Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The question is:

Develop a formula for $\delta_2(n)$, the sum of the squares of the positive divisors of $n$.

share|improve this question
    
    
$\delta_2$ is usually denoted $\sigma_2$. –  lhf Jul 4 '12 at 10:35

2 Answers 2

Hopefully, this can help. Let's say you wanted to find the sum of the positive divisors of $180=2^2\times3^2\times5$. The sum can be written as

$$(1+2^1+2^2)(1+3^1+3^2)(1+5)$$

Since each of the factors is a geometric series, this can be rewritten as

$$\dfrac{2^3-1}{2-1}\times\dfrac{3^3-1}{3-1}\times\dfrac{5^2-1}{5-1}$$

Can you see how to alter this method for $\delta_2(180)?$

share|improve this answer
    
Thank you. I ended up showing it and arrived at the formula. Here is it, $\delta_2(n)=(\frac{p_1^{2e_1+2}-1}{p_1^{2} - 1})(\frac{p_2^{2e_2+2}-1}{p_2^{2}-1})...(\frac{p_k^{2e_k+2}-1}{p_k^{2}-1})$ :) –  HowardRoark Jul 4 '12 at 10:23

Hint: prove it's multiplicative, then evaluate it on prime powers.

share|improve this answer
1  
I can prove that it is multiplicative. I have been on this problem for a while now and I finally decided to ask for help. I can't get the formula. –  HowardRoark Jul 4 '12 at 9:32
1  
So you can't find $\delta_2(p^k)$ for prime $p$? –  Gerry Myerson Jul 4 '12 at 9:47
    
I got it :) Thank you very much. I thought they meant something else by a formula. I don't know what I was thinking. Thank you, I really appreciate that you gave me the hint instead of the answer. It feels much rewarding that way when do finally arrive at the answer. –  HowardRoark Jul 4 '12 at 10:20

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.