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)



Symmetry in Hardy-Littlewood k-tuple conjecture
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 $N\_7$.




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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](https://hal.archives-ouvertes.fr/hal-02620310)


