In the sieve of Eratosthenes the erased composite numbers are identified by $ 5 $ equations:
$nc=2(a+1)$
$nc=3(2a+1)$
$nc=−1+6(6ab+a−b)$
$nc=1+6(6ab+a+b)$
$nc=1+6(6ab−a−b)$
with $ a\geq 1 $ and $b\geq 1$.
How a modified version of Eratosthenes sieve can be tested to find primes greater than $ 5 $, out of order, is as follows:
we write the numbers $-1\bmod6$ greater than $5$
11, 17, 23, 29, 35, 41, 47, 53, 59, 65, 71, 77, 83, 89, 95, 101, 107, 113, 119, 125, 131, 137, 143, 149, 155, 161, 167, 173, 179, 185, 191, 197, 203, 209, 215, 221, 227, 233, 239, 245, 251, 257, 263, 269, 275, 281, 287, 293, 299, 305, 311, 317, 323, 329, 335, 341, ...
note that if we divide the multiples of $5$ by $5$ we get the numbers $1\bmod6$ (indicated in brackets)
11, 17, 23, 29, 35 (7), 41, 47, 53, 59, 65 (13), 71, 77, 83, 89, 95 (19), 101, 107, 113, 119, 125 (25), 131, 137, 143, 149, 155 (31), 161, 167, 173, 179, 185 (37), 191, 197, 203, 209, 215 (43), 221, 227, 233, 239, 245 (49), 251, 257, 263, 269, 275 (55), 281, 287, 293, 299, 305 (61), 311, 317, 323, 329, 335 (67), 341, ...
if we delete multiples of $11, 17, 23, 29, 35, ...$
at the end delete the number $125$
and replace the muliples of $5$ with the same number divided by $5$,
only primes remain (considering also those in brackets).
Example of sieve procedure
The composite numbers eliminated with this method other than $125$ can be identified with the following Diophantine equation:
$$nc=-1+6 \cdot ( 6 \cdot a \cdot b +a-b)$$
with $ a\geq 2 $ and $b\geq 1$
QUESTION
Is it possible to proof by contradiction that there are infinitely many prime numbers?
For example consider $p_n$ the greatest prime number any finite list of prime numbers
let's set $k_n=\frac{5p_n+1}{6}$ if $p_n \equiv 1 \bmod 6$ or
$k_n=\frac{5p_n-1}{6}$ if $p_n \equiv -1 \bmod 6$
if we assume that there are no prime numbers greater than $ p_n $ then it is the same as saying
$\forall k> k_n$ $\quad \exists a_k \geq 2 ,b_k \geq 1$ $ |$ $k =6 \cdot a_k \cdot b_k+a_k-b_k$
Is it possible to prove that this cannot happen so that it is a contradiction?
Is it possible to find in minimum value of $k$ for which $ k \neq 6 \cdot a_k \cdot b_k + a_k -b_k \quad \forall a_k \geq 2 ,b_k \geq 1$ in order to determine an upper bound of the prime gap?
The sieve implies something stronger than the existence of infinitely many primes, namely $\sum_p 1/p=\infty$ as discovered by Euler. Indeed the asymptotic density of the integers not divisible by the primes $p_1,p_2,\ldots,p_k$ is $\prod_{i=1}^k (1-p_i^{-1})$. These are exactly the integers not erased in the first $k$ stages of the sieve. Since every integer greater than 1 is divisble by some prime, the product over all primes satisfies $$\prod_p (1-p^{-1})=0 \,. \quad (*)$$ To see this analytically, expand each factor in the product as a geometric series $1+1/p+1/p^2+\ldots$ and observe that the product $(*)$ yields the harmonic series $\sum_{n \ge 1} 1/n$ since each $n$ has a unique factorization as a product of prime powers. Finally $(*)$ implies $\sum_p 1/p =\infty$ because of the classical equivalence between infinite products and the sums of the differences of individual factors from 1. https://en.wikipedia.org/wiki/Infinite_product