Why does $3+ (-1)\left(\left\lfloor\sum_{k=1}^{|\{x\in P,\;x\le n\}|} \frac{P_k}{1-P_k}\right\rfloor\right) = \pi(n),\quad n\ge1223$?

Let $P$ denote $\text{primes}$, and $\pi(x)$ denote $|P| \le x$.

Here's my first question: Why does

$$3+ (-1)\left(\left\lfloor\sum_{k=1}^{|\{x\in P,\;x\le n\}|} \frac{P_k}{1-P_k}\right\rfloor\right) = \pi(n),\quad n\ge1223$$

And similarly:

$$2+ (-1)\left(\left\lfloor\sum_{k=1}^{|\{x\in P,\;x\le n\}|} \frac{P_k}{1-P_k}\right\rfloor\right) = \pi(n),\quad 11\le n\lt1223$$

I know that it is a fairly weak statement as of yet, but I can't find this (or a statment similar enough) anywhere, which seems weird, as it seems to be a interesting way of defining $\pi(x)$.

If anyone has any explanation on why this is the case, please share.

Edit:

If the statment is redefined as $j+ (-1)\left(\left\lfloor\sum_{k=1}^{|\{x\in P,\;x\le n\}|} \frac{P_k}{1-P_k}\right\rfloor\right) = \pi(n)$, is there a way to determine $j$?

Edit 2:

$$j + (-1)\left(\left\lfloor\sum_{k=1}^{|\{x\in P,\;x\le n\}|} \frac{P_k}{1-P_k}\right\rfloor\right),\quad j>0,\; e^{e^{j-1}} < n < e^{e^{j}} = \pi(n)$$

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What's the differnce between "Why does ...?" and "Is there a way to prove this?"? –  Hagen von Eitzen May 23 '13 at 4:55
@HagenvonEitzen: Good point. –  JohnWO May 23 '13 at 5:01

Note that $$-\frac {P_k}{1-P_k}=1+\frac1{P_k-1}>1+\frac1{P_k}$$ and the sum of prime reciprocals diverges. Therefore, for $n$ big enough, your $\sum$ will be bigger than $\pi(n)+r$ for any $r$, thus disproving your conjecture. (The divergence is slow, however, for the $r$th steps you have to go up to $n$ in the order of $e^{e^{r}}$).
@JohnWO .. but the estimate $e^{e^r}$ is not necessarily exact. What we have is that $\sum_{p<n}\frac1p-\ln\ln n$ approaches the value $0.26149\ldots$ as $n\to\infty$ –  Hagen von Eitzen May 23 '13 at 14:45
$e^{e^{r}}$ is actually correct up to at least $10^8$, when flooring the value. –  JohnWO May 24 '13 at 5:46