# Why would the roots of the generating functions of the number of k-almost primes less than x have negative real parts?

Specifically, I find it appealing to count only squarefree numbers having $k$ prime factors, so I define

$$\pi_k(x)=\#\{n\leq x: \omega(n)=k;\mu(n)\neq0 \}$$

and consider the generating functions

\begin{eqnarray}f(z,x)&=&\sum_{k=0}^{m(x)}\pi_k(x) z^k\\ &=&\sum_{n\leq x}|\mu(n)|z^{\omega(n)}. \end{eqnarray}

On one hand, these generating functions are polynomials in $z$ of degree $$m(x)=\max \{\omega(n): n\leq x;\mu(n)\neq 0\}\sim\log x/\log\log x.$$ On the other, they are the inverse Mellin transform of

$$F(z,s)=\prod_p1+zp^{-s}=H(z,s)\zeta^z(s)$$

where $H(z,s)$ is an analytic function of $s$ for fixed $z$ which is bounded above and away from zero in any half plane $\sigma\geq\sigma_0>1/2$.

I have checked that the roots of $f(z,x)$, as polynomials in $z$, do indeed have negative real parts for all $x\leq 10^6$.

Why would the roots have negative real parts? What would it say about $\zeta(s)$ or the numbers of $k$-almost primes less than $x$ if the roots were to have negative real parts?

On the Riemann hypothesis one would expect to find the roots---in some sense---closer to the non positive integers as $x\rightarrow\infty$ because these are the only values of $z$ for which $\zeta^z(s)$ is analytic throughout a neighbourhood of $s=1$ and, therefore, the only points at which $f(z,x)\in O(x^a)$ for some $a<1$. However, I am interested in the less-restrictive conjecture that they have negative real parts.

I am aware of the notion of 'stability' of linear translation invariant systems and in dynamical systems, and it's equivalence with the positivity of the principle minors of the associated Hurwitz matrices, Routh tables, Sylvester's criterion, etc. I find that these equivalences serve more as a test than to provide reasoning but, if you can say something in this regard, I would be pleased to hear about that too.

It appears this may be related to the fact that $\zeta^z(s)$ tends to infinity or zero as $s\rightarrow 0^+$ depending on whether $\Re z$ is positive or negative. However, for small $\Re z$ and $s=1$ the convergence is very slight, and making the distinction appears to be a tricky problem.

One can generalize this question and ask about the locations of the roots of $$f(z,x;s)=\sum_{n\leq x} \frac{|\mu(n)|z^{\omega(n)}}{n^s}.$$

Computationally, it appears that the same conjecture holds whenever $s$ is real and, moreover, there is a good reason for this to be true when $s>1$ is real because $$\lim_{x\rightarrow\infty}f(z,x;s)=F(z,s)$$ and this limit certainly does have it's zeros in the left half plane. One only needs then an approximation theorem on the distribution of the zeros of a uniformly convergent sequence whose limit has a prescribed distribution of zeros. One possible approach to the original problem might then be via partial summation $$f(z,x;s)=x^{-s}f(z,x)+s\int_{1}^{x}\frac{f(z,y)dy}{y^{s+1}}$$ but I have not had any success with this yet.

## 1 Answer

In any bounded region, for large $x$, the polynomial can only take on zeros very near the negative real axis (and indeed near the non-positive integers). This follows from the work of Selberg (Note on a paper by L.G. Sathe, see Theorem 2 there) which shows that, for bounded $z$ and large $x$, $$f(z,x) = x C(z) \frac{(\log x)^{z-1}}{\Gamma(z)} + O(x (\log x)^{z-2}),$$ with $$C(z) = \prod_{p} \Big(1+\frac{z}{p}\Big) \Big(1-\frac 1p\Big)^z.$$ If $z$ is bounded, but not too close to a non-positive integer, then the main term above dominates the error term, and shows that $f(z,x)$ cannot be zero.

Further note that $1/\Gamma(z)$ becomes zero at the non-positive integers, and the absolutely convergent Euler product $C(z)$ can only be zero when $z$ equals the negative of a prime number. Therefore, if $-k$ is a non-positive integer with $k$ not being prime, then we see from the asymptotics that $f(z,x)$ changes sign when passing from $z=-k+\epsilon$ to $z=-k-\epsilon$ producing a real zero in this interval. If $-k$ is a non-positive integer with $k$ a prime number, then take a small circle centered at $-k$ with radius $\epsilon$. Computing the change in argument around this circle for the main term in our asymptotic, we find that there are two zeros in this small circle. Note that these zeros may not be real, but just very close to the negative real axis.

Of course this asymptotic does not explain why all the zeros of the polynomial should have negative real part. I don't quite see why that holds; perhaps somewhat relevant to this would be the unimodality of the integers below $x$ with $k$ prime factors (see the work of Balazard establishing a conjecture of Erdos).

• Those results of Selberg and Deligne are incredible. I'd forgotten about them until you pointed out your observations here, which are in this new sense quite incredible too. I don't have copies of the papers at present but can you explain why this forces those roots in a bounded domain to be real for sufficiently large x? In answer to your question, non-real zeros do occur quite often in the range I've observed, but that doesn't count for much-the degree is still at most six. – Kevin Smith Feb 28 '15 at 23:44
• The librarian of the Indian mathematical society very kindly sent me a copy of Selberg's paper. I still don't see how the roots must be real for fixed $z$ and sufficiently large $x$. Would you explain why this is so please? – Kevin Smith Mar 4 '15 at 12:19
• @KevinSmith: I added some clarifications to the answer. – Lucia Mar 5 '15 at 5:30
• Thank you. I guess that the key to proving this is then finding enough control over the error term w.r.t the degree. – Kevin Smith Mar 5 '15 at 11:57
• Thank you. I guess that the key to proving this is then finding enough control over the error term w.r.t the degree. – Kevin Smith Mar 5 '15 at 11:57