# Limit kernel maybe interesting supposing opposite of twin prime conjecture (By way of contradiction).

Define. $$G_n =\{ x \in \Bbb{Z} : x^2 = 1 \mod n\}$$ for any $$n \geq 1$$. Clearly, moduloing everything by $$n$$, it is a subgroup of $$\Bbb{Z}/n^{\times}$$ and has size $$2^{\omega(d) - [2\mid n]}$$, where $$[2\mid n] = \begin{cases}1 \text{ if } 2 \mid n \\ 0 \text{ else }. \end{cases}$$, whenever $$n$$ is square-free.

Clearly, $$a \in [p_m + 2, \dots, x]$$ is a twin prime average if and only if $$a \notin G_n$$ for any $$n \mid p_m\#$$, where $$m = \pi(\sqrt{x + 1})$$. By the sieve of Eratosthenes.

We can count this explicity using some inequality-solving and inclusion-exclusion principle:

$$\rho(x) = -1 + \sum_{n \mid p_m\#} \mu(n) \sum_{r^2 = 1 \mod n } \left\lfloor \frac{x - r}{n}\right\rfloor$$

The $$-1$$ can be brought back in via, Dirac-delta:

$$\rho(x) = \sum_{n \mid p_m\#} \mu(n) \sum_{r^2 = 1 \mod n} \left( \lfloor\dots \rfloor - \delta_1(n)\right)$$

In other words. If the twin prime conjecture is false, the number of twin primes in $$[p_m + 2, x]$$ must eventually vanish as $$m = \pi(\sqrt{x + 1})$$ and $$x \to \infty$$.

To state this in usual mathy terms we say:

$$T = \sum_{n \mid p_m\#}\mu(n)\sum_{r^2 = 1\mod n} \bullet$$

to be a transformation from $$T: \{\Bbb{N}^2 \times \Bbb{R} \to \Bbb{Z}\} \to \{\Bbb{R}\to \Bbb{Z}\}$$.

And so if we define $$K = \lim \ker T := \{ f : \Bbb{N}^2 \times \Bbb{R} \to \Bbb{Z} : \lim_{x\to \infty} T(f)(x) = 0 \}$$, where it is understood that it's the set of all $$f$$ such that the limit exists and is $$0$$.

Then we can say that $$f(n,r,x) = \left\lfloor \frac{x - r}{n}\right\rfloor - \delta_1(n) \in K$$ the limit kernel of $$T$$.

But, I came up with this more general identity that is not found in the floor-ceiling article on Wikipedia.

It's closely related to one under "Quotient" (Properties).

Lemma 1.

$$\sum_{i = 0}^{kn - 1} \left\lfloor\frac{x - i}{n} \right\rfloor = k(x - \frac{(k + 1)}{2} n + 1)$$

But this is totally different from what they have on Wikipedia, and also I think their formula might be off, from experimenting in SymPy. If ound that it's not true that this sum of floors equals just $$x$$.. Anyway, I've code-checked the above formula.

Question 1.

How can we prove the above formula?

Motivation.

Well, if the limit kernel behaves like $$\Bbb{Z}$$-submodule of the domain function space, then there's no reason why we can't add in other related floor-functions such as subtracting $$i$$ and summing up, all the way to a multiple of the denominator, or in our case $$p_m\# - 1$$. That means in the formula $$k = \frac{p_m\#}{n}$$.

However, it's not that simple since we don't yet know what happens at negative $$x$$.

Question 2. What can we prove the bounds to be on $$\rho(-x)$$ for $$x \gt 0$$?

From code experiments we have:

$$-(\rho(x) + 2) \leq \rho(-x) \leq \rho(x)$$

but can we prove this and how?

## Alternative

Question 3. Is Question 2 even necessary since we're taking a limit as $$x \to \infty$$??? I think it still matters because the sum over $$x-i$$ will always delve into going past $$-x$$ and so on...

## Further Derivation & Question 4.

$$x \in \Bbb{Z}, x \geq 4, n = \pi(\sqrt{x + 1}) \implies x \in [p_n + 2, x] \text{ is a twin prime average } \\ \iff x^2 \neq 1 \mod m, \forall m \mid p_n \#.$$

Therefore, to discount the sets $$X_{m} = \{ x \in I_x := [0, x] : x^2 = 1 \mod m\}$$ for $$m\mid p_n\#$$, we can first solve:

$$0 \leq z = mz' + r\leq x$$ for any given $$r \in X_m \mod m$$, and then add the individual counts.

