# Why is $\{n=4r+1,r = {n-1\over 2}\}\subset \mathbb{P}$ true under these conditions?

Let $p=p_k$, $q=p_{k+1}$ and $r=p_{k+2}$, where $p_m$ denotes the $m$th prime.

I conjecture that whenever $n$ is prime, where $n$ is defined as follows: $$n = 1+\left(\left\lfloor{p\over q}\right\rfloor+r\right) \left\lfloor{(p+r)(q^2+pr)\over(pqr)}\right\rfloor$$ then: $$\frac{n-1}{4} \qquad\text{and}\qquad 4r+1$$ are both prime.

Although I've tried vigorously, I have no explanation as to why $n=4r+1,r = {n-1\over 2}$, although I heavily suspect that the solution rests on twin primes.

Alternative form:

$$n = 1+ z \left \lfloor{x\over y}+{z\over y}+{y\over z}+{y\over x}\right\rfloor+\left \lfloor{x\over y}\right\rfloor \left \lfloor{x\over y}+{z\over y}+{y\over z}+{y\over x}\right\rfloor$$

Example:

Terms $1\to 5$, where $t_m$ denotes the $m$th number $k$ which satisfies the conditions described in this question:

$$t_1 = 29 = 1+\left(\left\lfloor{3\over 5}\right\rfloor+7\right)\left\lfloor{(3+7)(5^2+3\cdot 7)\over(3\cdot 5\cdot 7)}\right\rfloor$$ $$\frac{29-1}{4} = 7,\qquad 4\cdot 7+1 = 29$$

$$t_2 = 53 = 1+\left(\left\lfloor{7\over 11}\right\rfloor+13\right)\left\lfloor{(7+13)(11^2+7\cdot 13)\over(7\cdot 11\cdot 13)}\right\rfloor$$ $$\frac{53-1}{4} = 13,\qquad 4\cdot 13+1 = 53$$

$$t_3 = 149 = 1+\left(\left\lfloor{29\over 31}\right\rfloor+37\right)\left\lfloor{(29+37)(31^2+29\cdot 37)\over(29\cdot 31\cdot 37)}\right\rfloor$$ $$\frac{149-1}{4} = 37,\qquad 4\cdot 37+1 = 149$$

$$t_4 = 173 = 1+\left(\left\lfloor{37\over 41}\right\rfloor+43\right)\left\lfloor{(37+43)(41^2+37\cdot 43)\over(37\cdot 41\cdot 43)}\right\rfloor$$ $$\frac{173-1}{4} = 43,\qquad 4\cdot 43+1 = 173$$

$$t_5 = 269 = 1+\left(\left\lfloor{59\over 61}\right\rfloor+67\right)\left\lfloor{(59+67)(61^2+59\cdot 67)\over(59\cdot 61\cdot 67)}\right\rfloor$$ $$\frac{269-1}{4} = 67,\qquad 4\cdot 67+1 = 269$$

• What do the notations $\mathbb{P}$ and $\mathbb{P}_k$ mean? May 15, 2013 at 2:59
• Is there a typo in the title? May 15, 2013 at 3:05
• The question does not make much sense. Do you realize you're claiming that if $r$ is prime (greater then 3) then $4r+1$ is also prime, which is clearly false? May 15, 2013 at 3:09
• @ViniciusM Agreed. That was going to be my next comment. The title is also confusing, apart from the contradictory equations, it also uses $r$. May 15, 2013 at 3:11
• @JohnWO, I recommend sharing the data you've collected and showing the pattern within that data. May 15, 2013 at 3:15

Note that $\lfloor \frac{p}{q} \rfloor=0$, since $0<\frac{p}{q}<1$. Thus $n=1+r\lfloor \frac{(p+r)(q^2+pr)}{pqr} \rfloor=1+r \lfloor \frac{q}{r}+\frac{r}{q}+\frac{p}{q}+\frac{q}{p} \rfloor$.
Note that for $1<x<2$, we have $2<x+\frac{1}{x}<\frac{5}{2}$, since $x+\frac{1}{x}>2 \Leftrightarrow x^2+1>2 \Leftrightarrow (x-1)^2>0$ and $x+\frac{1}{x}<\frac{5}{2} \Leftrightarrow 2x^2+2<5x \Leftrightarrow (2x-1)(x-2)<0$.
We clearly have $p<q<r$, and by Bertrand's postulate $q<2p, r<2q$, so $1<\frac{r}{q}, \frac{q}{p}<2$, so by above we have $4<\frac{q}{r}+\frac{r}{q}+\frac{p}{q}+\frac{q}{p}<5$, so $n=1+r \lfloor \frac{q}{r}+\frac{r}{q}+\frac{p}{q}+\frac{q}{p} \rfloor=1+4r$.
It is now straightforward to see that when $n$ is prime, $\frac{n-1}{4}=r$ and $4r+1=n$ are prime.