some ${\displaystyle M\in {\text{M}}_{\mathcal {H}}}$ let ${\displaystyle \forall 1\leq i,j\leq d_{\mathcal {H}}\colon M_{ij}\equiv _{d_{\mathcal {H}}}\left(i-1\right)\left(j-1\right)}$ be the entries of the modular multiplication table ${\displaystyle M}$. This multiplication table has under the left and right action of the symmetric group ${\displaystyle {\frac {d_{\mathcal {H}}!^{2}}{\varphi \left(d_{\mathcal {H}}\right)}}}$ symmetries.

### The connection with the discrete Fourier transformation

We may define the generalized discrete Fourier transformation (GDFT) as an unitary matrix ${\displaystyle F}$ such that ${\displaystyle F_{ij}={\frac {1}{\sqrt {d_{\mathcal {H}}}}}\exp \left({\text{i}}\theta _{ij}\right)}$ where ${\displaystyle \sum _{k=1}^{d_{\mathcal {H}}}F_{ki}F_{kj}^{*}={\frac {1}{d_{\mathcal {H}}}}\sum _{k=1}^{d_{\mathcal {H}}}\exp \left({\text{i}}\left(\theta _{kj}-\theta _{ki}\right)\right)=\delta _{ij}}$. The connection follows then by letting ${\displaystyle \theta ={\frac {2\pi }{d_{\mathcal {H}}}}M}$. This is also the complex representation of the multiplication of elements of the finite cyclic group ${\displaystyle {\frac {\mathbb {Z} }{d_{\mathcal {H}}\mathbb {Z} }}}$.

### The connection with the microcanonical ensemble

We establish here a connection to the microcanonical ensemble under the von Neumann measurement. Given some diagonal ${\displaystyle \left(1,1\right)}$-tensor ${\displaystyle D\in {\text{D}}_{\mathcal {H}}}$ and an unitary matrix ${\displaystyle U\in {\text{U}}_{\mathcal {H}}}$ such that the new measurement base is ${\displaystyle U{\mathcal {B}}_{\mathcal {H}}}$ the von Neumann measurement acts on the diagonal entries as the transition ${\displaystyle D_{ii}\rightarrow \left(U^{*}DU\right)_{ii}}$. If we demand now that ${\displaystyle \forall D\in {\text{D}}_{\mathcal {H}}\colon \forall 1\leq i\leq d_{\mathcal {H}}\colon \left(U^{*}DU\right)_{ii}={\frac {1}{d_{\mathcal {H}}}}{\text{tr}}D}$ we have ${\displaystyle U=F}$. If we impose ${\displaystyle {\text{tr}}D=1}$ and ${\displaystyle \forall 1\leq i\leq d_{\mathcal {H}}\colon D_{ii}\geq 0}$ this tells us also that it is always possible to push a quantum system into the state of maximal entropy. It is therefore pretty useful to investigate the space of all GDFTs.

### The rank of the modular multiplication table

${\displaystyle {\text{rk}}\left(M\right)=\left\lfloor {\frac {d_{\mathcal {H}}+1}{2}}\right\rfloor +\sigma _{0}\left(d_{\mathcal {H}}\right)-2}$

Proof:

We set ${\displaystyle n=d_{\mathcal {H}}}$. If ${\displaystyle M=\left(v_{i}|0\leq i is given in column notation, we build a new matrix ${\displaystyle N=\left(w_{i}|0\leq i with ${\displaystyle w_{i}=v_{i}+v_{n-i}}$ where we identified ${\displaystyle v_{n}\equiv v_{0}}$. We may now order the ${\displaystyle w_{i}}$ according to the divisors of ${\displaystyle n}$: every divisor ${\displaystyle d}$ of ${\displaystyle n}$ spawns an equivalence class ${\displaystyle \left[d\right]=\left\{w_{di}|\forall 1\leq i<{\frac {n}{d}}\colon i\perp {\frac {n}{d}}\right\}}$ of column vectors of ${\displaystyle N}$. Elements of different equivalence classes are linearly independent so that our rank is determined by the size ${\displaystyle \left|\left[d\right]\right|=\varphi \left({\frac {n}{d}}\right)}$ of the equivalence classes. The case where ${\displaystyle \left|\left[d\right]\right|\in 2\mathbb {N} }$ will decrease the rank by ${\displaystyle {\frac {\left|\left[d\right]\right|}{2}}-1}$ and ${\displaystyle \left|\left[d\right]\right|\in 2\mathbb {N} +1}$ will decrease the rank by ${\displaystyle {\frac {\left|\left[d\right]\right|+1}{2}}-1}$. In the case where ${\displaystyle n\in 2\mathbb {N} +1}$ all these classes except ${\displaystyle \left|\left[n\right]\right|}$ have even size and we get:

