Showing that for $p_k\ \ge 31, p_{k-1}\# > (p_{k+1})^2$

Question

I am trying to show that for $p_k \ge 31, p_{k-1}\# > (p_{k+1})^2$

Does the following work? Is there a stronger or more straight forward result that can be shown?

Let:

• $p_k$ be the $k$th prime
• $p\#$ be the primorial for $p$.

Base Case: $7\# = 210 > 13^2 = 169$

Assume that $p_{k-1}\# > (p_{k+1})^2$

If $p_k \ge 31$, there are at least $4$ primes $q$ where $p_{k-1} < q < 2p_{k-1}$ [See here for details]

It follows that $p_{k-1} < p_{k+1} < p_{k+2} < p_{k+3} < 2p_{k-1}$

Let $c = p_{k+2} - p_{k+1} < 2p_{k-1} - 2 - p_{k-1} = p_{k-1}-2$

$(p_k-1)(p_{k-1}\#) > (p_{k}-1)(p_{k+1})^2> c(p_{k+1})^2 = c(2p_{k+1}) + c(p_{k+1}-2)(p_{k+1}) > 2cp_{k+1} + c^2$

$p_k\# > (p_{k+1})^2 + 2cp_{k+1}+c^2 = (p_{k+1}+c)^2 = (p_{k+2})^2$

Answer 1

I think your solution if fine. But you might find this a bit tighter ?

We have \begin{eqnarray*} p_{k-1} < p_{k+1} < p_{k+2} < p_{k+3} < 2p_{k-1}. \end{eqnarray*} So \begin{eqnarray*} p_{k} \#= p_k p_{k-1} \# >p_k p_{k+1}^2 > 4p_k^2 > p_{k+2}^2. \end{eqnarray*}