For all (n ϵ ℕ), n divided by p, will equal 1, or p.
(∀n ϵ ℕ) [n|p ⟹ [(n = 1) ˅ (n = p)]].
For all (n ϵ ℕ) there exists a k element of ℕ that satisfies the equation, n = 2k + 1, where n is odd and is not divisible by 2.
(∀n ϵ ℕ) (∃k ϵ ℕ), n = 2k + 1, n is odd ⟹ (2 ∤ n).
For all (n ϵ ℕ) there exists a k element of ℕ that satisfies the equation, n = 2k + 1, where 5 or 7 or 11 is divisible by n.
(∀n ϵ ℕ) (∃k ϵ ℕ), n = 2k + 1, ⟹ (5∪7∪11| n).
Firstly starting at the first prime number of 2 the third prime is found by adding the first two prime numbers 2 and 3. The next successive prime is found by adding the resulting prime number 5 to the previous prime number 3 and taking the difference between the previous two prime numbers away from the summation of the higher two.
Example 1
Finding the third prime number P₃ given the first two prime numbers P₁ and P₂.
P₃ = P₁ + P₂
2 + 3 = 5
n|p = 5|5 = 1 and 5|1 = 5
Finding the fourth prime number P4.
P4 = (P₃ + P₂) – (P₂ - P₁)
= (5 + 3) – (3–2)
= 8–1
= 7
n|p = 7|7 = 1 and 7|1 = 7
Finding the fifth prime number P5.
P5 = (P₄ + P₃) – (P₂ - P₁)
(7 + 5) – (3–2)
= 12–1
= 11
n|p = 11|11 = 1 and 11|1 = 11.
Example 2
The following is a formula to determine the successive prime number after Px.
Finding each successive prime number > 11 using the formula.
P = (Px + P−2) – P− 3
Where;
P = the successive prime number.
Px = the highest prime number known.
P− 2 = the first previous prime number or multiple of 5∪7∪11 < Px.
P− 3 = the next previous prime number or multiple of 5∪7∪11 < P− 2.
This is the theoretical gap between all primes.
P = (Px + P−2) – P− 3
(11 + 7) – 5
= 18–5
= 13
n|p = 13|13 = 1 and 13|1 = 13.
Finding the next Successive Prime number or multiple of 5∪7∪11.
P = (Px + P−2) – P− 3
(13 + 11) – 7
= 24–7
= 17
n|p = 17|17 = 1 and 17|1 = 17.
Demonstrating a number that is a multiple of 5.
P = (Px + P−2) – P− 3
23 + 19) – 17 = 25
n|p = 25|25 = 1 and 25|1 = 25 and 25|5 = 5.
Finding the successive prime number or multiple of 5∪7∪11.
P = (Px + P−2) – P− 3
= (25 + 23) – 19
= 48–19
= 29
n|p = 29|29 = 1 and 29|1 = 29.
Demonstrating a number that is a multiple of 7.
P = (Px + P−2) – P− 3
(47 + 43) – 41 = 49
n|p = 49|49 = 1 and 49|1 = 49 and 49|7 = 7.
Finding the successive prime number or multiple of 5∪7∪11.
P = (Px + P−2) – P− 3
= (49 + 47) – 43
= 96–43
= 53
n|p = 53|53 = 1 and 53|1 = 53.
Demonstrating a number that is a multiple of 11.
P = (Px + P−2) – P− 3
(73 + 71) – 67 = 77
n|p = 77|77 = 1 and 77|1 = 77 and 77|11 = 7.
Finding the successive prime number or multiple of 5∪7∪11.
P = (Px + P−2) – P− 3
= (77 + 73) – 71
= 150–71
= 79
n|p = 79|79 = 1 and 79|1 = 79.
Q.E.D.
Example 3
Finding the difference between each successive prime number using the following formula.
dk = │Pa – Pb │
Where
Pa = (Px + P−2) – P− 3), for the highest prime number or multiple of 5∪7∪11.
Pb = (Pa + P−x) – P− 2), for the prime number previous to Pa.
dk = │((Px + P−2) – P− 3) – ((Pa + P−x) – P− 2) │ = │Pa – Pb │
The following demonstrates the gap between prime numbers or a multiple of 5∪7∪11 starting at any random prime number.
Starting at the prime number 73.
Pa = (Px + P−2) – P− 3)
= (73 + 71) – 67
= 77
Pb = (Pa + P−x) – P− 2)
= (77 + 73) – 71
= 79
dk = │Pa – P− b │
= │77–79 │
= │– 2│
= 2
Continuing from 79.
Pa = (Px + P−2) – P− 3)
= (79 + 77) – 73
= 83
dk = │Pa – P− b │
= │83–79 │
= │4│
= 4
Pb = (Pa + P−x) – P− 2)
= (83 + 79) – 77
= 85
dk = │Pa – P− b │
= │83–85 │
= │– 2│
= 2
This formula also demonstrates that the difference between Px and P− 2 added to the final answer Pa will give the answer to the next Pb or multiple of 5∪7∪11.
Example 4
Pa = (Px + P−2) – P− 3)
= (79 + 77) – 73
= 83
Pb = (Px – P− 2) + Pa
= (79–77) + 83
= 2 + 83
= 85
This shows the next successive number is 85 which is the next value for Pb. Continuing this procedure the difference between Px and P− 2 alternates between a value of two and four and does not change.