• ${\displaystyle \forall n\in 2\mathbb {N} +1\colon {\text{rk}}\left(M\right)=n-1-{\frac {1}{2}}-\sum _{d|n}{\frac {\left|\left[d\right]\right|}{2}}-1={\frac {n+1}{2}}+\sigma _{0}\left(n\right)-2}$

In the case where ${\displaystyle n\in 2\mathbb {N} }$ also ${\displaystyle \left[{\frac {n}{2}}\right]}$ consists of only one element and we get

• ${\displaystyle \forall n\in 2\mathbb {N} \colon {\text{rk}}\left(M\right)=n-1-1-\sum _{d|n}{\frac {\left|\left[d\right]\right|}{2}}-1={\frac {n}{2}}+\sigma _{0}\left(n\right)-2}$

The result follows then together with the identity ${\displaystyle \forall n\in \mathbb {N} \colon n=\left\lfloor {\frac {2n+1}{2}}\right\rfloor }$.

### The binary representation

If we pick as a representative for ${\displaystyle M_{ij}}$ the smallest natural number possible we have ${\displaystyle M_{ij}\equiv _{d_{\mathcal {H}}}\left(i-1\right)\left(j-1\right)-d_{\mathcal {H}}C_{ij}}$ so that ${\displaystyle C}$ counts the number of cycles ${\displaystyle C_{ij}=\left\lfloor {\frac {\left(i-1\right)\left(j-1\right)}{d_{\mathcal {H}}}}\right\rfloor }$ we passed already. E.g. ${\displaystyle \left(5-1\right)\left(4-1\right)\equiv _{7}5=12-7}$ so ${\displaystyle C_{54}=1}$. Now we build the column difference matrix ${\displaystyle B}$ with ${\displaystyle B_{ij}=C_{i+1j}-C_{ij}}$ which contains only zeros and ones. This is the binary representation of ${\displaystyle M}$.

Fig. 1: An example for the binary representations of 1 to 20.

The binary representation has many interesting properties and we will present here some of them. We will show here two derivations but first we introduce ${\displaystyle {\mathcal {H}}_{mn}\equiv {\mathcal {H}}_{m}\otimes {\mathcal {H}}_{n}^{*}}$ as a new ${\displaystyle \left(1,1\right)}$-tensor vector space where ${\displaystyle d_{{\mathcal {H}}_{m}}=m}$ and ${\displaystyle d_{{\mathcal {H}}_{n}}=n}$. In the first place we are interested in a special element ${\displaystyle R\left(m,n\right)\in {\mathcal {H}}_{mn}}$. It is a binary rectangular matrix with all zeros except where the geometric line from ${\displaystyle \left(0,0\right)^{\text{T}}}$ to ${\displaystyle \left(m,n\right)^{\text{T}}}$ intersects one of the ${\displaystyle mn}$ ${\displaystyle 1\times 1}$ squares. We find now that ${\displaystyle {\mathcal {H}}_{\infty \infty }\cong \oplus _{i=1}^{\infty }{\mathcal {H}}_{ii}\cong \times _{i=1}^{\infty }\times _{i=j}^{\infty }{\mathcal {H}}_{ij}}$ so we can look at all the elements ${\displaystyle R\left(m,n\right)}$ at once. The identification works because for some ${\displaystyle M\in {\mathcal {H}}_{\infty \infty }}$ and ${\displaystyle M\left(m,n\right)\in {\mathcal {H}}_{mn}}$ we identify ${\displaystyle M_{H_{-1}\left(m\right)H_{-1}\left(n\right)}=M_{mn}\left(m,n\right)}$ and recall that ${\displaystyle H_{-1}\left(n\right)=\sum _{i=1}^{n}i={\frac {n\left(n+1\right)}{2}}}$ which is the ${\displaystyle n^{\text{th}}}$ triangular number and we have ${\displaystyle H_{-1}\left(n\right)-H_{-1}\left(n-1\right)=n}$.

