# Proof of shifted primorial periodicity of aggregated prime signatures

This problem originates from some of my experiments looking at difference of prime numbers, i've removed a lot of the intuition to simplify and shorten the question, if requested I will include it at the end of the question.

Define a signature to be a list of boolean numbers $0,1$. Two signatures $S_1, S_2$ are said to be equal if there exists a signature $S_3$ such that $S_2$ and $S_1$ can be produced by concatenating a natural number of copies $a_2, a_1$ of $S_3$ respectively. Example: $S_1 = [1, 0 ,1], S_2 = [1, 0 ,1 ,1 , 0 , 1]$ then $S_1 = S_2$ since both are a natural number copies of $[1,0,1]$.

Define an elementary signature $S_e$ of a natural number $n \in \mathbb{N}$ as a sequence of binary numbers:

$$[ \underbrace{0, \ 0, \ ... 0}_{n-1 \ \text{times}} \ , 1 ]$$

Example, $S_e(2) = [0, 1]$ $S_e(3) = [0,0,1]$

Define the "overlap" $S_1 \oplus S_2$ of 2 signatures $S_1, S_2$ to be a new signature whose length is the least common multiple of the lengths of $S_1, S_2$ and whose entries are the pairwise boolean OR of each other corresponding elements in $S_1, S_2$.

Example $$S_e(2) \oplus S_e(3) = \\ [0,1] \oplus [0,0,1] =\\ [0, 1, 0, 1, 0, 1] \oplus [0, 0, 1, 0, 0, 1] = [0, 1,1,1, 0 ,1]$$

Now one last definition, we can define the "aggregation" $\aleph(S)$ of a signature $S$, to be a list of natural numbers, where for every sequence of non-zero values in the signature the aggregation carries a single element equal to their sum.

Example

$$\aleph(S_e(5)) = \aleph([0,0,0,0,1]) = [0,0,0,0,1]$$ $$\aleph(S_e(2) \oplus S_e(3) ) = \aleph([0,1,1,1,0,1]) = [0, 1+1+1, 0, 1] = [0,3,0,1]$$ $$\aleph(S_e(2) \oplus S_e(3) \oplus(S_e(5)) =\\ \aleph([0,1,1,1,0,1] \oplus [0,0,0,0,1]) = \\ \aleph([0,1,1,1,1,1,0,1,1,1,0,1,0,1,1,1,0,1,...]) = \\ [0,1+1+1+1+1, 0, 1+1+1, 0 ,1 ...] = \\ [0,5,0,3,0,1,0,3,0, 1,0,3,0,5,0,1]$$

If you look at the lengths of the signatures denoted by $\dim$ (to be formal the length is the length of smallest member of the equivalence class given by a signature)

$$\dim \aleph(S_e(2)) = 2$$ $$\dim \aleph(S_e(2) \oplus S_e(3)) = 4$$ $$\dim \aleph(S_e(2) \oplus S_e(3) \oplus S_e(5)) = 16$$ $$\dim \aleph(S_e(2) \oplus S_e(3) \oplus S_e(5) \oplus S_e(7)) = 96$$ $$\dim \aleph(S_e(2) \oplus S_e(3) \oplus S_e(5) \oplus S_e(7) \oplus S_e(11) ) = 960$$

Notice $$4 = (3-1)(2) \\ 16 = (5-1)(4) \\ 96 = (7-1)(16) \\ 960 = (11 - 1)(96) \\ \vdots$$

This leads to the natural conjecture, which I'm trying to prove (let $P_n$ denote the $n^{\text{th}}$ prime number)

$$\dim \aleph \left( \bigoplus_{i=1}^{n} S_e(P_i) \right) = 2\prod_{i=1}^{n} (P_i - 1)$$

What subject does this fall under? What type of machinery would be needed to do this.

• You might want to define two signatures "equaling" each other slightly differently so that $A=B\iff B=A:$ elsewise, you end up with $=$ denoting a relation that is not an equivalence relation, and doesn't act how $=$ is generally assumed to act. – Austin Weaver Oct 20 '17 at 18:16
• It doesn't seem coincidental that $\phi(\#_n)=\prod_{i=1}^n(P_i-1)$, where $\#_n$ is the primorial product. So likely this is calculating the order of the group $\mathbb{Z}\backslash \mathbb{Z}_{\#_n}$ in a different way, and in fact in this case by the chinese remainder thoerem: $\mathbb{Z}\backslash \mathbb{Z}_{\#_n}=\mathbb{Z}\backslash \mathbb{Z}_{P_1}\times\cdots\times \mathbb{Z}\backslash \mathbb{Z}_{P_n}$ – Alex R. Oct 20 '17 at 18:51
• @AustinWeaver noted! I modified the definition to adjust for that – frogeyedpeas Oct 20 '17 at 19:08

## 1 Answer

First, I will show that $\dim\aleph\left(\bigoplus_{i=1}^nS_e(p_i)\right)$ is equal to twice the number of $0$'s in $\bigoplus_{i=1}^nS_e(p_i)$. Now, this immediately follows from the fact that there are no adjacent $0$'s in the overlap, and the fact that the overlap starts with $0$ and ends in $1$. But how do we know there are no adjacent $0$'s in the overlap? Because every other item of the overlap is guaranteed to be $1$, coming from the $S_e(2)$.

We have established that the LHS of your conjecture is equal to twice the number of $0$'s in the overlap. So now all we have to show is that the number of $0$'s in the overlap is equal to $\prod_{i=1}^n(p_i-1).$ This product is also equal to $\varphi\left(\prod_{i=1}^np_i\right)$ where $\varphi$ is Euler's totient function. See here.

Let's think about what it means for the $k$th entry of the overlap to be a $0$. For this to be true, the $k$th entry of each $S_e(p_1),S_e(p_2),\dots,S_e(p_n)$ must be $0$. In other words, $k$ must be coprime to each $p_1,p_2,\dots,p_n$. This is equivalent to saying $k$ must be coprime to $\prod_{i=1}^np_i$. The number of all such $k< \prod_{i=1}^np_i$ is exactly $\varphi\left(\prod_{i=1}^np_i\right)$ by definition.

Therefore, we have shown that $$\dim\aleph\left(\bigoplus_{i=1}^nS_e(p_i)\right)=2\prod_{i=1}^n(p_i-1)$$