Let`s calculate primes! Part I - Representation of times tables

Fr 29 Dezember 2017 in Math
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#primes #number theory

Some days ago, I had several nice ideas for calculating primes recursively, which I want to share with you in a small series of posts.

I will use some insights of my work https://github.com/Samdney/primescalc and assume you already know they. If not, please read it sidewise to the following posts.

We have given our already known equation

$$ x_{i,j} = 2x_{i}x_{j} + x_{i} + x_{j}\\ = \left(2x_{j} + 1\right)x_{i} + x_{j}\\ = \left(2x_{i} + 1\right)x_{j} + x_{i} $$

with $x_{i}, x_{j} \in \mathbb{N}$. Remember that this equation gives us all $x_{i,j}$ for which $2x_{i,j} + 1$ is an integer divisible number.

So let`s look, for example on the numbers of $x_{i} = 1$, which are

$$x_{1,j} = 4, 7, 10, 13, 16, 19, 22, 25, 28, \dots$$

Now we will choose a simple way of representation of this numbers. Be given the general form of a number in decimal representation:

$$ \sum_{k = 0}^{n} 10^{k} $$

with $n+1$, $n \in \mathbb{N}$, digits. Now assume, in our example, the number 4 is represented by the number 1, at the $4+1$ digit, the number 7 by the number 1, at the $7+1$ digit and so on. So we can write (read from right to left) as represenation for $x_{1,j}$:

$$ \sum_{x_{j}} 10^{\left(2x_{i} + 1\right)x_{j} + x_{i}} = \sum_{x_{j}} 10^{3x_{j} + 1} $$

and

$$ \dots 10010010010010010010010010000 $$

In this way, we can write every of our times tables which are given by $x_{i,j} = \left(2x_{i} + 1\right)x_{j} + x_{i}$.

If we finally calculate this sum, we receive

$$ \sum_{x_{j} = l}^{u} 10^{\left(2x_{i} + 1\right)x_{j} + x_{i}} = - \frac{10^{\left(2x_{i} + 1\right)l + x_{i}} - 10^{\left(2x_{i} + 1\right)\left(u+1\right) + x_{i}}}{10^{2x_{i} + 1} - 1} $$