Fig. 2: The binary ${\displaystyle R}$ matrix for ${\displaystyle m=4,n=8}$.
Fig. 3: The binary ${\displaystyle R}$ matrix for ${\displaystyle m=5,n=8}$.
Fig. 4: The composed binary ${\displaystyle R}$ matrix for all ${\displaystyle 1\leq m,n\leq 8}$.

From figure 2 we can guess that ${\displaystyle R\left(m,n\right)}$ is composed as the direct sum of its irreducible element - that is we may write ${\displaystyle R\left(m,n\right)=\oplus _{i=1}^{{\text{gcd}}\left(m,n\right)}R\left({\frac {m}{{\text{gcd}}\left(m,n\right)}},{\frac {n}{{\text{gcd}}\left(m,n\right)}}\right)}$.

Proof: We draw a line along ${\displaystyle \left(x,{\frac {m}{n}}x\right)^{\text{T}}=\left(x,{\frac {m{\text{gcd}}\left(m,n\right)}{n{\text{gcd}}\left(m,n\right)}}x\right)^{\text{T}}}$. To intersect the lattice nodes we demand that ${\displaystyle x,{\frac {m}{n}}x\in \mathbb {N} }$ which is only possible if ${\displaystyle {\frac {n}{{\text{gcd}}\left(m,n\right)}}{\big \vert }x}$. Because ${\displaystyle 0\leq x\leq n}$ this will happen exactly ${\displaystyle {\frac {n}{\frac {n}{{\text{gcd}}\left(m,n\right)}}}={\text{gcd}}\left(m,n\right)}$ times.

Further from figure 3 we guess that the number of ones for two coprime natural numbers ${\displaystyle m\perp n}$ is given as ${\displaystyle {\text{tr}}R\left(m,n\right)^{\text{T}}R\left(m,n\right)=m+n-1}$.

Proof: We draw a line along ${\displaystyle \left(x,{\frac {m}{n}}x\right)^{\text{T}}}$. Because ${\displaystyle 0\leq x\leq n}$ this adds ${\displaystyle n}$ ones. Another one is added if we transit to the next natural number in ${\displaystyle y}$-direction meaning that we add another ${\displaystyle n-1}$ ones.

From that it follows that the number of ones in a generic ${\displaystyle R\left(m,n\right)}$ matrix is {\displaystyle {\begin{aligned}{\text{tr}}R\left(m,n\right)^{\text{T}}R\left(m,n\right)&=m+n-{\text{gcd}}\left(m,n\right)=\sum _{i=1}^{n}\left\lceil {\frac {m}{n}}i\right\rceil -\left\lfloor {\frac {m}{n}}\left(i-1\right)\right\rfloor \\&=mn-2\sum _{i=1}^{n-1}\left\lfloor {\frac {m}{n}}i\right\rfloor {\text{.}}\end{aligned}}}

The last equation arises by counting the number of zeroes and removing them from the maximal amount of ones which is ${\displaystyle mn}$ or by a little bit more complicated transformation of ceiling to floor. It is not new and a reference will be added. It is this trace formula which tells us that the diagonal and the spectrum of ${\displaystyle R\left(m,n\right)^{\text{T}}R\left(m,n\right)}$ carry some important information about the natural numbers. We may also interpret ${\displaystyle R\left(m,n\right)}$ as a bipartite pure quantum state and ${\displaystyle R\left(m,n\right)^{\text{T}}R\left(m,n\right)}$ as the reduced state in subsystem ${\displaystyle {\mathcal {H}}_{n}}$ and ${\displaystyle R\left(n,m\right)^{\text{T}}R\left(n,m\right)}$ as the reduced state in subsystem ${\displaystyle {\mathcal {H}}_{m}}$. Further we have the properties:

