6
$\begingroup$

It is well-known that the Euler $\phi$-function is multiplicative: that is, for co-prime positive integers $m,n$ we have $\phi(mn) = \phi(m)\phi(n)$. Thus it is defined by its values on prime powers. We know that $\phi(p^k) = p^{k-1} (p-1)$ for all primes $p$.

What about the multiplicative function $\psi$ defined on the primes by $\psi(p^k) = p^{k-1} (p+1)$? Does it have a name? If so, what's known about it?

For example, can one evaluate $\sum_{n \leq X} \psi(n)$?

Improve this question
$\endgroup$
3

2 Answers 2

Reset to default
12
$\begingroup$

$\psi$ is the multiplicative convolution of $\mu^2$ and the identity function, hence its Dirichlet series is $$\sum_{n=1}^\infty\frac{\psi(n)}{n^s}=\frac{\zeta(s)\zeta(s-1)}{\zeta(2s)},\qquad\Re(s)>2.$$ This implies by Perron's formula and standard bounds that $$\sum_{n \leq X} \psi(n)\sim\frac{\zeta(2)}{2\zeta(4)}X^2=\frac{15}{2\pi^2}X^2.$$

Improve this answer
$\endgroup$
5
  • 2
    $\begingroup$ I will just add this function appears in Exercise 11 in Chapter 3 in Apostol's book. The last part of this exercise is the asymptotic formula $\sum_{n\le x} \varphi_1(n)=\frac{\zeta(2)}{2\zeta(4)}x^2+O(x\log x)$. (Apostol denotes the function by $\varphi_1$.) $\endgroup$
    – Martin Sleziak
    Commented Oct 18, 2018 at 13:58
  • 2
    $\begingroup$ In the field of automorphic forms, this function is often denotes $\nu(n)$ (for example, in Iwaniec-Luo-Sarnak's seminal paper on low-lying zeroes of $L$-functions of holomorphic cuspidal eigenforms). $\endgroup$
    – Peter Humphries
    Commented Oct 18, 2018 at 14:07
  • $\begingroup$ @MartinSleziak: Thanks for your comment. Clearly, one can also prove this statement by elementary means, by making use of the convolution I mentioned. $\endgroup$
    – GH from MO
    Commented Oct 18, 2018 at 14:35
  • $\begingroup$ @PeterHumphries: Indeed the function looked familiar! It is also the index of $\Gamma_0(n)$ in $\Gamma_0(1)$ if I am not mistaken. $\endgroup$
    – GH from MO
    Commented Oct 18, 2018 at 14:36
  • 1
    $\begingroup$ @GHfromMO, yep, which is why it comes up in the field automorphic forms so often! $\endgroup$
    – Peter Humphries
    Commented Oct 18, 2018 at 15:03
8
$\begingroup$

This is also called the Dedekind $\psi$ function: $$ \psi(n):=n\prod_{p|n}(1+p^{-1}) $$

See also A001615 and A158523.

Improve this answer
$\endgroup$
4
  • $\begingroup$ This should be a comment in my opinion, but thanks for it anyways. (Usually answers should be true answers, which really answer the original question.) $\endgroup$
    – GH from MO
    Commented Oct 18, 2018 at 14:38
  • 11
    $\begingroup$ @GHfromMO You obviously know more about the etiquette of this site than me, but I do believe I answered the main two questions in the OP: "does this have a name", and "what is known about it". I didn't say anything about the asymptotics, because that was already covered by you, and because that seemed to me as an example question, and not the main one. But I appreciate the feedback anyway. $\endgroup$
    – AccidentalFourierTransform
    Commented Oct 18, 2018 at 14:42
  • 2
    $\begingroup$ I agree with you. I forgot about those parts of the question! (BTW it is usually discouraged to ask several questions in one post, exactly to avoid confusions like this.) $\endgroup$
    – GH from MO
    Commented Oct 18, 2018 at 14:44
  • 4
    $\begingroup$ I find it amazing that I used the same letter as the generally accepted one! $\endgroup$
    – Stanley Yao Xiao
    Commented Oct 18, 2018 at 15:09

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .