Mysterious multiplicative function?

Question

Consider $$f(n)=\prod_{p^k||n} p^{2k}(1+p^{-2})$$

Can this function be expressed by usual ones, as convolutions or directly? I do not know very well if convolution can be seen on decomposition in products like this, so any help will be welcome!

Answer 1

$f(n)$ is multiplicative $$1+\sum_{k= 1}^\infty f(p^k) p^{-s} = 1+\sum_{k= 1}^\infty p^{2k} (1+p^{-2})p^{-sk} = 1+p^{2-s}\frac{1+p^{-2}}{1-p^{2-s}}=\frac{1-p^{-s}}{1-p^{2-s}}$$

$$\sum_{n=1}^\infty f(n) n^{-s} = \prod_p \frac{1-p^{-s}}{1-p^{2-s}} = \frac{\zeta(s-2)}{\zeta(s)}$$ $$f(n) = \sum_{d | n} \mu(d) (n/d)^{2}$$

Answer 2

Hint: You can simplify it likes the following as $n = p_1^{n_1} p_2^{n_2} \cdots p_k^{n_k}$:

$$f(n) = (p_1^{n_1} p_2^{n_2} \cdots p_k^{n_k}) (p_1^{n_1} p_2^{n_2} \cdots p_k^{n_k})((1+p_1^{-2})(1+p_2^{-2})\cdots(1+p_k^{-2})) = n^2((1+p_1^{-2})(1+p_2^{-2})\cdots(1+p_k^{-2}))$$

Then you can think on Euler function.

Answer 3

f(n) is known as OEIS A065958, the Dedekind psi_2 function.