• ${\displaystyle \forall m,n\in \mathbb {N} \colon R\left(m,n\right)^{\text{T}}=R\left(n,m\right)}$
• ${\displaystyle \forall n\leq m\colon {\text{tr}}R\left(m,n\right)R\left(n,m\right)={\text{tr}}R\left(m-n,n\right)R\left(n,m-n\right)+n}$
• ${\displaystyle \forall m\in \mathbb {N} \colon \sum _{i=1}^{m}{\text{tr}}R\left(m,i\right)R\left(i,m\right)=3\sum _{d|m}d\sum _{e\in \left[d\right]}e-\sum _{d|m}\left(d+1\right)\varphi \left({\frac {m}{d}}\right)}$ with ${\displaystyle \left[d\right]\equiv \left\{i\in \mathbb {N} |1\leq i\leq {\frac {m}{d}}\wedge i\perp {\frac {m}{d}}\right\}}$
• ${\displaystyle \sum _{d|m}d\sum _{e\in \left[d\right]}e={\frac {m\left(m+1\right)}{2}}=H_{-1}\left(m\right)}$.
• Notice that we can generalize it e.g. to ${\displaystyle \forall z\in \mathbb {C} \colon \sum _{d|m}d^{-z}\sum _{e\in \left[d\right]}e^{-z}=H_{z}\left(m\right)}$ so that we get the Euler identity for free: ${\displaystyle H_{0}\left(m\right)=m=\sum _{d|m}d^{0}\sum _{e\in \left[d\right]}e^{0}=\sum _{d|m}\varphi \left({\frac {m}{d}}\right)}$.
• ${\displaystyle 2m-1\leq \sum _{i=1}^{m}{\text{gcd}}\left(m,i\right)=\sum _{d|m}d\varphi \left({\frac {m}{d}}\right)\leq m\sigma _{0}\left(m\right)}$
• The last inequality actually tells us that the most simple upper bound ${\displaystyle \sum _{i=1}^{m}{\text{gcd}}\left(m,i\right)\leq H_{-1}\left(m\right)}$ which we can choose is pretty worse because we have by far ${\displaystyle m\sigma _{0}\left(m\right)\leq H_{-1}\left(m\right)}$. The equality and upper bound can also easily extended to the slightly more general gcd-sum: ${\displaystyle \forall z\in \mathbb {C} \colon \sum _{i=1}^{m}{\text{gcd}}\left(m,i\right)^{z}=\sum _{d|m}d^{z}\varphi \left({\frac {m}{d}}\right)\leq m\sigma _{z-1}\left(m\right)}$. Again we get the Euler identity for free: ${\displaystyle m=\sum _{i=1}^{m}{\text{gcd}}\left(m,i\right)^{0}=\sum _{d|m}\varphi \left({\frac {m}{d}}\right)}$.

Now we are turning back to the matrix ${\displaystyle R\left(m,n\right)}$ and show how it is related to the binary representation ${\displaystyle B}$. As a hint we may look at the elements ${\displaystyle \left\{R\left(4,i\right)|1\leq i\leq 5\right\}}$ and we observe that the differences of the number of white squares in every row of the half spaces created through the black line of consecutive matrices belong to the binary representation ${\displaystyle {\begin{pmatrix}0&0&0&0\\0&0&1&0\\1&0&1&0\\1&1&1&0\end{pmatrix}}}$. This is because we have for the number of white squares in every row ${\displaystyle j}$ in the left half space ${\displaystyle \left\lfloor {\frac {i}{m}}\left(j-1\right)\right\rfloor =C_{i+1j}}$.

Let ${\displaystyle H_{mn}\equiv R\left(m,n\right)R\left(n,m\right)}$ and ${\displaystyle \rho _{mn}={\frac {H_{mn}}{{\text{tr}}\left(H_{mn}\right)}}}$. We note that ${\displaystyle H_{n-1n}}$ describes the homogeneous commutative two-point nearest neighbor and self interaction of a linear chain[1]. Then the spectrum [2] is given as

${\displaystyle {\text{sp}}H_{n-1n}=\left\{2\cos \left({\frac {\pi }{n}}i\right)+2|1\leq i with the corresponding normalized eigenbasis [2] ${\displaystyle {\mathcal {B}}_{H_{n-1n}}=\left\{\left({\sqrt {\frac {2}{n}}}\sin \left({\frac {\pi }{n}}ij\right){\big \vert }1\leq j. Because ${\displaystyle {\text{tr}}\left(H_{n-1n}\right)=2\left(n-1\right)}$ we can calculate the entanglement information {\displaystyle {\begin{aligned}I\left(\rho _{n-1n}\right)&=\ln \left(n-1\right)-S\left(\rho _{n-1n}\right)=\ln \left(n-1\right)+\sum _{i=1}^{n-1}{\frac {\cos \left({\frac {\pi }{n}}i\right)+1}{n-1}}\ln {\frac {\cos \left({\frac {\pi }{n}}i\right)+1}{n-1}}\\&={\frac {1}{n-1}}\left(\sum _{i=1}^{n-1}\cos \left({\frac {\pi }{n}}i\right)\ln \left(\cos \left({\frac {\pi }{n}}i\right)+1\right)+\sum _{i=1}^{n-1}\ln \left(\cos \left({\frac {\pi }{n}}i\right)+1\right)\right)\end{aligned}}}

