# Problem about the sum of the divisors of a number

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
Mentor
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
Gold Member
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?

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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
Gold Member
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##?

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
Mentor
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.

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
Mentor
$$3n = 3 \cdot p_1^{a_1}\cdots p_j^{a_j}\leq \dfrac{p_1^{a_{1}+1}-1}{p_1-1} \cdots \dfrac{p_j^{a_{j}+1}-1}{p_j-1}= \sigma(n)$$