The starting point of this post is an earlier question, where I conjectured (and GH from MO confirmed) that the von Mangoldt function is the limit at $s=1$ of a certain Dirichlet series, $$\Lambda(m)=\lim_{s\to 1+}\zeta(s)\sum_{d\mid m}\frac{\mu(d)}{d^{s-1}}=\lim_{s\to 1+}\;\sum_{n=1}^\infty\frac{1}{n^s}\left(\sum_{d\mid\gcd(n,m)}d\mu(d)\right),\qquad m>1.\tag{1}$$ The prime $k$-tuple conjecture for $k=2$ states the asymptotic formula $$\sum_{m \leq X} \Lambda(m+h) \Lambda(m) \sim {\mathfrak S}(h) X.\tag{2}$$ In the light of $(1)$, we have $$\Lambda(m+h) \Lambda(m) = \lim_{s\to 1+}\;\sum _{n=1}^{\infty}\frac{T_h(n,m)}{n^s},\qquad m>1\tag{3},$$ where $$T_{h}(n,m):=\sum_{k\mid n}\left(\sum_{c \mid \gcd (k,m+h)} c \mu (c)\right) \left(\sum_{d \mid \gcd \left(\frac{n}{k},m\right)} d \mu (d)\right) .\tag{4}$$

**Conjecture 1.** For any positive integers $h$, $m$, $n$, we have
$$T_{h}(n,m) = T_{h}(n,m+n).\tag{5}$$

Denote the sum of the mod $n$ repeating entries by $$S_h(n) := \sum\limits_{m=1}^n T_{h}(n,m).$$

**Conjecture 2.** We have the asymptotic formula
$$\sum\limits_{m \leq X} T_{h}(n,m)\sim\frac{S_h(n)}{n}X. \tag{6}$$

Observe that by $(3)$ we have

$$\sum_{m \leq X} \Lambda(m+h) \Lambda(m) = \lim_{s\to 1+}\;\sum\limits_{n=1}^{\infty} \frac{1}{n^s} \sum\limits_{m \leq X} T_{h}(n,m). \tag{7}$$

Main question.Is it true for all $n$ that $$S_2(n) = S_4(n)\ ? \tag{8}$$To me this would mean that the asymptotic density of twin primes $h=2$ is equal to the asymptotic density of cousin primes $h=4$.

Associated Mathematica program

In general this should be a classification of asymptotic densities of prime gaps and not only twin primes and cousin primes. I should maybe add that it appears that $$\sum\limits_{m=1}^{m=n} T_{h_{1}}(n,m)=\sum\limits_{m=1}^{m=n} T_{h_{2}}(n,m)$$ whenever: $$\varphi^{\ast -1}(h_{1})=\varphi^{\ast -1}(h_{2})$$ where $\varphi^{\ast -1}(h)$ is the Dirichlet inverse of the Euler totient.