which converges to ${\displaystyle \lim _{n\to \infty }I\left(\rho _{n-1n}\right)=1-\ln 2\approx 0.306853}$ which is less than the expected information of ${\displaystyle {\frac {1}{2}}}$ if we would take a pure bipartite quantum state at random under the unitarily invariant Haar measure. Anyway we conjecture here that ${\displaystyle I\left(\rho _{mn}\right)=O\left({\frac {\min \left(m,n\right)}{\max \left(m,n\right)}}\right)}$ meaning that it behaves similar like the information content of a random ensemble. Further if ${\displaystyle d|n}$ then ${\displaystyle I\left(\rho _{dn}\right)=0}$.

## Bipartite factorization and the Lorentz scalar

Let ${\displaystyle n,m_{1},m_{2}\in \mathbb {N} +1}$ such that ${\displaystyle n=m_{1}m_{2}}$. If we define ${\displaystyle x_{1}\equiv {\frac {1}{2}}\left(m_{1}+m_{2}\right)}$ and ${\displaystyle x_{2}\equiv {\frac {1}{2}}\left(m_{1}-m_{2}\right)}$ then ${\displaystyle x_{1}^{2}-x_{2}^{2}=\left(x_{1}+x_{2}\right)\left(x_{1}-x_{2}\right)=m_{1}m_{2}=n}$. But ${\displaystyle x^{\text{T}}{\text{M}}x=x_{1}^{2}-x_{2}^{2}}$ is the squared semi-norm of the ${\displaystyle 2}$-dimensional Minkowski space where ${\displaystyle {\text{M}}={\text{diag}}\left(1,-1\right)}$. Therefore ${\displaystyle x}$ is just another parametrization of ${\displaystyle n}$ and we have ${\displaystyle x^{\text{T}}{\text{L}}_{t}^{\text{T}}{\text{M}}{\text{L}}_{t}x=x^{\text{T}}{\text{M}}x=n}$ where ${\displaystyle {\text{L}}_{t}\equiv {\begin{pmatrix}\cosh t&\sinh t\\\sinh t&\cosh t\end{pmatrix}}}$ is the Lorentz boost. It has the nice property that ${\displaystyle {\text{L}}_{t}={\text{L}}_{1}^{t}}$ and forms therefore a one-parameter group. This ultimately shows that ${\displaystyle n}$ is really Lorentz scalar under bipartite factorization. Now ${\displaystyle n}$ is invariant under the dilation of ${\displaystyle n}$ with ${\displaystyle {\text{T}}_{2}={\text{diag}}\left(s,{s^{-1}}\right)}$. We find that a boost of ${\displaystyle x}$ with ${\displaystyle t}$ dilates ${\displaystyle m}$ with ${\displaystyle s=\cosh t+\sinh t=\exp \left(t\right)}$. This makes it now easy for us to find the boost for certain lattice points.

Example (Boost of integer points): Let ${\displaystyle m_{1}=2*2}$ and ${\displaystyle m_{2}=3}$. Then there exists a boost from the point ${\displaystyle x={\frac {1}{2}}\left(7,1\right)}$ to the point ${\displaystyle x=\left(4,2\right)}$. This is because if we choose ${\displaystyle s={\frac {3}{2}}}$ then ${\displaystyle t=\ln {\frac {3}{2}}}$.

Lemma (Boost of integer points): There exists a boost from the point ${\displaystyle x=\left({\frac {n+1}{2}},{\frac {n-1}{2}}\right)}$ to the point ${\displaystyle x=\left({\frac {n+1}{2}},-{\frac {n-1}{2}}\right)}$.

