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Prime Work:
(1) There are infinitely many more positive integers (even or odd) than there are prime numbers, or prime numbers have a zero density relative to the positive integers,
and
(2) prime numbers generate the positive even integers so efficiently that gaps between two consecutive prime numbers increase without bound.
(3) Prime Parity Law (PPL):
Pi(e = mg = 1 + p2n) = 2 * Pi(g = 1 + pn) = 2n where Pi(*) is the prime counting function,
and pn > 2, p2n are odd prime numbers,
2 < m ≤ 3;
and as g goes to infinity, m goes to 2.
Where does PPL come from?
It came from a model entitled, 'A Prime Model', which I formulated about the properties of prime numbers several years ago.
And the sound Goldbach Conjecture (GC) is central to its formulation.
A Prime Model Formulation:
I. (GC): e = p + q for positive even integers, e, where p, q are odd prime numbers. I include one as prime.
II. Sum of Two Primes Axiom (Distinct Goldbach partitions (p,q)) for e > 4:
e = p1 + q1 = p2 + q2 = ... where p1, p2, ... are distinct odd primes ≤ e/2, and q1, q2, ..., are distinct odd primes ≥ e/2, respectively.
III. (From I and II, we have PPL):
Prime Parity Law (PPL):
Pi(e = mg = 1 + p2n) = 2 * Pi(g = 1 + pn) = 2n where Pi(*) is the prime-counting function,
and pn > 2, p2n are odd prime numbers;
2 < m ≤ 3, and as g goes to infinity, m goes to 2.
PPL Corollary 1:
As g → [tex]\infty[/tex], m → 2. This result is stated above. This means asymptotically (as e→[tex]\infty[/tex]), there are infinitely many primes in
(0, e/2] as there are infinitely many primes in (e/2, e].
Hence, we have prime parity, and Lord GOD knows how to double-down most gloriously!
PPL Corollary 2:
Prime numbers have a zero density relative to the positive integers.
Pi[e1 = m1 * g = 1 + p2n] = 2 * Pi[g = 1 + pn] = 2n ≥ 6 with 2 < m1 ≤ 3 and where
pn > 2 and p2n are the nth prime and 2nth prime, respectively. We count one as prime.
Pi[*] is the prime-counting function.
Pi[e2 = m2 * e1] = 2 * Pi[e1] = 2^2 *n with 2 < m2 ≤ 3;
Pi[e3 = m3 * e2] = 2 * Pi[e2] = 2^3 *n with 2 < m3 < 3;
...
Pi[ej = mj * esub(j-1)] = 2 * Pi[esub(j-1)]] = 2^j *n with mj → 2 as ej → [tex]\infty[/tex].
Note: 'sub' indicates a long subscript of a variable.
Therefore.
Pi[ ej = g * Product[i = 1 to j of mi] ] / (ej = g * Product[i = 1 to j of mi] )
= (2^j * n ) / (m1 * m2 * ...*mj * g) where Pi[g] = n ≥ 3.
= (n/g) * Product[i = 1 to j of ri] where ri = 2/mi < 1 since 2 < mi ≤ 3.
Therefore, as j → [tex]\infty[/tex], Product[i = 1 to j of ri] → 0, and thus,
Pi[ ej = g * Product[i = 1 to j of mi] ] / (ej = g * Product[i = 1 to j of mi] ) → 0 as j → [tex]\infty[/tex].
Note: There are six more corollaries to 'A Prime Model'. Can you guess what they are?
P.S. Keep the faith (effort and hope) and keep an open mind. Thank Lord GOD!
David Cole
aka primework123
Please support my research work at: https://www.gofundme.com/david_cole
Thank you!