Bounds on the average of the divisors of natural numbers.

Question

I managed to find an interesting inequality containing the sum of the divisors of a number and the number of them, using AM-GM between each 2 of the divisors of it: $\sigma(n)\geq\sqrt n.d(n)+(\sqrt n-1)^2$. I tried to use this to find a bound for the average of divisors of $n$, and I got: $\frac{\sigma(n)}{d(n)}\geq\sqrt n+\frac{n-2\sqrt n+1}{d(n)}$(I) and we know that $d(n)\leq2\sqrt n$ so $\frac{1}{d(n)}\geq\frac{1}{2\sqrt n}$. Using this on inequality (I) we get: $\frac{\sigma(n)}{d(n)}\geq\sqrt n+\frac{n-2\sqrt n+1}{d(n)}\geq\sqrt n+\frac{n-2\sqrt n+1}{2\sqrt n}=\frac{3}{2}\sqrt{n}-1+\frac{1}{2\sqrt n}\geq\frac{3}{2}\sqrt{n}-1$
Now we have a lower bound for $\frac{\sigma(n)}{d(n)}$. According to this question, we can also say $\frac{\sigma(n)}{d(n)}\leq\frac{3}{8}n+1$ for composite $n>4$.
To sum it up, for composite $n>4$ have $\frac{3}{8}n+1\geq\frac{\sigma(n)}{d(n)}\geq\frac{3}{2}\sqrt{n}-1$
Can this bound be improved? I realized the higher bound is very good, but the lower bound will not be good for large $n$. I would appreciate any help.

Answer 1

$$ \frac{\sigma(n)}{d(n)} \; \geq \; \frac{14}{3} \; \left(\frac{n}{12} \right)^{\left( \frac{2}{3} \right)} $$

$$ \frac{\sigma(n)}{d(n)} \; \geq \; \frac{9762}{5} \; \left(\frac{n}{55440} \right)^{\left( \frac{3}{4} \right)} $$

$$ \color{magenta}{ \frac{\sigma(n)}{d(n)} \; \geq \; 340156800 \; \left(\frac{n}{321253732800} \right)^{\left( \frac{4}{5} \right)} } $$

These are fairly difficult. The procedure was invented by Ramanujan and is what defines the superior highly composite numbers and the colossally abundant numbers. There was a bit of a trick needed here because of nontrivial number theoretic multiplicative functions in both numerator and denominator.

I am going to submit the following sequence to the OEIS as the Miserable Average Divisor numbers. Given real $\delta > 0,$ such a number minimizes $$ \frac{n^\delta \sigma(n)}{n d(n)} = \frac{ \sigma(n)}{n^{1-\delta} \, d(n)}. $$ Let's see, for a fixed prime $p$ and exponent $k,$ the first (largest) $\delta$ that causes $p$ to get exponent $k$ is $$ \delta = \frac{\log(k+1) + \log p + \log \left(p^k -1 \right) - \log k - \log \left(p^{k+1} -1 \right) }{\log p}. $$ This is enough information to find the first MAD numbers by placing some bound on $\delta$ and ordering them. Now, the other thing we really want is to solve for $k$ once given a prime $p$ and a real $\delta.$ So far, I have been unable to get a closed form for $k$ using the floor function, which is how it usually works. Can't win them all.

Miserable Average Divisor numbers

Given real delta > 0, minimize

 n^delta * sigma(n) / ( n * d(n)   )



delta     prime bumped   n
1.0000000000000000    1  1 = 1
0.4150374992788438    2  2 =  2
0.3690702464285426    3  6 =  2 3
0.3625700793847085    2  12 =  2^2 3
0.3173938055140146    5  60 =  2^2 3 5
0.3155018257279289    2  120 =  2^3 3 5
0.2962122340986647    3  360 =  2^3 3^2 5
0.2875856257839555    7  2520 =  2^3 3^2 5 7
0.2746223801090051    2  5040 =  2^4 3^2 5 7
0.2527782636907857   11  55440 =  2^4 3^2 5 7 11
0.2413455870407011   13  720720 =  2^4 3^2 5 7 11 13
0.2399507927207518    2  1441440 =  2^5 3^2 5 7 11 13
0.2388142451834079    3  4324320 =  2^5 3^3 5 7 11 13
0.231556173994648     5  21621600 =  2^5 3^3 5^2 7 11 13
0.22447612992302     17  367567200 =  2^5 3^3 5^2 7 11 13 17
0.2179885169004592   19  6983776800 =  2^5 3^3 5^2 7 11 13 17 19
0.2109876580641981    2  13967553600 =  2^6 3^3 5^2 7 11 13 17 19
0.2074912346627816   23  321253732800 =  2^6 3^3 5^2 7 11 13 17 19 23
0.1992720636135298    7  2248776129600 =  2^6 3^3 5^2 7^2 11 13 17 19 23
0.1957789461245263   29  65214507758400 =  2^6 3^3 5^2 7^2 11 13 17 19 23 29
0.1955601177190355    3  195643523275200 =  2^6 3^4 5^2 7^2 11 13 17 19 23 29
0.1926036536716005   31  6064949221531200 =  2^6 3^4 5^2 7^2 11 13 17 19 23 29 31
0.186976327855704     2  12129898443062400 =  2^7 3^4 5^2 7^2 11 13 17 19 23 29 31
0.1845732718387193   37  448806242393308800 =  2^7 3^4 5^2 7^2 11 13 17 19 23 29 31 37
0.1801633621115231   41  18401055938125660800 =  2^7 3^4 5^2 7^2 11 13 17 19 23 29 31 37 41
0.1781765504948862   43  791245405339403414400 =  2^7 3^4 5^2 7^2 11 13 17 19 23 29 31 37 41 43
0.1747511848382693    5  3956227026697017072000 =  2^7 3^4 5^3 7^2 11 13 17 19 23 29 31 37 41 43
0.1745631183239829   47  185942670254759802384000 =  2^7 3^4 5^3 7^2 11 13 17 19 23 29 31 37 41 43 47
0.1698754303797211   53  9854961523502269526352000 =  2^7 3^4 5^3 7^2 11 13 17 19 23 29 31 37 41 43 47 53