Proof: This is because if we choose ${\displaystyle s={\frac {1}{n}}}$ then ${\displaystyle t=-\ln n}$. ${\displaystyle \square }$

If we choose ${\displaystyle t=\ln \left(\pm {\frac {\sqrt {m_{2}}}{\sqrt {m_{1}}}}\right)=\ln \left(\pm {\frac {\sqrt {x_{1}-x_{2}}}{\sqrt {x_{1}+x_{2}}}}\right)}$ then ${\displaystyle {\text{L}}_{t}x=\left(\pm {\sqrt {n}},0\right)}$. This shows that if we want to resolve a bipartite factorization of ${\displaystyle n}$ it is sufficient to know the fraction ${\displaystyle {\frac {m_{2}}{m_{1}}}}$ of the factors. This can be seen also by ${\displaystyle {\sqrt {n{\frac {m_{2}}{m_{1}}}}}={\sqrt {m_{1}m_{2}{\frac {m_{2}}{m_{1}}}}}=m_{2}}$. However without a lot of work there is no chance to infer this fraction with our correspondence principle between dilation and Lorentz boost. But it allows us to state this as an inverse problem by finding ${\displaystyle {\text{L}}_{t}^{-1}={\text{L}}_{-t}}$ so that ${\displaystyle {\text{L}}_{-t}\left(\pm {\sqrt {n}},0\right)\in \mathbb {Z} ^{\times 2}}$. We find a bipartite factorization for odd numbers for example if we determine first ${\displaystyle k_{\min }=\min \left\{k>1|m2^{k}\equiv _{2n+1}m\right\}}$ and then ${\displaystyle m_{\min }=\min \left\{m|m2^{k_{\min }}\equiv _{2n+1}m\right\}}$ which gives us then the greatest non-trivial factor of ${\displaystyle 2n-1}$. If ${\displaystyle 2n-1}$ is not a prime we have ${\displaystyle 2n-1=m_{\min }\left(k_{\min }+1\right)}$. It is unclear how the formalism above would help us here.

Because ${\displaystyle 1=\left(\cosh t\right)^{2}-\left(\sinh t\right)^{2}=\left(\cosh t+\sinh t\right)\left(\cosh t-\sinh t\right)}$ we may think also of a different parametrization for ${\displaystyle x}$. We might choose ${\displaystyle {\tilde {x}}'={\sqrt {n}}\left(\cosh t,\sinh t\right)}$. Then ${\displaystyle {\tilde {x}}'^{\text{T}}{\text{M}}{\tilde {x}}'=n}$ and to make ${\displaystyle {\tilde {x}}'}$ independent of ${\displaystyle t}$ we introduce ${\displaystyle {\tilde {\text{L}}}_{t}={\text{diag}}\left(\cosh t,\sinh t\right)}$ so that ${\displaystyle {\tilde {x}}'={\tilde {\text{L}}}_{t}{\tilde {x}}}$ with ${\displaystyle {\tilde {x}}={\sqrt {n}}\left(1,1\right)}$. Here we cannot revover ${\displaystyle m}$ from ${\displaystyle {\tilde {x}}}$ (actually we have ${\displaystyle {\tilde {x}}^{\text{T}}{\text{M}}{\tilde {x}}=0}$) but we may think of recovering ${\displaystyle m}$ from ${\displaystyle {\tilde {x}}'={\text{L}}_{t}\left({\sqrt {n}},0\right)}$ because we have the identity ${\displaystyle {\text{L}}_{t}\left(1,0\right)={\tilde {\text{L}}}_{t}\left(1,1\right)}$. From the identity it follows that there is not a ${\displaystyle t_{x}}$ such that ${\displaystyle {\tilde {\text{L}}}_{t_{x}}=1}$ so ${\displaystyle {\tilde {\text{L}}}_{t}}$ acts not continuous on the diagonal vector.

### Compatibility of bipartitions

We did not talk about compatibility between two vectors ${\displaystyle x}$ and ${\displaystyle y}$ under a boost ${\displaystyle t}$. We want to calculate the boost parameter ${\displaystyle t_{y}}$ such that ${\displaystyle x^{\text{T}}{\text{L}}_{t_{x}}^{\text{T}}{\text{M}}{\text{L}}_{t_{y}}y=x^{\text{T}}{\text{M}}x\equiv n_{x}}$. Under the boosts ${\displaystyle t_{x},t_{y}}$ we say that ${\displaystyle y'\equiv {\text{L}}_{t_{y}}y}$ is compatible with ${\displaystyle x'\equiv {\text{L}}_{t_{x}}x}$.

Corollar (Distance of compatible bipartitions): ${\displaystyle \left(x'-y'\right)^{\text{T}}{\text{M}}\left(x'-y'\right)=n_{y}-n_{x}}$.

