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The elements of the reduced residue system modulo $30$ are $\{1, 7, 11, 13, 17, 19, 23, 29\}$

If we order them as $e_1, e_2, e_3, \dots$ so that $e_1 = 1, e_2 = 7, \dots$, it follows that $3.75(i-1) < e_i < 3.75i$.

We can generalize this.

If $\gcd(x,30)=1,$ then $x = 30a + b$ where $b \in \{1, 7, 11, 13, 17, 19, 23, 29\}$.

If we order $\{1,7, 11, \dots, 29, 31, \dots, 59, \dots, 30a-1, \dots, x, \dots, 30a+31, \dots, 30a+59 \}$ as $e_1, e_2, e_3, \dots$ there exists $j$ with $e_j = x$ and $3.75(j-1) < x < 3.75j$. (This is true for $x < 30$. Assume it is true for $x < 30c$ where $c \ge 1$. It is clearly also true for each $e_j=x$ where $x < 30(c+1)$).

For $4 \le i \le 15$, $p_i = e_{i-2} > (i-3)*3.75$.

For $16 \le i \le 21$, $p_i = e_{i-1} > (i-2)*3.75$

For $22 \le i \le 24$, $p_i = e_i > (i-1)*3.75$

For $i \ge 25$, $p_i \ge e_{i+1} > 3.75i$

Is this reasoning valid? If so, what would be a more concise way of making the same argument?

Larry Freeman
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  • An overkill approach : $ p_n \sim n \log n$ from prime numbers theorem . –  Dec 25 '15 at 19:15

1 Answers1

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Essentially, $p_n > n\log{n}$ (Rosser's Theorem) for sufficiently large $n$. And since $\log{n}$ is unbounded (in particular, is not bounded by 3.75), such a result is to be expected.

To limit the number of particular cases that must be checked manually, we can invoke a refinement of Rosser's Theorem according to which $$p_n > n(\log{n} + \log{\log{n}} - 1),\; \forall n \ge 6.$$

It turns out that the function $f: x \mapsto \log{x} + \log{\log{x}} - 1$ is increasing for $x > 1$, and that $f(34) > 3.7866268 > 3.75$. Thus we obtain that $$ p_n > 3.75n,\; \forall n \ge 34.$$

Now check the remaining cases $p_{25}, p_{26},\ldots, p_{33}$ with a small script :)

dohmatob
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