# Symmetry in Hardy-Littlewood k-tuple conjecture

Assuming Hardy-Littlewood $$k$$-tuple conjecture, do the "dual" prime constellations $$(0,h_1, h_2,\cdots, h_i,\cdots, h_{k-1}=d)$$ and $$(0, h_{k-1}-h_{k-2}, h_{k-1}-h_{k-3},\cdots,h'_i=h_{k-1}-h_{k-i},\cdots,h_{k-1})$$ corresponding to reversed sequences of prime gaps have the same distribution?

If yes, does it imply that the function $$f(n):=\dfrac{\log g_n}{\log\log p_n}$$ and the function $$f'(n)$$ obtained through the substitution $$g_n\mapsto g'_n:=\dfrac{\log^{2} p_n}{g_n}$$ reach the same values an asymptotically equal number of times? Is it related to the functional equation of zeta with which it would then share the same type of symmetry?

(Edited after Lagrida's answer and accordingly)

• Yes there is a symmetry, i will add an heuristic argument for that ! – LAGRIDA Sep 13 at 11:07

Let $$k \in \mathbb{N}, k \geqslant 2$$.

Let $$q \in \mathbb{P}, \ q \geqslant 5$$ and : $$N_q := \displaystyle{\small \prod_{\substack{p \leqslant q \\ \text{p prime}}} {\normalsize p}}$$ Let : $$1 \leqslant b \leqslant N_q$$.

We have : $$\gcd(b, N_q) = 1 \iff \gcd(N_q-b, N_q)=1 \tag{1}$$

Then the numbers coprime to $$N_q$$ and less than $$N_q$$ are symetric to $$\dfrac{1}{2}N_q$$.

Consider the k-tuple : $$\mathcal{H}_k := (0,h_1,h_2,\cdots,h_{k-1})$$, with $$0 < h_1 < \cdots < h_{k-1}$$.

Using $$(1)$$, if $$(b,b+h_1,b+h_2,\cdots,b+h_{k-1})$$ is coprime to $$N_q$$ then we have also $$(N_q-b-h_{k-1}, N_q-b-h_{k-2}, \cdots,N_q-b-h_2, N_q-b-h_1, N_q-b)$$ is coprime to $$N_q$$, (name that property 1).

Consider the k-tuple : $$\mathcal{H}^{'}_k := (0,(h_{k-1}-h_{k-2}),(h_{k-1}-h_{k-3}),\cdots,(h_{k-1}-0))$$

Using property1, you can see that : $$b+\mathcal{H}_k \text{ is coprime to } N_q \iff N_q-b-h_{k-1}+\mathcal{H}^{'}_k \text{ is coprime to } N_q$$

Example : Let $$\mathcal{H}_3=(0,2,6)$$ and $$q=7$$, for $$b=11$$ we have $$11+(0,2,6)=(11, 13, 17)$$ is coprime to $$N_7=210$$.

We have $$N_7-b-h_{k-1}=210-11-6=193$$ and $$\mathcal{H}^{'}_3 = (0, 4, 6)$$.

Then we have $$193+(0, 4, 6) = (193, 197, 199)$$ it coprime too to $$N_7$$.

Using Chineese Romander theorem we can prove that : $$\#\{(b,b+h_1,b+h_2,\cdots,b+h_{k-1})\in\mathbb{N}^{k} \, | \, 1 \leqslant b \leqslant N_q \ , \gcd(b, N_q)=\gcd(b+h_i, N_q)=1\} = \displaystyle{\small \prod_{\substack{p \leqslant q \\ \text{p prime}}} {\normalsize (p-w(\mathcal{H}_k, p))}}$$ Where $$w(\mathcal{H}_k, p)$$ is the number of distinct residues $$\pmod p$$ in $$\mathcal{H}_k$$.

Let $$x \in \mathbb{R}$$.

Let $$q(x)$$ be the largest prime number verifiying $$x \geqslant \displaystyle \Big({\small \prod_{\substack{p \leqslant q(x) \\ \text{p prime}}} {\normalsize p}}\Big)$$.

Using prime number theorem we have $$q(x) \sim \log(x)$$.

Consider : $$I_{\mathcal{H}_k}(x) := \#\{(b,b+h_1,b+h_2,\cdots,b+h_{k-1})\in\mathbb{N}^k \, | \, b \leq x, \ \gcd(b, N_{q(x)}) = \gcd(b+h_i, N_{q(x)})=1 \}$$ And : $$\pi_{\mathcal{H}_k}(x) := \#\{(p,p+h_1,p+h_2,\cdots,p+h_{k-1})\in\mathbb{P}^k \, | \, p \leq x\}$$ We can prove as $$x \to +\infty$$ that:

$$I_{\mathcal{H}_k}(x) \sim \mathfrak{S}(\mathcal{H}_k) \, e^{-\gamma k} \, \dfrac{x}{\log(\log(x))^k}$$ With $$\mathfrak{S}(\mathcal{H}_k) := \displaystyle\prod_{\text{p prime}}\frac{1-\frac{w(\mathcal{H}_k, p)}{p}}{(1-\frac1p)^{k}}$$.

If $$p \in \mathbb{P}, \ p > q(x)=(1+o(1)) \log(x)$$ then $$p$$ is coprime to $$N_{q(x)}$$, this is the relation trivial between prime numbers less than $$x$$ and numbers coprime to $$2,3,\cdots,q(x)$$ and less than $$x$$. I give a non-trivial relation as : $$I_{\mathcal{H}_k}(x) \sim \pi_{\mathcal{H}_k}(x) \big( \pi(q(x)) e^{-\gamma} \big)^k$$ If we prove this conjecture then we have : $$\pi_{\mathcal{H}_k}(x) \sim \mathfrak{S}(\mathcal{H}_k) \dfrac{x}{\log(x)^k}.$$ We can find the same results with Goldbach's conjecture or primes of the form $$n^2+1$$, you can see my article : here

• I accepted your answer to my first question but I'm of course still interested in answers to the remaining ones. – Sylvain JULIEN Sep 13 at 19:42