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In my previous post, using the fact that $$(6npm 1)in mathbb P iff nne 6abpm a pm b$$

I proposed a sieve to identify suitable candidates for $n$, and by extension twin primes of the form $6npm 1$, by serially eliminating integers of the form $5bpm1$,then $7bpm1$, $11bpm2$, $13bpm2, dots , (6kpm 1)b pm k$. This sieve is a series of periodic operations, each removing every $p^th$ integer from a determinable starting point.

By way of illustration, eliminating integers which are multiples of primes (the well known Sieve of Eratosthenes) can be accomplished by finding a function which maps multiples of primes to $0$. The $sin$ function can be adapted for this purpose. $$textFor nge 4, F(n):=prod_p le sqrtn sin(fracnpcdot 2pi)$$

$F(n)=0 Rightarrow n textis composite; F(n)ne 0 Rightarrow n textis prime$. The infinitude of primes means that $$not exists n_max: forall n>n_max Rightarrow F(n)=0$$

I note that with regard to $F(n)$, the various $sin$ terms are multiplied, not added, so I do not think that Fourier analysis of the function would be productive. However, since I am inexpert in that area, I am open to being educated by more knowlegable persons. I make one further observation. If we let $m=n-1$, then $F(n)$ identifies integers $m$ which are $1$ greater than primes, and they are also infinite in number. So 'displacing' the argument by some constant value changes the nature of the numbers identified, but not their abundance.

In my sieve, the object is not to identify primes, but to identify $nne 6abpm a pm b$. The sieve eliminates not multiples of primes, but integers which have a particular remainder with respect to various moduli. The removed integers occur periodically, at modular distances from a detrminable starting point. I formulate an analogous periodic function: $$textFor p(k)=(6kpm 1), G(n):=prod_k bigl( sin(fracn-kp(k)cdot 2pi)cdot sin(fracn+kp(k)cdot 2pi) bigr)$$

If $nequiv pm k modp(k)$, that term (and hence the entire product) is $0$. Thus far, $p(k)$ is not defined to be prime, only to be of the form $6kpm 1$. If $p(k)$ is composite, its (prime) factors are also of the form $6k'pm1$.

Assume $(6k_1pm1)=(6k_2pm1)(6k_3pm1)$. Then $(6k_1pm1)=36k_2k_3pm 6k_2 pm 6k_3 pm 1$. Noting that the $1$ must have the same sign on each side, we get $k_1=6k_2k_3pm k_2 pm k_3$.

Gathering like terms (say for $k_2$), we see $k_1=k_2(6k_3pm 1)pm k_3$; thus $k_1equiv k_3 mod 6k_3pm 1$. Also, $(6k_3pm1)$ is a factor of $(6k_1pm1)$, so $(6k_1pm1)b equiv 0 mod(6k_3pm1)$.

In toto, $(6k_1pm1)b pm k_1 equiv k_3 mod(6k_3pm1)$. $(6k_1pm1)b pm k_1 = (6k_3pm1)b'pm k_3$. Any number to be sieved out by virtue of having the form $(6k_1pm1)b pm k_1$ will already have been sieved out by virtue of having the form $(6k_3pm1)b' pm k_3$. So it is appropriate restrict $p(k)in mathbb P$. This increases the similarity between $F(n)$ and $G(n)$.

The function $F(n)$ does not go to $0$ for all arbitrarily large $n$, and that does not change if the argument in each term is displaced by some constant amount. In the very similar function $G(n)$, the argument in each term is displaced, but by a non-constant amount. The properties of perioidic functions is wildly outside of my familiarity, so I can go no further than this set up.

Here is my question: Given the behavior of $F(n)$, and the similarity of $G(n)$, is it possible to show that $$not exists n_max: forall n>n_max Rightarrow G(n)=0$$ If $G(n)$ does not go to $0$ for all arbitrarily large $n$, then there are arbitrarily large numbers $nne 6abpm a pm b$, and the twin prime conjecture is proved.