Determine whether $\sigma(n)<e^\gamma n \omega(n)$ for all $n$ not of the form $2^x$

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.

        2 = 2                      1.5
3 = 3        1.333333333333333
4 = 2^2                     1.75
5 = 5                      1.2
7 = 7        1.142857142857143
8 = 2^3                    1.875
9 = 3^2        1.444444444444444
12 = 2^2 * 3        1.166666666666667
16 = 2^4                   1.9375
24 = 2^3 * 3                     1.25
25 = 5^2                     1.24
27 = 3^3        1.481481481481481
32 = 2^5                  1.96875
36 = 2^2 * 3^2        1.263888888888889
40 = 2^3 * 5                    1.125
48 = 2^4 * 3        1.291666666666667
49 = 7^2        1.163265306122449
54 = 2 * 3^3        1.111111111111111
64 = 2^6                 1.984375
72 = 2^3 * 3^2        1.354166666666667
80 = 2^4 * 5                   1.1625
81 = 3^4        1.493827160493827
96 = 2^5 * 3                   1.3125
108 = 2^2 * 3^3        1.296296296296296
112 = 2^4 * 7        1.107142857142857
121 = 11^2        1.099173553719008
125 = 5^3                    1.248
128 = 2^7                1.9921875
144 = 2^4 * 3^2        1.399305555555556
160 = 2^5 * 5                  1.18125
162 = 2 * 3^4         1.12037037037037
192 = 2^6 * 3        1.322916666666667
200 = 2^3 * 5^2                   1.1625
216 = 2^3 * 3^3        1.388888888888889
224 = 2^5 * 7                    1.125
243 = 3^5        1.497942386831276
256 = 2^8               1.99609375
288 = 2^5 * 3^2                 1.421875
320 = 2^6 * 5                 1.190625
324 = 2^2 * 3^4        1.307098765432099
343 = 7^3        1.166180758017493
384 = 2^7 * 3                 1.328125
400 = 2^4 * 5^2                  1.20125
432 = 2^4 * 3^3        1.435185185185185
448 = 2^6 * 7        1.133928571428571
486 = 2 * 3^5        1.123456790123457
512 = 2^9              1.998046875
576 = 2^6 * 3^2        1.433159722222222
625 = 5^4                   1.2496
640 = 2^7 * 5                1.1953125
648 = 2^3 * 3^4        1.400462962962963
720 = 2^4 * 3^2 * 5        1.119444444444444
729 = 3^6        1.499314128943759
768 = 2^8 * 3        1.330729166666667
784 = 2^4 * 7^2        1.126913265306122
800 = 2^5 * 5^2                 1.220625
864 = 2^5 * 3^3        1.458333333333333
896 = 2^7 * 7        1.138392857142857
972 = 2^2 * 3^5        1.310699588477366
1000 = 2^3 * 5^3                     1.17
1024 = 2^10             1.9990234375
1080 = 2^3 * 3^3 * 5        1.111111111111111
1152 = 2^7 * 3^2        1.438802083333333
1280 = 2^8 * 5               1.19765625
1296 = 2^4 * 3^4        1.447145061728395
1331 = 11^3         1.09992486851991
1440 = 2^5 * 3^2 * 5                   1.1375
1458 = 2 * 3^6        1.124485596707819
1536 = 2^9 * 3               1.33203125
1568 = 2^5 * 7^2        1.145089285714286
1600 = 2^6 * 5^2                1.2303125
1728 = 2^6 * 3^3        1.469907407407407
1792 = 2^8 * 7                 1.140625
1800 = 2^3 * 3^2 * 5^2        1.119444444444444
1944 = 2^3 * 3^5        1.404320987654321
2000 = 2^4 * 5^3                    1.209
2048 = 2^11            1.99951171875
2160 = 2^4 * 3^3 * 5        1.148148148148148
2187 = 3^7        1.499771376314586
2304 = 2^8 * 3^2        1.441623263888889
2401 = 7^4        1.166597251145356
2560 = 2^9 * 5              1.198828125
2592 = 2^5 * 3^4        1.470486111111111
2880 = 2^6 * 3^2 * 5        1.146527777777778
2916 = 2^2 * 3^6        1.311899862825789
3072 = 2^10 * 3        1.332682291666667
3125 = 5^5                  1.24992
3136 = 2^6 * 7^2        1.154177295918367
3200 = 2^7 * 5^2               1.23515625
3240 = 2^3 * 3^4 * 5         1.12037037037037
3456 = 2^7 * 3^3        1.475694444444444
3584 = 2^9 * 7        1.141741071428571
3600 = 2^4 * 3^2 * 5^2        1.156759259259259
3888 = 2^4 * 3^5        1.451131687242798
4000 = 2^5 * 5^3                   1.2285
4096 = 2^12           1.999755859375
4320 = 2^5 * 3^3 * 5        1.166666666666667
4374 = 2 * 3^7         1.12482853223594
4608 = 2^9 * 3^2        1.443033854166667
5000 = 2^3 * 5^4                   1.1715
5120 = 2^10 * 5             1.1994140625
5184 = 2^6 * 3^4        1.482156635802469
5400 = 2^3 * 3^3 * 5^2        1.148148148148148
5488 = 2^4 * 7^3        1.129737609329446
5760 = 2^7 * 3^2 * 5        1.151041666666667
5832 = 2^3 * 3^6        1.405606995884774
6048 = 2^5 * 3^3 * 7        1.111111111111111
6144 = 2^11 * 3             1.3330078125
6272 = 2^7 * 7^2        1.158721301020408
6400 = 2^8 * 5^2              1.237578125
6480 = 2^4 * 3^4 * 5        1.157716049382716
6561 = 3^8        1.499923792104862
6912 = 2^8 * 3^3        1.478587962962963
7168 = 2^10 * 7        1.142299107142857
7200 = 2^5 * 3^2 * 5^2        1.175416666666667
7776 = 2^5 * 3^5        1.474537037037037
8000 = 2^6 * 5^3                  1.23825
8064 = 2^7 * 3^2 * 7        1.096230158730159
8192 = 2^13          1.9998779296875
8640 = 2^6 * 3^3 * 5        1.175925925925926
8748 = 2^2 * 3^7        1.312299954275263
9000 = 2^3 * 3^2 * 5^3        1.126666666666667
9072 = 2^4 * 3^4 * 7        1.102586713697825
9216 = 2^10 * 3^2        1.443739149305556
9600 = 2^7 * 3 * 5^2        1.097916666666667
9720 = 2^3 * 3^5 * 5        1.123456790123457
10000 = 2^4 * 5^4                  1.21055
jagy@phobeusjunior:~\$

• Thanks. I got most of this, but didn't realize this was all that was needed. I thought you might make an appearance. – Jaycob Coleman Oct 31 '13 at 17:32