Arithmetic properties of a sum related to the first Hardy-Littlewood conjecture 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.
 A: Nice question! As I explained in the comments, Conjectures 1-2 follow easily. So let me answer your main question much more generally, and also the question that you implicitly asked at OEIS A298825. I will denote by $\tau$ the divisor function.
Theorem. Let $r=\operatorname{rad}(h)$ be the radical of $h$. Then
$$S_h(n)=n\sum_{\substack{e\mid r\\e^2\mid n}}e\sum_{\substack{n=e^2fg\\\gcd(e,f)=1}}\mu(f)\tau(f)\tau(g).\tag{$\ast$}$$
Corollary 1. If $\operatorname{rad}(h)=\operatorname{rad}(j)$, then $S_h(n)=S_j(n)$.
Corollary 2. $S_h(n)$ is divisible by $n$, and it vanishes unless $n$ is powerful.
Proof of Theorem. Starting from the definitions,
\begin{align*}S_h(n)&=\sum_{m=1}^n\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)\\[6pt]
&=\sum_{n=ab}\sum_{\substack{c\mid a\\d\mid b}}\mu(c)\mu(d)cd\sum_{\substack{1\leq m\leq n\\m\equiv -h\pmod{c}\\m\equiv 0\pmod{d}}} 1\\[6pt]
&=n\sum_{n=ab}\sum_{\substack{c\mid a\\d\mid b}}\mu(c)\mu(d)\gcd(c,d)\sum_{\substack{1\leq m\leq\operatorname{lcm}(c,d)\\m\equiv -h\pmod{c}\\m\equiv 0\pmod{d}}} 1\\[6pt]
&=n\sum_{n=ab}\sum_{\substack{c\mid a\\d\mid b\\\gcd(c,d)\mid h}}\mu(c)\mu(d)\gcd(c,d).
\end{align*}
Only square-free $c$'s and $d$'s contribute to this sum, hence $e:=\gcd(c,d)$ is also square-free. Therefore, $e\mid h$ is equivalent to $e\mid r$, while $e\mid c\mid a$ and $e\mid d\mid b$ imply that $e^2\mid n$. Using the notation 
$$a':=a/e,\qquad b':=b/e,\qquad c':=c/e,\qquad d':=d/e,$$ 
we get
\begin{align*}S_h(n)&=n\sum_{\substack{e\mid r\\e^2\mid n}}e\sum_{n=e^2a'b'}
\sum_{\substack{c'\mid a'\\d'\mid b'\\\gcd(c',d')=1}}\mu(ec')\mu(ed')\\[6pt]
&=n\sum_{\substack{e\mid r\\e^2\mid n}}e\sum_{n=e^2a'b'}
\sum_{\substack{c'\mid a'\\d'\mid b'\\\gcd(c'd',e)=1}}\mu(c'd').
\end{align*}
Now we introduce $f:=c'd'$ and $g:=a'b'/f=(a'/c')(b'/d')$, and we obtain $(\ast)$ readily.
Proof of Corollary 1. The statement is clear, since the RHS of $(\ast)$ only depends on $r$ and $n$.
Proof of Corollary 2. It is clear from $(\ast)$ that $S_h(n)$ is divisible by $n$. For a given $e\mid r$ in $(\ast)$, let us decompose $n/e^2$ as $st$, where $\gcd(s,e)=1$ and $t\mid e^\infty$. Then, we can rewrite $(\ast)$ as
$$S_h(n)=n\sum_{\substack{e\mid r\\e^2\mid n}}e\tau(t)\sum_{f\mid s}\mu(f)\tau(f)\tau\left(\frac{s}{f}\right).$$
The inner sum is the convolution of $\mu\tau$ and $\tau$ at $s$. It is multiplicative in $s$, and for $s=p^k$ a prime power, it equals
$$\tau(p^k)-\tau(p)\tau(p^{k-1})=(k+1)-2k=1-k.$$
In particular, for $s=p$ a prime, it is zero, hence the inner sum is supported on powerful numbers. It follows that $S_h(n)$ vanishes unless $n$ has a square divisor $e^2$ such that the coprime-to-$e$ part of $n/e^2$ is powerful. This condition on $n$ implies that $n$ itself is powerful, hence we are done.
Added. Of course the upshot of the question, namely that the density of prime pairs for $h=2$ should be the same as for $h=4$, is not new. Indeed, by the familar product formula for $\mathfrak{S}(h)$ (see e.g. here), it is clear that $\mathfrak{S}(h)$ only depends on the radical of $h$, hence in particular $\mathfrak{S}(2)=\mathfrak{S}(4)$. The original post and my answer above should be regarded as a variant (or refinement) of this observation. Indeed, it seems likely that (cf. $(2)$, $(6)$, $(7)$ in the original post)
$$\mathfrak{S}(h)=\sum_{n=1}^\infty\frac{S_h(n)}{n^2},$$
but I have not attempted to prove this (the proof seems tedious but straightforward).
