# Are there infinitely many zeroes of $\sum_{r = 1}^{n-1} \mu(r)\gcd(n,r)$?

Let $$\mu(n)$$ be the Möbius function and $$S(x)$$ be the number of positive integers $$n \le x$$ such that

$$\sum_{r = 1}^{n-1} \mu(r)\gcd(n,r) = 0$$

My experimental data for $$n \le 6 \times 10^5$$seems to suggest that the number of solutions $$\le x$$ is growing at about

$$S(x) \sim (\zeta(2)-1)\sqrt{x}$$ Is there any explanation for this? The square may come from the growth rate of the Mertens function $$M(n) = \sum_{r \le n}\mu(r)$$ while the appearance $$\zeta(2)$$ may be due to the fact that it appears in many sums involving the Pillai function $$P(n) = \sum_{r \le n} \gcd(n,r)$$.

I also observed that $$200$$ out of the $$230$$ zeroes for $$n \le 1.3 \times 10^5$$, were primes which indicate that the zeroes might be dominated by primes. For a prime $$p$$, $$\sum_{r = 1}^{p-1} \mu(r)\gcd(p,r) = M(p-1)$$ so I guess it is more likely that a sequence of $$\pm 1$$ adds to to $$0$$ than a sequence of integers of higher absolute value. Update: Increased graph for the number of zeroes from $$1.5 \times 10^5$$ to $$6 \times 10^5$$

• I deleted my answer. I stated this inequality $$\sum_{r = 1}^n \mu(r) \leq \sum_{r = 1}^n \mu(r)\gcd(n,r) \leq n \sum_{r = 1}^n \mu(r).$$ It were true then the fact that Mertens function has infinitely many integral zeroes would prove your claim. But the inequality is false. I was careless. – Parthiv Basu Jul 18 at 15:20
• I doubt we can answer it, your sequence $a_n = \sum_{r = 1}^{n-1} \mu(r)\gcd(n,r)$ isn't directly related to $\zeta(s)$ and up to approximations we can only estimate the changes of signs, not the number of times it is $0$ – reuns Jul 18 at 20:14
• I guess I should post it in MO. – Nilotpal Kanti Sinha Jul 19 at 4:49

Here's an extended comment with some observations.

Quick summary is, I don't think the hypothesis is correct as $$S(x)/\sqrt{x}$$ starts to drop off as $$x$$ increases beyond $$1,000,000$$.

One way of thinking of this is in terms of $$\mu(r)$$ being "randomly" $$\pm1$$ for square-free $$r$$, so the cumulative sum of $$\mu(r)$$ would be like a random walk ... except the $$\mu(r)$$ will tend to cancel more frequently as, eg, $$\mu(r)$$ and $$\mu(2r)$$ will cancel for odd $$r$$. This would explain that $$S(x)$$ increases at a rate proportional to $$\sqrt{x}$$: the variance of the sum $$\sum_{r=1}^n \mu(r)$$ would have variance proportional to $$n$$ and thus a likelihood proportional to $$1/\sqrt{n}$$ of being zero. However, using this in a naive manner does not seem to work as I'll explain later, however it may still give some indications as to what is going on.

As is noted in the question, a large portion of the zeroes are for $$n$$ prime. This seems natural as all the terms in the sum then has $$\gcd(r,n)=1$$ and thus adds less variance. If $$n$$ has large proper divisors $$d|n$$, values of $$r$$ which are multiples of $$d$$ add more variance as they are multiplied by $$d$$.

However, if we look at the zeroes which are not prime, many of them are multiples of $$3$$: often $$n=3p$$ for some prime $$p$$. This is also natural, as the terms for multiples of $$p$$, $$r=p$$ and $$r=2p$$, cancel out.

There are more zeroes than just for $$n=p$$ and $$n=3p$$, but they tend to have in common that the partial sums for each divisor (ie for each different value of $$\gcd(r,n)$$) tends to be zero or close to zero.

It is not clear to me how the relative density of the different kinds of zeroes---primes, 3 times primes, and other composites---changes as $$n$$ increases. It seems to be fairly constant, perhaps with the "other composites" class falling slightly, although I find the numbers hard to judge. If that is the case, the number of zeroes for $$n$$ prime should be an indication of how fast $$S(x)$$ increases. However, if the likelihood of a prime $$n$$ being a zero is proportional to $$1/\sqrt{n}$$ and the density of primes is $$~1/\ln n$$, this should make $$S(x)$$ increase proportional to $$\sqrt{x}/\ln x$$.

I've run computations up to $$n=5,000,000$$ using the same formula as provided by Greg Martin at MathOverflow which allows partial sums per divisor to be cached for efficient computation: went reasonably fast even in Sage/Python.

What seems clear is that the ratio $$S(x)/\sqrt{x}$$ starts to drop as $$x$$ increases past $$1,000,000$$: for $$x>2,500,000$$, the ratio seems to stay below $$0.5$$, and I suspect it will keep dropping.

• Thanks for the in-depth heuristics and I agree with your analysis. I have stopped computing after $n = 600,000$ so did not observed this fall after $n > 1,000,000$. In my Sagemath implementation it took be a few days to compute for $n = 600,000$ so your code is definitely faster. Would it be possible for you to share your code and the data so that I can run some computations for higher $n$. – Nilotpal Kanti Sinha Jul 31 at 4:47
• Using your heuristics, I fitted a curve of the form $\frac{a\sqrt{x}}{\log x}$ the data for $n \le 600000$ and the best fit within only this much data is with $a = 7.7569$ and the fit is only very slightly worse than the curve mentioned in the question. So I guess for larger values of $n$ an estimate of this form with a sharper value of $a$ would me more accurate. – Nilotpal Kanti Sinha Jul 31 at 5:48
• @NilotpalKantiSinha: Yes, the simple heuristic doesn't work very well. I suspect one reason is that $\mu$ isn't "random" enough, but instead cancels itself out more systematically. Realise I didn't really write much about that...will revisit my posting when I get a second look at the results. – Einar Rødland Jul 31 at 8:17