# Multiplicative Functions and Totient Function

I have two questions.

1. If $f$(n) is a multiplicative function defined on the positive integers, is

$g(n)=$$\frac{f(n)}{n} multiplicative as well? I think the answer is yes, but I don't know how to prove it. 1. Evaluate$$\sum_{d|2016} \frac{\phi(d)}{d}$$where \phi(n) is the totient function. • The product of two multiplicative functions is multiplicative. – Lord Shark the Unknown May 16 '17 at 6:03 ## 2 Answers As regards the second question, note that$$g(n):=\sum_{d|n} \frac{\phi(d)}{d}$$is multiplicative (it is the Dirichlet convolution of two multiplicative functions). Moreover, n=2016=2^5\cdot3^2\cdot 7, and for any prime p,$$g(p^k)=\sum_{d|p^k} \frac{\phi(d)}{d}=1+\sum_{j=1}^k\frac{p^j-p^{j-1}}{p^j}=1+k\left(1-\frac{1}{p}\right).$$Hence$$g(2016)=g(2^5)\cdot g(3^2)\cdot g(7)=\frac{7}{2}\cdot\frac{7}{3}\cdot \frac{13}{7}=\frac{91}{6}.$$• That helps a lot. I'm getting 323/42 with that formula. – shrindle May 16 '17 at 6:14 • It should be 91/6. Check your computations. – Robert Z May 16 '17 at 6:16 1) Show that for positive integers$m,n$, one has$g(mn) =g (n)g(m) $. Note this is precisely the definition of a function being 'multiplicative.' 2) We have$d$as any divisor of 2016. Now observe that$2016=2^5×3^2×7\$.

Use the fact that the Euler Totient function is multiplicative for coprime numbers to simplify your counting process.

• That's the part I'm confused about. How do I separate this sum in such a way that I can "cleanly" solve it utilizing the multiplicative functions you discuss. I do realize that Euler's totient function is multiplicative when the numbers are relatively prime, but I'm still confused. – shrindle May 16 '17 at 6:04
• You should have g(mn) = f(mn)/mn. Then because f is multiplicative by assumption, write f (mn) =f(n)f(m). Can you see it now? – thedilated May 16 '17 at 6:07
• I understand the first question. What I'm confused by is the second one with the sum. – shrindle May 16 '17 at 6:09
• Let us see an e.g. Note that 63 is a divisor of 2016. So one of the terms in the summand is phi(63)/63. But 63 = 7×9 and gcd(7,9) =1. So we can write phi(63) = phi(7)phi (9) which is v easy to compute. – thedilated May 16 '17 at 6:15