Proof: ${\displaystyle \left(x'-y'\right)^{\text{T}}{\text{M}}\left(x'-y'\right)=x'^{\text{T}}{\text{M}}x'+y'^{\text{T}}{\text{M}}y'-x'^{\text{T}}{\text{M}}y'-y'^{\text{T}}{\text{M}}x'=y^{\text{T}}{\text{M}}y-x^{\text{T}}{\text{M}}x=n_{y}-n_{x}}$.${\displaystyle \square }$

But how does the parameter ${\displaystyle t_{y}}$ looks like? We solve ${\displaystyle x^{\text{T}}{\text{L}}_{t_{x}}^{\text{T}}{\text{M}}{\text{L}}_{t_{y}}y=n_{x}}$ for ${\displaystyle t_{y}}$ in our dilation coordinates ${\displaystyle m_{x}}$ and ${\displaystyle m_{y}}$. Remember that ${\displaystyle m_{x1}=x_{1}+x_{2}}$, ${\displaystyle m_{x2}=x_{1}-x_{2}}$ and ${\displaystyle m_{y1}=y_{1}+y_{2}}$, ${\displaystyle m_{y2}=y_{1}-y_{2}}$. We find that there exists two solutions: ${\displaystyle t_{y}^{\pm }=\ln {\frac {n_{x}\pm {\sqrt {n_{x}^{2}-n_{x}n_{y}}}}{m_{x1}m_{y2}\left(\cosh t_{x}-\sinh t_{x}\right)}}}$ which are only real if ${\displaystyle n_{y}\leq n_{x}}$ under the assumption that ${\displaystyle m_{x1}m_{y2}\left(\cosh t_{x}-\sinh t_{x}\right)>0}$. If we exclude a complex valued ${\displaystyle t_{y}^{\pm }}$ we can say alternatively that lower valued factorizations are only compatible with higher valued ones. However a complex valued boost parameter shifts our points also into the complex plane which means that we can now find a boost which transforms between the complex integers to the real ones! Consider e.g. the factorization ${\displaystyle 25=3^{2}+4^{2}=\left(3+{\text{i}}4\right)\left(3-{\text{i}}4\right)}$. If we take ${\displaystyle m_{x}=\left(25,1\right)}$ and ${\displaystyle m_{y}=\left(3+{\text{i}}4,3-{\text{i}}4\right)}$ then ${\displaystyle x=\left(13,12\right)}$ and ${\displaystyle y=\left(3,{\text{i}}4\right)}$. This means that ${\displaystyle t_{y}=\ln \left(\left(3-{\text{i}}4\right)\exp t_{1}\right)}$. Essential we can reach any factorization of some integer ${\displaystyle n}$ under any quadratic field which means that their integer lattices are connected under hyperbolic rotation. Every bipartite factorization of a number ${\displaystyle n}$ in any quadratic field lies on the orbit ${\displaystyle {\text{L}}_{t}\left({\sqrt {n}},0\right)}$. In our example we used a decomposition in the Gaussian integers which is the prototype for an imaginary quadratic field. Note that in our setting the physical interpretation of ${\displaystyle x_{2},y_{2}}$ is that of time or space (dependent on the convention) so we would need here imaginary time or space to reach the Gaussian integers (or in general complex numbers).

### A connection to Goldbach's conjecture

Our considerations may seem simple but there is already a connection to a really deep problem called the Goldbach conjecture (GC). If we consider the sum ${\displaystyle m_{1}+m_{2}=2x_{1}}$ of the factors then we see that this yields an even number if ${\displaystyle m_{1},m_{2}\in 2\mathbb {N} }$ or ${\displaystyle m_{1},m_{2}\in 2\mathbb {N} +1}$. For the GC to be true we need the latter property (because we have with ${\displaystyle 2+2=4}$ the only occurence of the first prime ${\displaystyle 2}$) so we fix ${\displaystyle x_{1}+x_{2}\in 2\mathbb {N} +1}$ with ${\displaystyle x_{1},{\tilde {x}}_{2}\in \mathbb {N} }$ such that ${\displaystyle x_{2}=2{\tilde {x}}_{2}+1-x_{1}}$. The GC states now that ${\displaystyle \forall x_{1}\in \mathbb {N} \colon \exists {\tilde {x}}_{2}\in \mathbb {N} \colon 3\leq \left|\left\{{\text{T}}|{\text{T}}\left(x_{1},2{\tilde {x}}_{2}+1-x1\right)\in \mathbb {Z} ^{\times 2}\right\}\right|\leq 4}$. Please note that the lower bound seems necessary only for the case ${\displaystyle 3+3=6}$. The upper and lower bound arises from the fact that the number of divisors of a number composed of two primes is ${\displaystyle \sigma _{0}\left(n\right)=3}$ in the case that ${\displaystyle n=p^{2}}$ with ${\displaystyle 2 and ${\displaystyle \sigma _{0}\left(n\right)=4}$ otherwise. The transformations we need to count take ${\displaystyle \left(x_{1},2{\tilde {x}}_{2}+1-x1\right)}$ to ${\displaystyle {\text{T}}\left(x_{1},2{\tilde {x}}_{2}+1-x1\right)=x_{1}\exp t\left(1,-1\right)+\left(2{\tilde {x}}_{2}-1\right)\left(-\sinh t,\cosh t\right)}$.

