Determine whether $\sigma(n)<e^\gamma n \omega(n)$ for all $n$ not of the form $2^x$. In words (to define the symbols), the sum of the divisors of $n$ is less than the product of Euler's number to the power of the Euler-Mascheroni constant, $n$, and the number of distinct prime factors of $n$. Those familiar with the definition of $\dfrac{\sigma(n)}n$ as the "abundancy" of $n$ might notice that this equivalently says the ratio of $n$'s abundancy to its number of distinct prime factors is $<e^\gamma\approx 1.78107$.
It is known that $$\lim_{n \to \infty} \sup\sigma(n)=e^\gamma n \log\log n,$$ $$\omega(n) \sim \log\log n,$$ and Robin's theorem says $$\sigma(n)<e^\gamma n \log\log n$$ for all $n>5040$ is equivalent to the Riemann Hypothesis. That is mainly what interests me about the inequality. Clearly the inequality is stronger than Robin's inequality whenever $\omega(n)<\log\log n$, so for any number of distinct prime factors the inequality will be stronger for all $n>e^{\large{e^{\large{\omega(n)}}}}$. The fundamental theorem of arithmetic implies every integer $>1$ has at least one prime factor, but $e^{\large{e}}\approx 15.15426$, so $n<16 \implies \omega(n)>\log\log n$. It's easy to show that sufficiently large powers of two violate the inequality. Indeed every power of two greater than or equal to $16$ violates the inequality, but consider that for a prime $p$, $$\lim_{a \to \infty}\dfrac{\sigma(p^a)}{p^a}=\dfrac{p}{p-1},$$ so we can see that no power of any other prime will violate the inequality, since $3/(3-1)=(2+1)/2=\sigma(2)/2$, which we have already verified satisfies the inequality, and clearly $$\dfrac{n+1}{(n+1)-1}<\dfrac{n}{n-1}$$ for all $n>1$, since $$\dfrac{n}{n-1}-\dfrac{n+1}n=\dfrac{n^2-(n^2-1)}{n^2-n}=\dfrac{1}{n^2-n}.$$
Now we have seen what the absolute largest possible abundancy is for any given prime power, and the exponents on the primes chosen will not affect the number of distinct prime factors, so it is obvious that for any given number of distinct prime factors $\omega(n)=k$,
$$\dfrac{\sigma(n)}{n}<2\dfrac{3}2\dfrac{5}4...\dfrac{p_k}{p_k-1},$$
where $p_k$ is the $k$th prime. It is then obvious that if R.H.S. $<ke^\gamma$ for all $k>1$, then our central inequality holds for all non-powers of two, but otherwise there are certain to be counterexamples given sufficiently large exponents on the first $k$ primes.
Recall that Merten's $3$rd theorem tells us the R.H.S is asymptotically $e^\gamma \log p_k$ as $k$ approaches infinity, which brings us to the inequality
$$p_k<e^k,$$
but not rigorously, and this inequality easily follows from Bertrand's postulate. That is, we can show by induction that $p_k<2^k<e^k$.
This essentially demonstrates that the inequality will be true for all $n$ such that $\omega(n)\geq A$ for some constant $A$. The question is whether we can take $A=2$.
Can you show me how to prove the inequality from the beginning? I'm not sure if what I've done so far is correct, and I've skipped over some steps.
A side note: it appears that $\omega(n)\geq \log\log n$ consistently for smaller colossally abundant numbers, but I'm not sure how to prove this in general. I suppose a start would be to show that only finitely many colossally abundant numbers exist with any given number of distinct prime factors. I'm interested in this question because we should not expect to get a much better estimate than Robin's inequality for the colossally abundant numbers, since denoting $a_r$ the $r$th colossally abundant number, $$\lim_{r \to \infty}\sigma(a_r)=e^\gamma a_r\log\log a_r.$$
You do not need all the machinery for this one. For a fixed $\omega(n),$ we need to do calculations for only the consecutive primes up to $\omega(n),$ the same collection of exponents on larger primes would shrink your ratio.
For $2^a,$ your ratio is less than $2$. For $2^a 3^b$ your ratio is less than $3/2.$ For $2^a 3^b 5^c$ your ratio is less than $5/4.$ And For $2^a 3^b 5^c 7^d$ your ratio is less than $35/32.$
Indeed, if I am using $k-1$ primes (with nonzero exponents) and increase to $k$ primes, the upper bound for the ratio is multiplied by $$ \left( \frac{k-1}{k} \right) \left( \frac{p_k}{p_k - 1} \right) $$ Since $p_k > k,$ this product is less than $1.$ As a result, adding a new prime shrinks the upper bound, always. We had already reached $3/2 < e^\gamma$ by the time we added the prime $3.$ So that's it, only the powers of $2$ work.
The business about using consecutive primes is the fact that $$ \left( \frac{p}{p - 1} \right) $$ decreases as $p$ increases. So, for any sinle prime the upper bound is that $\frac{p}{p - 1},$ so any choice of prime other than $2$ gives below $3/2.$ Any choice of two primes gives below $3/2.$ Any choice of three primes gives below $5/4.$
So, when I ask the computer for ratios above $35/32,$ I do get a mix of things; the thing that has been proved is that at most three primes will be used. These primes need not be consecutive, when searching in this manner.