A remarkable(?) condition on sequences of natural numbers There is a remarkable condition on increasing sequences of natural numbers $(a_n)_n$:
$$\bigg\lfloor\frac{a_n^2}{a_{n+1}}\bigg\rfloor=2a_n-a_{n+1}\tag 1$$
that - when $n$ is big enough - seems to hold for all increasing sequences $a_n\lesssim p_n$, where $p_n$ is the $n$-th prime. Whether it holds for 
$a_n=p_n$ depends on a weaker form of Oppermann's conjecture:
$$(p_{n+1}-p_n)^2<p_{n+1}\tag 2$$
Trivially $\;p_n^2=p_{n+1}(2p_n-p_{n+1})+(p_{n+1}-p_n)^2$, why $\;(2)\;$ implies $\;(1)\;$ for $a_n=p_n$.
In fact, I can't prove much of this and my claims rely on computations. I can't even prove that 
$$\bigg\lfloor\frac{n^2}{n+1}\bigg\rfloor=n-1,\;n>0$$
Except from proofs or counterexamples of certain examples I would like to know if there are something published on this subject.

Looks like I was totally wrong about that every smaller sequences did fulfill the conditions $(1)$.
 A: Let us see what does it take for your condition hold:
$$
\left\lfloor \frac{a_n^2}{a_{n+1}}\right\rfloor = 2a_n - a_{n+1} \quad \Leftrightarrow \quad 2a_n - a_{n+1} \leq \frac{a_n^2}{a_{n+1}} < 2a_n - a_{n+1} + 1
$$
The left inequality implies $a_n^2 - 2a_na_{n+1} + a_{n+1}^2 = (a_n - a_{n+1})^2 \geq 0$, which provided no information, but the right inequality is more interesting:
$$
(a_n - a_{n+1})^2 < a_{n+1}.
$$
This implies $a_{n+1} - a_{n} = O(\sqrt{a_n})$.
Let us try to provide an example by considering a sequence of this flavor and grow as fast as it can be, then the first terms will look as follows ($a_1 = 1$ will enforce a constant sequence, so let $a_1 = 2$):
$$
2, 3, 5, 7, 10, 13, 17, 21, 26, 31, 37, 43, 50, 57, 65, \cdots
$$ 
By observation, you see that $a_{2n-1} = n^2 + 1$ and $a_{2n} = n^2 + n + 1$. In general, sequences satisfying your conditions can grow as fast as $O(n^2)$.
In particular, prime number sequence $p_n$ grows asymptotically as fast as $O(n\log n)$, by the prime number theorem, so heuristically it is expected that your conditions are satisfied after get through enough terms.
Remark: The converse does not hold. The sequence can grow extremely slow while your condition is not satisfied. Consider the sequnce $a_n = 2^{\lfloor\ln\ln n\rfloor}$, then the sequence has asymptotically logarithmic growth, but your conditions fails at the jumps. If you need a strictly increasing sequence, think about $a_n = n2^{\lfloor\ln\ln\ln n\rfloor}$.
