How would I find a number where $\sum_{d\mid n}d > 100n$?

Question

How would I find an $n$ such that $$ \sum_{d\mid n} d > 100 n $$ I have absolutely no idea as to how to begin this question.

Answer 1

Claim: If $n = \prod p_i ^{a _i} $, then $\sum_{d \mid n} d = \prod \frac { p_i^{a_i+1} - 1 } { p_i - 1} $.

You should be aware of this from Number Theory.

Hence, if we want $\frac { \sum _{d \mid n} d} {n} > 100$, it is equivalent to $\prod\frac { p_i ^{a_i + 1} -1 } { (p_i -1 )p_i ^{a_i} }> 100$.

Fact 1: As $a_i \rightarrow \infty$, then $\frac { p_i ^{a_i + 1} -1 } { (p_i -1 )p_i ^{a_i}} \rightarrow \frac { p_i}{(p_i - 1)}$.

Fact 2: $$ \prod_{p < n, p \mbox{ prime}} \frac {p}{p-1} \rightarrow \infty.$$

These two facts combined guarantee that such an $n$ exists. However, it will be very large.

Answer 2

From a theorem of Robin we have an unconditional result (1984) that says that for $n \geq 13,$ we have the bound $$ \frac{\sigma(n)}{n} < \; e^\gamma \log \log n + \frac{0.64821364942...}{\log \log n},$$ with the constant in the numerator giving equality for $n=12.$ Note $$ e^\gamma = 1.7810724179901979852365\ldots $$ Let's see, the Riemann Hypothesis is equivalent to the statement that, for $n \geq 5041 = 1 + 7!,$ the number $0.64821364942...$ can be replaced by $0.$

In order for your ratio to reach 100, you need $$ n > e^{\left( e^{56.14594836} \right)} $$ which is a very, very big number. See http://en.wikipedia.org/wiki/Colossally_abundant_number and, for that matter, my answer at https://mathoverflow.net/questions/79927/which-n-maximize-gn-frac-sigmann-log-log-n

EEEDDDDIIITTT: The smallest number for which your ratio is at least 100 is called a "superabundant number" by Alaoglu and Erdos (1944). The way I like to say it is that a superabundant number $n$ sets a new world record for $\sigma(n)/n.$ On journal numbered page 450 we find Theorem 1, such numbers have nonincreasing exponents, and Theorem 2, all the exponents can be found within an error of $\pm 1$ by finding the exponent of 2. They give more theorems. The upshot is that superabundant numbers have factorizations very similar to those of a subsequence called the Colossally Abundant Numbers. Given a positive real number they call $\epsilon,$ the exponent of a prime $q$ is given in Theorem 10 on journal page 455. Finally, on pages 468-469, they display all superabundant numbers up to $10^{18}.$ Those numbers in this table that are also colossally
abundant are labelled. The largest of these, with $\sigma(n)/n \approx 6.407,$ can be found with $\epsilon = 0.007.$

EDIT: The software still needs work, but I found a small number with your ratio exceeding 10:

n 3087412123011300495851995214301626485649741980169578200745289299789026306722292583565231315659291930544327540696937285310080000

sigma 30941752989588328135737108482280837416279110376522697522825909747010685327091094608601433751156507323392138936320000000000000000

mpf 10.0219

This number $n$ is a CA for $\epsilon = 0.00064.$

Why not, the number displayed is $$ n = 2^{10} \, 3^6 \, 5^4 \, 7^3 \, 11^2 \, 13^2 \, 17^2 \, 19^2 \cdot 23 \cdot 29 \cdots 271 \cdot 277 $$

I found the first million superabundant numbers at a link at http://oeis.org/A004394 and excerpted just a few lines below. It turns out I did not do too badly with ratio 10. It is the 853'd SA number that first gives $\sigma(n) \geq 10n.$ I gave the 860th SA number. Note that the 1,000,000th SA number still has $\sigma(n)/n \approx 22.04,$ so 100 requires a really big SA number, probably so big that one should just be satisfied with the first CA number that succeeds. I threw in the first number with ratio at least 20. I also put in the first CA number with ratio at least 20, the relevant $\epsilon = 0.00000118087 = 1.18087 \cdot 10^{-6}$

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// This file contains information about the first 1000000 superabundant
// numbers (SA numbers).  For each SA number n, we give its abundance,
// which is Sigma(n)/n, its base-10 logarithm, and its factorization.
// In column "*", a "C" indicates that the number is also a colossally
// abundant number  The factorization is given in a very compact form.
// For example, {13,5,0,2} means 13 * 11 * 7 * 5^2 * 3^2 * 2^4.
//
// Created by Tony D. Noe, [email protected] on 15-Oct-2009.
// Algorithm developed with help from Devin Kilminster.
//
// Corrected 30-Oct-2009: some SA numbers were erroneously marked with a "C".
//
// position       abundance                log10     *   factorization
//
         1   1.0000000000000000         0.0000000000 S {0}
         2   1.5000000000000000         0.3010299957 C {2}
         3   1.7500000000000000         0.6020599913 S {0,2}
         4   2.0000000000000000         0.7781512504 C {3}
         5   2.3333333333333333         1.0791812460 C {3,2}
         6   2.5000000000000000         1.3802112417 S {3,0,2}
         7   2.5277777777777778         1.5563025008 S {0,3}
         8   2.5833333333333333         1.6812412374 S {3,0,0,2}
         9   2.8000000000000000         1.7781512504 C {5,2}
        10   3.0000000000000000         2.0791812460 C {5,0,2}

