Problem about the sum of the divisors of a number

[phymath7]

Homework Statement

The sum of divisor function σ(n) returns the sum of all positive divisors d of the number n. We denote $N_k$ any number that fulfils the following condition:
σ($N_K$) ≥$k.N_K$.
Find examples for $N_3$;$N_4$;$N_5$ and prove that they fulfil this condition.

Relevant Equations

$σ(n)=\frac{p_{1}^{a_1+1}-1}{p_1-1}.\frac{p_{2}^{a_2+1}-1}{p_2-1}.......\frac{p_{k}^{a_k+1}-1}{p_k-1}$

I've found that $N_1$ is 1. But it's really tiresome to find them one by one. I also tried to use the equation but couldn't. Please help me out.

 

[fresh_42]

I do not understand your condition. What is $k$, what $N_K$, what $K$? Can you say what the condition for $\sigma(N_3)$ is?

 

[WWGD]

Are the $p_i$ in your first equation the prime divisors of $N$? Like fresh said, can you give as one worked example? Also, if '.' stands for product, please use \cdot instead? Edit: If you want a number that is less than the sum of its proper divisors, then 12, 18, 20 are examples. Is this what you're looking for?

 

[GwtBc]

I do not understand your condition. What is $k$, what $N_K$, what $K$? Can you say what the condition for $\sigma(N_3)$ is?

I believe the condition stated in words is that the number of positive divisors for $N_k$ is greater than or equal to $k$ times $N_k$, for example, if $n$ is $N_3$ then it must have $\sigma(n) \geq 3n$. So basically the $N_k$ are best thought of as classes.

 

[WWGD]

I believe the condition stated in words is that the number of positive divisors for $N_k$ is greater than or equal to $k$ times $N_k$, for example, if $n$ is $N_3$ then it must have $\sigma(n) \geq 3n$. So basically the $N_k$ are best thought of as classes.

Yes, but what are the k's about? What is , e.g.$N_3$?

 

[GwtBc]

Yes, but what are the k's about? What is , e.g.$N_3$?

If $n$ satisfies $\sigma(n) \geq 3n$ then $n \in N_3$, so $N_1 \subseteq N_2 \subseteq N_3 \subseteq ...$. OP denotes $n$ by $N_3$ instead but that would just be confusing since there is clearly more than one such number, and it doesn't make a difference to the problem so...

What's not clear to me is what the $a_i$ are. I assume the $p_i$ are the prime divisors of $n$? @phymath7

 

[fresh_42]

The $a_i$ are the powers in the prime decomposition of $n=p_1^{a_1}\cdots p_k^{a_k}$.

So for $N_3$ we have all numbers $n$ with three prime factors plus the condition that $\sigma(n)\geq 3n$? $\sigma(N_3)\geq 3N_3$ still makes no sense: You defined it by using it! The closest I come from all what you wrote is
$$
N_3 = \{\,n\,|\,\exists \,p_1,p_2,p_3\text{ prime and }\exists\,a_1,a_2,a_3\in \mathbb{Z}\, : \,n=p_1^{a_1}p_2^{a_2}p_3^{a_3}\,\wedge\,\sigma(n)\geq 3n\,\,\}
$$
but that is a guess.

 

[phymath7]

The $a_i$ are the powers in the prime decomposition of $n=p_1^{a_1}\cdots p_k^{a_k}$.

So for $N_3$ we have all numbers $n$ with three prime factors plus the condition that $\sigma(n)\geq 3n$? $\sigma(N_3)\geq 3N_3$ still makes no sense: You defined it by using it! The closest I come from all what you wrote is
$$
N_3 = \{\,n\,|\,\exists \,p_1,p_2,p_3\text{ prime and }\exists\,a_1,a_2,a_3\in \mathbb{Z}\, : \,n=p_1^{a_1}p_2^{a_2}p_3^{a_3}\,\wedge\,\sigma(n)\geq 3n\,\,\}
$$
but that is a guess.

I'm sorry as there is some problem in the first post. The homework equation would be like this,if $n=p_1^{a_1}\cdots p_j^{a_j}$ then, $σ(n) = \frac {p_1^{a_1+1}} {P_1-1}\cdots \frac {p_t^{a_j+1}} {P_j-1}$.I just mixed up with j and k.Now please do what you can.

 

[fresh_42]

So let's start with $N_3$. We thus have
$$
3n = 3 \cdot p_1^{a_1}\cdots p_j^{a_j}\leq \dfrac{p_1^{a_{1}+1}-1}{p_1-1} \cdots \dfrac{p_j^{a_{j}+1}-1}{p_j-1}= \sigma(n)
$$
You are requested to give an example. So choose some primes and play with this inequality until you find $p_i$ and $a_i$ for which it holds.

There is also a trivial upper bound: a number $n$ cannot have more divisors than all numbers $1,\ldots, \dfrac{n}{2}$. So $\sigma(n) \leq 1+\ldots + \dfrac{n}{2}=\dfrac{1}{8}(n^2+2n)$. What does that mean for $\sigma(n)\geq kn$?