Brun's proof about the sum over the reciprocal twin primes, unclear formula I try to understand Brun's original paper about the convergence of the sum over the reciprocal twin primes (see https://gallica.bnf.fr/ark:/12148/bpt6k486270d.image.f110 p. 110-114 and p. 124-128).
On p. 127 there is a formula (formula $(13)$) that I don't understand. Unfortunally I don't speak french, but google translator says to that part "As one easily can see...".
More precisely, $(13)$ is the equation
$$Z(n)+Z\biggl(\frac{n}{2}\biggr)+Z\biggl(\frac{n}{3}\biggr)+Z\biggl(\frac{n}{4}\biggr)+...+Z\biggl(\frac{n}{\lambda}\biggr) =
\biggl\lfloor \frac{n}{5} \biggr\rfloor + \biggl\lfloor \frac{n}{7} \biggr\rfloor + \biggl\lfloor \frac{n}{11} \biggr\rfloor +\cdots + \biggl\lfloor \frac{n}{p_{\mu}} \biggr\rfloor$$
where $Z(n)$ denotes the numbers of twin primes under $n$ (maybe the same as $2*\pi_2(n)$?) and $p_{\mu}$ are the twin primes under $n$ without $3$. Because the arguments in both series are falling, the terms tended to zero.
I counted this as examples for some values of $n$, but I can't verify this equation. That's why I believe, that I don't really understood the equation. It would be nice, if one can help.
ps: I counted now systematically with a little program and I could verify the formula. But why is it true?
 A: A better translation would be "We use a known method, after which we have $(13)$". And on page 128, we find a short hint as to what that known method is, namely

One easily deduces formula $(13)$ by drawing the hyperbola $y = \frac{n}{x}$.

Thus, the method is to count things in two ways. Consider the region $R$ below the hyperbola $xy = n$ in the first quadrant. We want to count the lattice points $(k,m)$ in $R$ whose first coordinate is a twin prime (a member of a twin-prime pair, but we are neglecting the pair containing $3$).
First, counting the points horizontally, we have $Z(n)$ such points with second component $1$, generally $Z(n/m)$ such points with second component $m$. Clearly the largest $m$ for which there is any such point with second component $m$ is $\lambda = \bigl\lfloor \frac{n}{5}\bigr\rfloor$. Thus, this method of counting shows that there are
$$\sum_{m = 1}^{\lambda} Z(n/m) = Z(n) + Z\biggl(\frac{n}{2}\biggr) + Z\biggl(\frac{n}{3}\biggr) + \ldots + Z\biggl(\frac{n}{\lambda}\biggr)$$
such points in the region.
On the other hand, counting the points vertically, we see that there are $\bigl\lfloor \frac{n}{k}\bigr\rfloor$ such points whose first coordinate is $k$ if $k$ is a twin prime not exceeding $n$, and of course there are no such points with first coordinate $k$ if $k$ isn't a twin prime. Thus this way of counting shows there are
$$\sum_{\substack{p \text{ twin prime} \\ 5 \leqslant p \leqslant n}} \biggl\lfloor \frac{n}{p}\biggr\rfloor = \biggl\lfloor \frac{n}{5}\biggr\rfloor + \biggl\lfloor \frac{n}{7}\biggr\rfloor + \ldots + \biggl\lfloor \frac{n}{p_{\mu}}\biggr\rfloor$$
such points, where $p_{\mu}$ is the largest twin prime not exceeding $n$.
Since the number of points doesn't depend on the order in which we count them, equation $(13)$ follows.