       100   6.2948845987914689        16.5059308532 S {37,5,0,3,0,2}

       853  10.0039459446460710       125.4088426721 S {271,23,7,5,0,3,0,0,0,2}

       860  10.0219056467943640       126.4895946051 C {277,19,7,5,0,3,0,0,0,2}

      1000  10.3332076452686825       149.9416130802 S {337,23,7,5,0,3,0,0,2}

     10000  14.3329541785168048      1378.4010046121 S {3163,79,19,7,5,0,0,0,3,0,0,0,2}

    100000  18.2270550779625505     12146.0718137217 S {27893,229,41,17,0,7,5,0,0,3,0,0,0,0,0,0,0,2}

    297146  20.0000073661334977     32800.4952412562 S {75401,373,53,19,11,7,0,5,0,0,0,3,0,0,0,0,0,0,2}

    297210  20.0002475182905199     32804.8955086267 C {75403,373,53,19,11,7,0,5,0,0,3,0,0,0,0,0,0,0,2}

   1000000  22.0419355366430523    103082.8167397490 S {237173,673,83,31,13,0,7,5,0,0,0,0,3,0,0,0,0,0,0,2}

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Sunday, 7 April 2013: It took a while, but I found the first Colossally Abundant number with $\sigma(n) / n \geq 30. $ It is easy to show that such numbers are superabundant. $$ 2^{27} 3^{17} 5^{11} 7^9 11^7 13^7 17^6 19^6 23^5 \cdots 37^5 41^4 \cdots 89^4 97^3 \cdots 383^3 389^2 \cdots 6301^2 \cdot 6311 \cdots 20663801. $$ This number is about $$ 10^{8974734.77703953} $$ The Alaoglu-Erdos parameter is $$ \epsilon = 0.000000002873077 = 2.873077 \cdot 10^{-9}. $$ Oh, $$ \sigma(n)/n \approx 30.00000029489 $$

Answer 3

Consider $n=p_1^{k_1}p_2^{k_2} \cdots p_l^{k_l}$. We then have $$\sum_{d \mid n} d = \left(\dfrac{p_1^{k_1+1}-1}{p-1} \right) \left(\dfrac{p_2^{k_2+1}-1}{p_2-1} \right)\cdots \left(\dfrac{p_l^{k_l+1}-1}{p_l-1} \right) \tag{$\star$}$$ We will do this for the following case: $$\sum_{d \mid n} d > 10n$$ You can get the idea and convince yourself that there exists a $n$ such that $\sum_{d \mid n} d > m \times n$ is true for any $m \in \mathbb{N}$.

We want $(\star)$ to be greater than $10p_1^{k_1}p_2^{k_2} \cdots p_l^{k_l}$, i.e., we want $$\left(\dfrac{p_1^{k_1+1}-1}{p_1-1} \right) \left(\dfrac{p_2^{k_2+1}-1}{p_2-1} \right)\cdots \left(\dfrac{p_l^{k_l+1}-1}{p_l-1} \right) > 10p_1^{k_1}p_2^{k_2} \cdots p_l^{k_l}$$ Hence, we want $$\left(1 - \dfrac1{p_1^{k_1+1}}\right)\left(1 - \dfrac1{p_2^{k_2+1}}\right) \cdots \left(1 - \dfrac1{p_l^{k_l+1}}\right) > 10\left(1 - \dfrac1{p_1}\right)\left(1 - \dfrac1{p_2}\right) \cdots \left(1 - \dfrac1{p_l}\right) \tag{$\dagger$}$$ Our task is to choose $p_j$'s, $l$ and $k_j$'s such that we obtain the inequality. First note that since $$\sum_{p = \text{ primes}} \dfrac1p$$ diverges, we can choose primes $p_j$'s such that $\left(1 - \dfrac1{p_1}\right)\left(1 - \dfrac1{p_2}\right) \cdots \left(1 - \dfrac1{p_l}\right)$ is arbitrarily small. You can read here, why this is the case. $\left(\displaystyle \sum_{p = \text{ primes}}^{p \leq n} \dfrac1p \sim \log \left(\log (n)\right) \right)$. Note that the product on the left of $(\dagger)$ is always less than $1$. Hence, let us first make the product on the right in $\dagger$ less than $1$. To get the minimum $l$ such that $$10\left(1 - \dfrac1{p_1}\right)\left(1 - \dfrac1{p_2}\right) \cdots \left(1 - \dfrac1{p_l}\right) < 1$$ we need to choose the first $l$ primes. We soon find that choosing $l=55$ gives us $$10\left(1 - \dfrac1{p_1}\right)\left(1 - \dfrac1{p_2}\right) \cdots \left(1 - \dfrac1{p_{55}}\right) \approx 0.9996 < 0.9997$$ Now once we have this we have made the right hand side of $\dagger$ less than $0.9997$. We have our $p_1,p_2 , \ldots, p_{55}$. Now note that the left hand side of $\dagger$ is $$\left(1 - \dfrac1{p_1^{k_1+1}}\right)\left(1 - \dfrac1{p_2^{k_2+1}}\right) \cdots \left(1 - \dfrac1{p_{55}^{k_{55}+1}}\right)$$ tends to $1$ as $k_1,k_2,\ldots, k_{55} \to \infty$, i.e., we have $$\lim_{k_1,k_2,\ldots,k_{55} \to \infty}\left(1 - \dfrac1{p_1^{k_1+1}}\right)\left(1 - \dfrac1{p_2^{k_2+1}}\right) \cdots \left(1 - \dfrac1{p_{55}^{k_{55}+1}}\right) = 1$$ Hence, there exists $k_1,k_2,\ldots,k_{55}$ such that $$0.9997 < \left(1 - \dfrac1{p_1^{k_1+1}}\right)\left(1 - \dfrac1{p_2^{k_2+1}}\right) \cdots \left(1 - \dfrac1{p_{55}^{k_{55}+1}}\right)$$ This choice of $k_1,k_2,\ldots,k_{55}$ gives us what we want. This same procedure will work if you replace $10$ by any other number $m$. The corresponding $l$ will be large and will be a function increasing faster than an exponential in $m$.