Wwe have $$\frac{-r}{m} \leq z' \leq \frac{x-r}{m}$$ which means a total possible $$z'$$ count of:

$$\left\lfloor\frac{x - r}{m} \right\rfloor-\left\lceil\frac{-r}{m} \right\rceil +1$$

Then to count:

$$|[0,x]\setminus \bigcup_{m \mid p_n\#}(m\Bbb{Z} +r)|$$

what we're after (twin prime average count), then it's inclusion-exclusion:

$$F(x) = -1 +\sum_{m \mid \sqrt{x + 1}\#}\mu(m)\sum_{r^2 =1 \mod m}\left\lfloor\frac{x-r}{m} \right\rfloor$$

Define $$f(m,r,x) := \left\lfloor\frac{x - r}{m} \right\rfloor - \delta_1(m)$$.

We can say that if the twin prime conjecture is false, then $$f$$ sits nicely in $$f \in K = \lim \ker T = \{ f : \Bbb{N}^2 \times \Bbb{R} \to \Bbb{Z} : \lim_{x \to \infty} T(f)(x) = 0 \}$$ where $$T$$ is the apparent double-summation operator w.r.t. $$x$$.

Now, by nature of limits, for each fixed $$i \in \Bbb{N}$$ we may not have:

$$\lim_{x \to \infty} T(f)(x + i) = 0$$

as this augments the interval range and allows in twin pseudoprime averages or averages that are not provably anything because they are outside of the Eratosthenes bound $$x$$. On the other hand, certainly as $$x \to \infty$$ we have that $$x - i \in [p_n + 2, x]$$ so that since there are no twin prime averages in the larger interval, there can't possibly be any in the smaller, and the same thing happens for sufficiently large $$x$$, as $$x \to \infty$$.

Therefore, we have:

$$0 = \sum_{i = 0}^{\sqrt{x + 1}\#-1}\lim_{x\to\infty} T(f)(x-i) = \lim_{x \to \infty}\sum_{i = 0}^{\sqrt{x + 1}\# - 1}\sum_{m \mid \sqrt{x + 1}\#}\mu(m)\sum_{r^2 = 1 \mod m} f(m,r,x-i) \\$$

Question 4. Is this a valid rearrangement: I commuted a sum of $$0$$-valued limits with the sum which is bounded in the limiting variable?

Then there is a floor identity:

$$\sum_{i = 0}^{km - 1} \left\lfloor\frac{x - i}{m} \right\rfloor = k(x - \frac{(k + 1)}{2} m + 1)$$

So since at each $$x \in \Bbb{R}$$ we have that the limit expression is a finite sum of sums, we can rearrange those sums and move the outer $$\sum_{i}$$ to be innermost:

$$g(m,r,x) := \sum_{i = 0}^{\sqrt{x + 1}\# - 1}\left(\left\lfloor\frac{x - r - i}{m} \right\rfloor- \delta_1(m)\right) \in K = \lim \ker T$$

by closure of a $$\Bbb{Z}$$-submodule (the limit kernel) of $$\{\Bbb{N}^2 \times \Bbb{R} \to \Bbb{Z}\}$$.

But by the formula:

$$g(m,r,x) = \frac{\sqrt{x + 1}\#}{m}\left(x - \frac{\frac{\sqrt{x + 1}\#}{m} + 1}{2}m + 1 \right) - \sqrt{ x + 1}\# \delta_1(m) \\ = \frac{\sqrt{x + 1}\#}{m}(x -\frac{\sqrt{x + 1}\#}{2} -\frac{m}{2} +1) - \sqrt{x + 1}\#\delta_1(m) \\ \ \\ \ = \frac{x\sqrt{x + 1}\#}{m} - \frac{(\sqrt{x + 1}\#)^2}{2m} -\frac{\sqrt{x + 1}\#}{2} +\frac{ \sqrt{x + 1}\#}{m} -\sqrt{x + 1}\#\delta_1(m)$$

Under the $$T$$-transformation, constants such as $$\sqrt{x + 1}\#/2$$ (i.e. constant with respect to the summation) will vanish, so we have:

$$T(g)(x) =-\sqrt{x+1}\# + \left(\frac{(x + 1)\sqrt{x + 1}\#}{1} - \frac{(\sqrt{ x + 1}\#)^2}{2}\right) \sum_{m \mid \sqrt{x + 1}\#}\frac{\mu(m)}{m}\sum_{r^2 = 1 \mod m}1$$

This is clearly a Merten's factor (which is always a positive number - easy to prove), times a much faster negatively growing number. That means $$\lim_{x \to\infty}(g)(x) \neq 0$$ which is a contradiction.

• what does the notation $\#$ mean? Commented Aug 29 at 13:21
• @Vincent I think it means product of primes up to ... but I am unsure.
– mick
Commented Sep 4 at 23:49
• @mick yes correct Commented Sep 5 at 1:07