How to prove than $p_{2n}-(p_{2n}\mod p_{n}) = 2p_{n}$ ? where $p_{n}$ is the $_{n}$th prime number ? (for $n$ > 1) Let the prime function $p_n$ be the $n$th prime number.
For example $p_1$ = 2, $p_2$ = 3, $p_3$ = 5, $p_4$ = 7, $p_5$ = 11 etc.
I noticed something with the prime function : it seems than $p_{2n}-(p_{2n}\mod p_{n}) = 2p_{n}$, for $n$ > 1
For example :

*

*$p_{4}-(p_{4}\mod p_{2})$ = 7 - (7 mod 3) = 6 = 2*3

*$p_{6}-(p_{6}\mod p_{3})$ = 13 - (13 mod 5) = 10 = 2*5

*$p_{8}-(p_{8}\mod p_{4})$ = 19 - (19 mod 7) = 14 = 2*7

*$p_{100}-(p_{100}\mod p_{50})$ = 541 - (541 mod 229) = 458 = 2*229

It seems than $p_{2n}$ is connected with $p_n$. This is something counter-intuitive for me because this is the first time I see a link with $p_{2n}$ and $p_n$ and I thought there was no link about these two numbers.
Is there a way to explain that ? I don't know how to start for proving it.
 A: To complete the proof, we must show that $2p_n < p_{2n} < 3p_n$ for all $n>1$.
Let's restate this in terms of the prime counting function $\pi(x)$. If $x = p_n$, then we want $2x < p_{2n}$ or $\pi(2x) < 2n = 2 \pi(x)$, as well as $3x > p_{2n}$ or $\pi(3x) > 2 \pi(x)$.
For this, I am going to borrow from Wikipedia a relatively clean bound from Dusart's "Estimates of Some Functions Over Primes without R.H": for $x \ge 60184$,
$$
   \frac{x}{\log x - 1} < \pi(x) < \frac{x}{\log x -1.1}.
$$
This inequality means that for $x$ at least that large,
$$
    \frac{2x}{\log(2x)-1.1} < 2 \cdot \frac{x}{\log x - 1} \implies \pi(2x) < 2\pi(x)
$$
and the first inequality holds for all $x$ for which the denominators are positive, since the difference between $1$ and $1.1$ is less than $\log (2x) - \log x = \log 2$. Using the same estimate, we have
$$
    \frac{3x}{\log(3x)-1} > 2 \cdot \frac{x}{\log(x)-1.1} \implies \pi(3x) > 2\pi(x).
$$
The first inequality here says $3(\log x - \frac{11}{10}) > 2(\log(3x)-1)$ or $\log x > \frac{33}{10} + 2\log 3 - 2$, which in particular holds for $x \ge 34$.

So this says that when $p_n > 60184$, the bound we want holds. The largest prime below this range is $p_{6076} = 60169$, so we check the rest with brute force in Mathematica:
And @@ Table[2 Prime[n] < Prime[2 n] < 3 Prime[n], {n, 2, 6076}]

We get True i.e. the bound holds for all $n>1$.

Looking at the upper bound on $\pi(x)$ more carefully, it appears that $1.1$ is actually a uniquely bad value - in the sense that the requirements on $x$ are not worth it. Just from numerical tests (which we can limit to $x \le 60184$) it appears that
$$
   \pi(x) < \frac{x}{\log x - 9/8}
$$
for all $x \ge 4$, but if we want to improve $\frac98$ to $\frac{10}{9}$, then we need to assume $x \ge 24140$.
Using Dusart's bounds to limit our tests to finitely many $x$, we can verify for all $x \ge 59$ the bounds
$$
\frac{x}{\log x - 1/2} < \pi(x) < \frac{x}{\log x - 9/8}. 
$$
I stand by these bounds as being a good compromise between approximating $\pi(x)$ well, and not asking a lot from $x$.
We can use these in the same way as earlier to prove that $\pi(2x) < 2\pi(x)$ for all $x \ge 59$  and $\pi(3x) > 2\pi(x)$ for all $x \ge 97$. Therefore provided $p_n \ge 97$, we have $2p_n < p_{2n} < 3p_n$; now we only need to check $n=2, 3, \dots, 24$ separately.
A: This does not really say anything special about the relation between $p_{2n}$ and $p_n$. Consider the following. Let $p$ be a prime, and let $q$ be any integer in the interval $2p<q<3p$. Then $q=2p+k$ with $0<k<p$ and
$$
q-(q\bmod p)=q-k=2p.
$$
The prime number theorem implies (I think) that more often than not $$2p_n<p_{2n}<3p_n$$
and that's all you are seeing.

If we take the prime number theorem super seriously, then $p_n\approx n\ln n$. Therefore we have every right to expect
$$p_{2n}\approx(2n)\ln(2n)=2n\ln n+2n\ln 2\approx 2p_n+2n\ln2$$
to be in the interval $(2p_n,3p_n)$.

I tested the analogous conjecture relating $p_{3n}$ and $p_n$. For all $n>34$ we seem to have
$$
p_{3n}-(p_{3n}\bmod p_n)=3p_n.
$$
It fails, possibly for the last time, when $n=34$, $p_{34}=139$, $p_{102}=557\equiv1\pmod{139}$. In this case $p_{3n}$ exceeds $4p_n$ by a whisker.
A: Let's begin with an argument that doesn't quite work:
From the Wikipedia entry for PNT, we have
$$\log n+\log\log n-1\lt{p_n\over n}\lt\log n+\log\log n$$
for $n\ge6$. It follows that
$${\log2n+\log\log2n-1\over\log n+\log\log n}\lt{p_{2n}\over2p_n}\lt{\log2n+\log\log2n\over\log n+\log\log n-1}$$
(again for $n\ge6$).  Now it's not too hard to show that the upper bound is less than $3/2$ for $n\gt32$:
$$\begin{align}
{\log2n+\log\log2n\over\log n+\log\log n-1}\lt{3\over2}
&\iff2(\log2+\log n+\log\log2 n)\lt3(\log n+\log\log n-1)\\
&\iff3+2\log2\lt\log n+\log\left((\log n)^3\over(\log n+\log2)^2\right)
\end{align}$$
establishes the inequality eventually holds, and a little numerical experimentation confirms that
$$\log32+\log\left((\log32)^3\over(\log32+\log2)^2\right)\lt3+2\log2\lt\log33+\log\left((\log33)^3\over(\log33+\log2)^2\right)$$
This shows that $p_{2n}\lt3p_n$ for $n\ge33$, which is easily extended to $p_{2n}\lt3p_n$ for all $n$.
Now it would be nice if the lower bound were eventually greater than $1$. Unfortunately, it's not. To get a lower bound that works requires a closer look at the paper by Dusart that is the source of the bounds. Dusart states (without proof) a more exact upper bound
$${p_n\over n}\lt\log n+\log\log n-0.9484$$
for $k\ge39017$.  This gives us
$${\log2n+\log\log2n-1\over\log n+\log\log n-0.9484}\lt{p_{2n}\over2p_n}$$
and this lower bound is greater than $1$, since $\log\log n$ is clearly less than $\log\log2n+\ln2-0.0516$ for all $n\ge1$.
In sum, we have $2p_n\lt p_{2n}\lt3p_n$ for all $n\ge39017$, from which it follows that
$$p_{2n}-(p_{2n}\text{ mod }p_n)=p_{2n}-(p_{2n}-2p_n)=2p_n$$
for $n\ge39017$. Thus if there are any counterexamples to the OP's observation, they would have to occur with $n\lt39017$. Presumably none exist; perhaps someone can confirm this.
