How do you multiply in Primorial number system?

Question

Primorial number system is a number sytem that uses primorials which are defined as follows :

Let $p_1=2, p_2=3,p_3=5,p_4=7,p_5=11,...$ the primes. The sequence of primorials, noted $p_n\#$ is $$(p_n\#)_{n\geq 1}=(2,2.3=6,2.3.5=30,2.3.5.7=210,210.11=2310,...$$

For example, we write $15$ and $28$ as $$15=2\times 6+1\times 2+1=(2:1:1)$$ $$28=(4:2:0)$$

I'm trying to think with your help hopefully about a multiplication algorithm of positive integers written in Primorial number system(two steps 1 and 2).

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1.- Calculate congruences modulo any prime number via this number system, is very easy : for example, any integer like $(\color{green}{...:1:2:5:2:1:6:9:1:1:}\color{red}{0:0:2:1:1})$ is such that $$(\color{green}{...:6:9:1:1:}\color{red}{0:0:2:1:1})\equiv \color{green}{0+}\color{red}{0\times1+0\times 8+2\times 6+1\times 2+1\times 1 }\equiv 4 (\mod 11) $$since $2310\equiv 0 , 210\equiv 1 , 30\equiv 8 ,6\equiv 6, 2\equiv 2, 1\equiv 1 (\mod 11)$

So, $$\begin{cases} (2:1:1)\equiv \color{blue}{1,0,0,1,4} \mod 2,3,5,7,11\\ \land \\ (4:2:0)\equiv \color{blue}{0,1,3,0,6} \mod 2,3,5,7,11 \end{cases}$$

I took small integers but the same method works for integers with $\color{green}{\text{several hundred digits}}$ in base two.

2.- My idea to multiply $(2:1:1)$ and $(4:2:0))$ is then to use Chinese Remainder theorem once we have realized that $$P:=(2:1:1)\times (4:2:0)<2310\color{red}{(*)}$$

We obtain $$P\equiv \color{blue}{0,0,0,0,2} \mod 2,3,5,7,11$$ Then $\exists k, P=210k \land 210k=2 \text{ in } \mathbb Z/11\mathbb Z \iff 1k=2 \iff k=2$ And finally $$\boxed {P=(2:1:1)\times (4:2:0)=(2:0:0:0:0)}$$

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a. Is that correct ? In particular $\color{red}{(*)}$ should probably be clarified

b. Switching from primorial numeral to base ten or two is easy. $$P=(2:1:1)\times (4:2:0)\to 15\times 28=420 \to (2:0:0:0:0)$$

is a procedure which, at first glance, seems sufficient. Nevertheless, I think that the question arises as to whether this basic procedure is really sufficient, since I have mentioned $\color{green}{\text{several hundred digits}}$(A bit random I admit) and wondered whether we could not rather say several thousand or even more.

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Edit: To partially reply to the comment by @Steven Clark, here is another example $P=ab$, with $a=(...:c:b:a:0:1:9:6:2:2:1)\equiv 1,2,2,1,8,3,11$ and $b=(...:\gamma:\beta:\alpha:0:3:4:4:0:1:1)\equiv 1,0,3,4,6,2,5\mod 2,3,5,7,11,13,17$.

$P=\color{blue}1+2k_0. 1+2k_0=0 \text{ in }Z/3\implies k_0=\color{blue}1+3k_1. P=1+2(1+3k_1)=3+6k_1. P=3+1k_1 \text{ in }Z/5\implies k_1=3+5k_2. P=\color{blue}1\times 1+\color{blue}1\times 2+\color{blue}3\times 6 +30k_2. P=0+2k_2 \text{ in }Z/7\implies k_2=2+7k_3. P=\color{blue}1\times 1+\color{blue}1\times 2+\color{blue}3\times 6 +\color{blue}2 \times 30+210k_3...$

So, $$P=(...:\color{blue}{2:3:1:1})$$ (example constructed with $4\,397\times 7\,893=34\,705\,521$)

Answer 1

Your algorithm isn't clear to me from your description, so I'll give an algorithm and hopefully you can clarify if your algoritrhm is the same or how it differs.

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Assuming

$$a=\sum\limits_{j=0}^A a_j\, p_j\#=\{a_0, a_1, ..., a_A\}\tag{1}$$

and

$$b=\sum\limits_{j=0}^B b_j\, p_j\#=\{b_0, b_1, ..., b_B\}\tag{2}$$

then I believe

$$c=a\times b=\sum\limits_{j=0}^A \left(a_j\, p_j\#\, \sum\limits_{k=0}^B b_k\, p_k\#\right)\tag{3}.$$

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Formula (3) above gives a normal numerical result, but this can easily be converted back into the primorial representation

$$c=\sum\limits_{j=0}^C c_j\, p_j\#=\{c_0, c_1, ..., c_C\}\tag{4}$$

if desired.

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Could you please give a more generic description of your algorithm for $a\times b$ where $a$ and $b$ are defined in formulas (1) and (2) above? Perhaps the simple case $\left\{a_0, a_1, a_2\right\} \times \left\{b_0, b_1, b_2\right\}$ would provide some clarification because in your numeric example I don't understand where/how you're getting your numbers or applying the Chinese remainder theorem.