# On the relationship between $\max(p_i)$ and $\omega(b)$, if $\sigma(b^2)/b^2$ is bounded above by a specific number $U$

Let $\omega(x)$ denote the number of distinct prime factors of $x$, and let $\sigma(x)$ be the sum of the divisors of $x$. Denote the abundancy index of $x$ by $I(x) = \sigma(x)/x$.

Let the number $b \in \mathbb{N}$ be given in the form $$b = \prod_{i=1}^{\omega(b)}{{p_i}^{\alpha_i}},$$ where the $p_i$'s are primes with $p_1 < p_2 < \ldots < p_{\omega(b)}$, and $\alpha_i \geq 1 \hspace{0.05in} \forall i \in \left[1,\omega(b)\right]$.

Then suppose that I have the upper bound $$\dfrac{\sigma(b^2)}{b^2} < U.$$

Note that a lower bound for $I(b^2) = \sigma(b^2)/b^2$ is $$\prod_{i=1}^{\omega(b)}{I({p_i}^2)},$$ which, in turn, is bounded below by $$\prod_{i=1}^{\omega(b)}{I({p_{\omega(b)}}^2)} = \left(1 + \dfrac{1}{p_{\omega(b)}} + \left(\dfrac{1}{p_{\omega(b)}}\right)^2\right)^{\omega(b)}$$ where $p_{\omega(b)} = \max(p_i)$ is the largest prime dividing $b$.

Letting $P = p_{\omega(b)}$ and $W = \omega(b)$, and asking Wolfram Alpha to solve the resulting inequality, gives a "Standard computation time exceeded" error message.

Since $I(p_{\omega(b)}) < I({p_{\omega(b)}}^2)$, we can simplify our computations at the expense of a somewhat crude estimate: $$\left(1 + \dfrac{1}{p_{\omega(b)}}\right)^{\omega(b)} < U,$$ $$1 + \dfrac{1}{p_{\omega(b)}} < U^{1/\omega(b)},$$ $$\dfrac{1}{p_{\omega(b)}} < U^{1/\omega(b)} - 1,$$ $$p_{\omega(b)} > \dfrac{1}{U^{1/\omega(b)} - 1}.$$

Here are my questions:

(1) Is it possible to do significantly better than this?

(2) If (1) is not possible, how can one analytically compute the lower bound for $p_{\omega(b)}$ obtained from $$\left(1 + \dfrac{1}{p_{\omega(b)}} + \left(\dfrac{1}{p_{\omega(b)}}\right)^2\right)^{\omega(b)} < U?$$

It turns out that it is possible to analytically compute the lower bound for $p_{\omega(b)}$ obtained from $$\left(1 + \dfrac{1}{p_{\omega(b)}} + \left(\dfrac{1}{p_{\omega(b)}}\right)^2\right)^{\omega(b)} < U$$ by a simple application of the quadratic formula: $$1 + \dfrac{1}{p_{\omega(b)}} + \left(\dfrac{1}{p_{\omega(b)}}\right)^2 < U^{1/\omega(b)}$$ Letting $p$ denote $1/{p_{\omega(b)}}$, we obtain: $$1 + p + p^2 < U^{1/\omega(b)}$$ $$p^2 + p + \left(1 - U^{1/\omega(b)}\right) < 0$$ The critical points are at $$p_1 = \dfrac{-1 - \sqrt{1 - 4\left(1 - U^{1/\omega(b)}\right)}}{2}$$ and $$p_2 = \dfrac{-1 + \sqrt{1 - 4\left(1 - U^{1/\omega(b)}\right)}}{2}.$$ (Note that $1 - U^{1/\omega(b)} < 0$.) Therefore, the region of validity for the inequality is $$p_1 < p < p_2,$$ so that $$\dfrac{-1 - \sqrt{1 - 4\left(1 - U^{1/\omega(b)}\right)}}{2} < \dfrac{1}{p_{\omega(b)}} < \dfrac{-1 + \sqrt{1 - 4\left(1 - U^{1/\omega(b)}\right)}}{2}.$$ In particular, $$\dfrac{-1 - \sqrt{1 - 4\left(1 - U^{1/\omega(b)}\right)}}{2\left(1 - U^{1/\omega(b)}\right)} < p_{\omega(b)}$$ since the lower bound for $p = 1/{p_{\omega(b)}}$ is trivial, as $$\dfrac{-1 - \sqrt{1 - 4\left(1 - U^{1/\omega(b)}\right)}}{2} < 0.$$
Consequently, we obtain $$p_{\omega(b)} > \max\left(\dfrac{1}{U^{1/\omega(b)} - 1}, \dfrac{1 + \sqrt{1 + 4\left(U^{1/\omega(b)} - 1\right)}}{2\left(U^{1/\omega(b)} - 1\right)}\right).$$