Answer 4

You want lots of distinct prime divisors to have more divisors in total. A duplicate $p$ generally makes your $n$ larger without contributing as much to the sum as the next unused prime would. Try looking at $2\cdot3\cdot5\cdots p_k$.

In fact, if $n=2\cdot3\cdot5\cdots p_k$, then $$\sum_{d\mid n}d=\prod_{i=1}^k(p_i+1)$$ which we would like to be greater than $100\prod\limits_{i=1}^kp_i$. So we want $$\prod_{i=1}^k(1+1/p_i)>100$$

Now $$ \begin{align} \prod_{i=1}^k(1+1/p_i)& >1+\sum_{i=1}^k\frac{1}{p_i}\\ \end{align} $$

So it would suffice to have a value of $k$ large enough for the $k$th partial sum of the harmonic prime series to surpass $99$. This will eventually happen since the series diverges, but it does so even slower than the harmonic series. So it will take quite a while. Since $\sum\limits_{i=1}^k\frac{1}{p_i}$ is asymptotic to $\log(\log(k))$, $k$ will need to be in the neighborhood of $e^{e^{99}}$. But this is an overestimate with a lot lost when the product was replaced with the sum. You could get away with much smaller (although still enormous) $k$.

If you replace the $100$ by $10$, and if you trust WolframAlpha, then $k=553$ would work.

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Added later: I believe that if $k$ is large enough to make the harmonic prime series's $k$th partial sum larger than 13.13 then you will get a ratio above 100.

As above, I am working with $n=2\cdot3\cdots p_k$. As above, we want $\prod_{i=1}^k(1+1/p_i)>100$. By expanding the product to its second-order terms, we can get this result.

Let $S_k=\frac12+\frac13+\cdots\frac{1}{p_k}$. Then $$S_k^2=\overbrace{\frac{1}{2^2}+\frac{1}{3^2}+\cdots\frac{1}{p_k^2}}^{T_k}+2\overbrace{\left(\frac1{2\cdot3}+\frac1{2\cdot5}+\cdots+\frac1{p_{k-1}\cdot p_k}\right)}^{U_k}$$

Note that $T_k$ is bounded above by $c=\zeta(2)-1=\frac{\pi^2}{6}-1\approx0.645$. Now $$ \begin{align} \prod_{i=1}^k(1+1/p_i) & >1+S_k+U_k\\ & =1+S_k+\left(\frac{S_k^2-T_k}{2}\right)\\ & >1+S_k+\left(\frac{S_k^2-c}{2}\right)\\ \end{align} $$

Applying the quadratic formula, this expression exceeds $100$ when $S_k>-1+\sqrt{199+c}\approx13.1295765707\ldots$. So if you can get $S_k$ this large, the corresponding $n$ should work. That roughly reduces $k$ to about $e^{e^{13.13}}$. There are obvious ways to lower this bound: $c$ can be shaved closer to the true sum of prime square reciprocals, or we could move on to a third-order expansion which would involve $\zeta(3)$ and bring the $13.13$ figure down to something closer to $\sqrt[3]{6\cdot100}$. I applied these and unless I made an algebra error during the third-order expansion, it brought $k$ down to $e^{e^{7.3}}$.