2

As the term "summatory function" is somewhat ambiguous, the function I mean is $\Phi_2(n):=\sum _{d|n}\phi(d)^2$, where $\phi$ is Euler's totient. It is well-known that $\Phi_1(n):=\sum _{d|n}\phi(d)=n$. All powers of $\phi$ are multiplicative, hence their summatory functions (over divisors) $\Phi_k(n)$ are also multiplicative, and using that one can get a product formula for them after computing their value on prime powers. I am curious if anything is known beyond that, like Ramanujan's averaged large $n$ asymptotics for the squared divisor sum $\sigma_2(n)$. For a different kind of summatory function, $\sum_{k=1}^n \phi(k)^2$, the asymptotics are derived in Average order of Eulers totient function squared.

$\Phi_2(n)$ comes up as the dimension of a certain algebra, but explaining what it is would take too much space. Was $\Phi_2(n)$ studied, does it 'oscillate' around $cn^2$ for some constant $c$? References to literature are welcome.

Conifold
  • 12,093
  • 1
    There is a claim in https://oeis.org/A029939 that $\sum_n \Phi_2(n)$ approaches a constant times $n^3$. – R. J. Mathar May 15 '25 at 10:31
  • 1
    Your function is tabulated at https://oeis.org/A029939 – I think it jumps around too much to have any useful asymptotics. E.g., $\Phi_2(9660)=9151950$, $\Phi_2(9661)=93315601$ is more than ten times as big. But $\sum_1^n\Phi_2(k)$ is asymptotic to $cn^3$, where $c$ is a rather complicated constant, approximately $0.171593$ – Gerry Myerson May 15 '25 at 10:33
  • 2
    Since $\Phi_2$ is multiplicative, one can show that $$\Phi_2(n) = \phi(n)^2 \prod_{p\mid n} \biggl(1-\frac1{p^2}\biggr)^{-1} \prod_{p^k|n} \biggl(1-\frac2{p^{2k}(p-1)}\biggr).$$ Evaluating asymptotics for $\sum_{n\le x} \Phi_2(n)$ (and pointwise lower bounds for $\Phi_2(n)$) can therefore be done with the same tools used to address $\phi(n)^2$. – Greg Martin May 15 '25 at 23:05

0 Answers0