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Consider $q_n$ such that

$$q_n = \sum_{k=2}^{2^n} k \left \lfloor \frac{1}{1 + \text{Abs} \left (n-\left \lfloor \frac{1}{a(k)} \right \rfloor b (k) \right )} \right \rfloor$$

where

$$a(x) = \sum_{k=1}^{x-1} \left \lfloor \frac{k \left \lfloor \frac{x}{k} \right \rfloor }{x} \right \rfloor $$

and

$$b(x) = \sum_{k=2}^{x} \left \lfloor \frac{1}{a(k)} \right \rfloor$$

$\left \lfloor x \right \rfloor$ denotes $\text{Floor}(x)$.

How could one prove that $n \geq 3 \implies q_n = p_{n+1}$, where $p_n$ is the $n$th prime number?

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    $\begingroup$ A first step: your big Floor($\cdots$) is equal to $1$ when Abs($\ast)=0$ and $0$ otherwise. I suppose that Abs stands for "absolute value". On the other hand, Abs($\ast)=0$ iff $\ast=0$. $\endgroup$ – Matemáticos Chibchas Sep 16 '13 at 4:06
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Your big Floor($\cdots$) is equal to $1$ when Abs($\ast)=0$ and $0$ otherwise. I suppose that Abs stands for "absolute value". On the other hand, Abs($\ast)=0$ iff $\ast=0$.

Now for any integer $x\geq2$ each summand in the defining formula for $a(x)$ is the floor of a nonnegative rational number, so it is a natural number, and the first summand ($k=1$) is equal to $1$. Therefore $a(x)$ is a positive integer, so $\lfloor 1/a(x)\rfloor=1$ if $a(x)=1$ and $0$ if $a(x)>1$. But our previous reasoning shows that $a(x)=1$ iff for $k=2,\dots, x-1$ the corresponding summand in the formula for $a(x)$ is $0$, that is $k\lfloor x/k\rfloor/x<1$, which is equivalent to $\lfloor x/k\rfloor<x/k$.

Consequently $a(x)=1$ iff no number from $\{2,3,\dots,x-1\}$ divides $x$. In other words, $a(x)=1$ iff $x$ is prime and $a(x)>1$ otherwise. Using this fact in the formula for $q_n$ with $n\geq3$, we see that Abs$(\ast)=n$ when $k$ is composite, and so $\lfloor1/(1+$Abs$(\ast))\rfloor=\lfloor1(n+1)\rfloor=0$. Therefore the only (possibly) nonzero summands in the defining formula for $q_n$ are those with $k$ prime.

Now we ask: for $k\geq2$ prime, what is the value of $n-\lfloor1/a(k)\rfloor b(k)=n-b(k)$? our previous observation about the values of $a(x)$ shows that $b(k)$ is precisely the number of primes less or equal than $k$. Thus we have $n-b(k)>0$ iff the number of primes less or equal than $k$ is strictly less than $n$, that is $k=p_j$, with $j<n$. But $n-b(k)>0$ implies that Abs$(*)>0$, which in turn implies that the corresponding summand in $q_n$ is equal to $0$. Therefore the only nonzero summand in the defining formula for $q_n$ is the corresponding to $k=p_n$, and in this case Abs$(\ast)=0$, so Floor$(\cdots)=1$, which shows that $q_n=p_n$ (and not $p_{n+1}$, unless some of my calculations were wrong).

Well, one thing remains to be proved: you must prove that $p_n\leq 2^n$ for all $n\geq3$ in order to ensure that the value $k=p_n$ lies in the range $\{2,3,\dots,2^n\}$. But this can be proved using Chebyshev's theorem.

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