### Sums of bipartite factorizations

We can easily complicate the things further. Let us generalize our form for ${\displaystyle n}$ as ${\displaystyle n=\sum _{i=1}^{d}m_{1i}m_{2i}}$ where ${\displaystyle m_{1},m_{2}\in \mathbb {N} ^{\times d}}$. For ${\displaystyle d=1}$ we get back our initial case from above. We can also write ${\displaystyle n=m_{1}^{\text{T}}m_{2}}$ by taking ${\displaystyle m_{1},m_{2}}$ elements from the Hilbert space ${\displaystyle \mathbb {R} ^{\times d}}$. Here we encounter already the orthogonal symmetry because ${\displaystyle m_{1}^{\text{T}}{\text{O}}^{\text{T}}{\text{O}}m_{2}=m_{1}^{\text{T}}m_{2}}$. But this is just a special kind of dilation because ${\displaystyle {\text{O}}^{-1}={\text{O}}^{\text{T}}}$. So our complete symmetry is again defined over the space ${\displaystyle \left(m_{1},m_{2}\right)\in \mathbb {R} ^{\times 2d}}$. We have symmetry of ${\displaystyle n}$ under the transformation ${\displaystyle {\text{T}}_{2}={\text{S}}^{\text{T}}\oplus {\text{S}}^{-1}}$ with ${\displaystyle {\text{S}}\in {\text{GL}}_{d}\left(\mathbb {R} \right)}$. Now we do the same again and define ${\displaystyle x_{i1}={\frac {m_{1i}+m_{2i}}{2}}}$, ${\displaystyle x_{i2}={\frac {m_{1i}-m_{2i}}{2}}}$ and ${\displaystyle x\equiv \left(x_{i}|1\leq i\leq d\right)}$. Then ${\displaystyle n=\sum _{i=1}^{d}x_{i1}^{2}-x_{i2}^{2}=x^{\text{T}}{\text{M}}x}$ where ${\displaystyle {\text{M}}\equiv \oplus _{i=1}^{d}{\text{diag}}\left(1,-1\right)}$. We may infer for the finite dimensional case the symmetry group for the elements ${\displaystyle x}$. It will turn out that they are symmetric under ${\displaystyle {\text{O}}_{d,d}\left(\mathbb {R} \right)}$ because we have ${\displaystyle {\text{M}}\cong 1_{d}\oplus -1_{d}}$. However if we want to establish such an isomorphism we need a permutation matrix ${\displaystyle {\text{A}}}$ which maps the odd components ${\displaystyle x_{i1}}$ to the first half and the even components ${\displaystyle x_{i2}}$ to the second half. E.g. ${\displaystyle x_{i1}\rightarrow x'_{i}}$ and ${\displaystyle x_{i2}\rightarrow x'_{d+i}}$. Sadly this is not well defined for ${\displaystyle d\rightarrow \infty }$. Anyway we may conclude that ${\displaystyle {\text{O}}_{d}\left(\mathbb {R} \right)\oplus {\text{O}}_{d}\left(\mathbb {R} \right)\subset {\text{O}}_{d,d}\left(\mathbb {R} \right)}$ and an additional boost ${\displaystyle {\text{L}}_{t}}$ between e.g ${\displaystyle x'_{1}}$ and ${\displaystyle x'_{d}}$. We may write ${\displaystyle \forall \Lambda \in {\text{O}}_{d,d}\left(\mathbb {R} \right)\colon \Lambda ={\text{R}}_{11}\oplus {\text{R}}_{21}{\text{L}}_{t}{\text{R}}_{12}\oplus {\text{R}}_{22}}$