When taking a set of consecutive numbers we need to take into consideration that there are multiples of prime numbers that compromise most of the numbers:
0.5 of all numbers will be multiples of 2
0.333333 of all numbers will be multiples of 3, however 0.5 of these will be also multiples of 2 and thus only 0.16666666 will be multiples of only 3
0.2 of all numbers will be multiples of 5, however 0.33333333333333 of these will be also multiples of 3 and thus only 0.1333333333333 will be multiples of 5, however 0.5 of these will be also multiples of 2 and thus only 0.0666666666666 be multiples of only 5
and so on ....
As you can see here: http://numbersprime.com/newtz.php, the total seems to never add up to 1, thus meaning there will always be a percentage of numbers that are multiples of primes less than 100%, leading to the fact that the other numbers must be prime
However in my experiment, I take all multiples of 2 and all multiples of all odd numbers (not only prime numbers) and as you can see here: http://numbersprime.com/newtz2.php, it seems the total still will never add to 1, leading to the fact that the other numbers must be prime.
Since my computational power has limits as you can see in the links above, I was wondering whether to expect the total to always be under 1?
It seems that you are constructing the Sieve of Eratosthenes.
This is an efficient way to generate a list of primes. First write down the natural numbers starting at $2$ and as far as your paper and pencil (or computer) allows ($30$ in my case).
$2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30$
Now cross out all the multiples of $2$ except for $2$ itself.
$\require{cancel} 2, 3, \cancel{4}, 5, \cancel{6}, 7, \cancel{8}, 9, \cancel{10}, 11, \cancel{12}, 13, \cancel{14}, 15, \cancel{16}, 17, \cancel{18}, 19, \cancel{20}, 21, \cancel{22}, 23, \cancel{24}, 25, \cancel{26}, 27, \cancel{28}, 29, \cancel{30}$
Naively, nearly half of the numbers have gone.
Now cross out all the multiples of $3$ except for $3$ itself (which are not already crossed out).
$\require{cancel} 2, 3, \cancel{4}, 5, \cancel{6}, 7, \cancel{8}, \cancel{9}, \cancel{10}, 11, \cancel{12}, 13, \cancel{14}, \cancel{15}, \cancel{16}, 17, \cancel{18}, 19, \cancel{20}, \cancel{21}, \cancel{22}, 23, \cancel{24}, 25, \cancel{26}, \cancel{27}, \cancel{28}, 29, \cancel{30}$
Now, nearly a third of the remaining numbers have gone.
And now $5$,
$\require{cancel} 2, 3, \cancel{4}, 5, \cancel{6}, 7, \cancel{8}, \cancel{9}, \cancel{10}, 11, \cancel{12}, 13, \cancel{14}, \cancel{15}, \cancel{16}, 17, \cancel{18}, 19, \cancel{20}, \cancel{21}, \cancel{22}, 23, \cancel{24}, \cancel{25}, \cancel{26}, \cancel{27}, \cancel{28}, 29, \cancel{30}$
$8$ of $30$ remain, this is convenient, see below.
Etc.
What is left, are the primes up to the square of your last starting number. After that, you will also have non-primes whose factors are all greater than your last crossing out.
So, if at each stage, you cross out the number itself, you will only have primes larger than your last crossing out or composite numbers whose factors are all such primes.
I say naively above since although, intuitively, half of all natural numbers are even, some effort is needed to make a precise statement. Natural density is one way to do this.
Above, I say that $8$ out of $30$ was convenient. If you want to program the sieve then you can exploit this. In each consecutive block of $30$ natural numbers (beyond $30$), $22$ are certainly not prime. So, you can record the primeness of the $8$ candidates using a bitmap in one byte. In this way, with $1$GiB of memory for your sieve, you could generate a list of primes up to $8,000,000,000$ and a bit.