This question was asked earlier at MSE .

Let $\omega$(n) denote the number of distinct primes dividing $n$. The Mobius function is defined as $\mu(n) = (-1)^{\omega(n)}$ if $n$ is squarefree and $\mu(n) = 0$ otherwise . Next let $S(n) = \sum_{k=1}^n \frac{\mu(k)}{k}$. It is known that $S(n)$ approaches zero as $n$ approaches infinity and that this is equivalent to the prime number theorem (von Mangoldt, Landau).

What happens if we replace powers of $(-1)$ in the Mobius function with other roots of unity? To focus on a specific case, let's use fourth roots and define $f(n) = i^n$ if $n$ is squarefree and $f(n) = 0$ otherwise. Then $f(n)$ is a multiplicative function whose initial values are $(1,i,i,0,i,-1,i,0,0,-1,...)$. Finally, let $T(n) = \sum_{k=1}^n \frac{f(k)}{k}$. Does $T(n)$ have properties analogous to those of $S(n)$?

Questions: $(1)$ Does $\sum_{k=1}^{\infty}$ $\frac{f(k)}{k}$ converge? [The corresponding infinite product $\prod_p (1 + i/p)$ does not converge since $\sum\frac{1}{p^2} < \infty$ while $\sum \frac{1}{p} = \infty$. ]

$(2)$ The partial sums $S(n)$ are known to satisfy $|S(n)| \leq 1$ for all $n$. Are the partial sums $|T(n)|$ also bounded by some constant independent of $n$? [Over the initial stretch $1 \leq n \leq 20$ , one finds that the max $T$ value is $|T(19)| = 1.57 ...$ ].


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    $\begingroup$ I suppose that there is a typo and you wanted to define $f(n)=i^{\omega(n)}$ instead of $f(n)=i^n$? $\endgroup$ Nov 8, 2018 at 9:00
  • $\begingroup$ @ChristianBernert Yes, edited appropriately. The first 10,000 approx. values of $|T(n)|$ for $f(n)=i^{\omega(n)}$ are here: pastebin.com/kdKas96M. The values seem to be increasing generally. For third roots of unity (first 3000 approx. values of $|T(n)|$ are here: pastebin.com/ismMhVv8) the behavior is different, agreeing more with part (2) of your question. $\endgroup$ Nov 8, 2018 at 13:04

1 Answer 1


Theorem 1 in Section 6.1 of Tenenbaum's Introduction to Analytic and Probabilistic Number Theory states (after taking its $N=0$ for simplicity) that for any $|z|<2$, $$ \sum_{n\le x} z^{\omega(n)} \sim \frac1{\Gamma(z)} \prod_p \bigg( 1 + \frac z{p-1} \bigg) \bigg( 1-\frac1p \bigg)^z \cdot x\, (\log x)^{z-1}. $$ The same method would show that $$ \sum_{n\le x} \mu^2(n) z^{\omega(n)} \sim \frac1{\Gamma(z)} \prod_p \bigg( 1 + \frac zp \bigg) \bigg( 1-\frac1p \bigg)^z \cdot x\, (\log x)^{z-1}. $$ From here, partial summation gives (for $z\ne 0$) $$ \sum_{n\le x} \mu^2(n) \frac{z^{\omega(n)}}n = \frac1{\Gamma(z)} \prod_p \bigg( 1 + \frac zp \bigg) \bigg( 1-\frac1p \bigg)^z \cdot \frac{(\log x)^z}z + c(z) + o(1) $$ for some constant $c(z)$. So when $|z|<2$ and $\Re z<0$, the sum converges to this $c(z)$; when $|z|<2$ and $\Re z>0$, the sum grows to infinity in modulus. When $|z|<2$ and $\Re z=0$, the sum oscillates asymptotically around a circle in the complex plane. So the interesting case $z=i$, strangely, the answer to your question (1) is no while the answer to your question (2) is yes!


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