Let $d_k(n)$ denote the number of ways of expressing $n$ as a product of $k$ factors, and let $$D_k(x)=\sum_{n\leq x}d_k(n)$$ be the summatory function. During a study of Mertens' function I was lead to consider the zero distribution of the sequence of polynomials $$P(z,x)=\sum_{j=0}^{m}\left(\sum_{k=j}^{m}{m+1\choose k-j}(-1)^{m-k}D_{m+1-k}(x)\right)z^j$$ of degree $m=m(x)=\max \{\Omega(n):n\leq x\}\sim \log_2 x$.

I am interested in the question of whether the zeros of the sequence of polynomials $P(z,x)$ have strictly negative real parts-the truth of which depends, among other things, on the non-negativity of the coefficients $$c(j,x)=\sum_{k=j}^{m}{m+1\choose k-j}(-1)^{m-k}D_{m+1-k}(x).$$

I have checked this up to $x=50$ and for each such $x$ the $c(j,x)$ are positive and unimodal. For instance, for $1\leq x \leq 20$, the coefficients are: [1], [1, 2], [1, 3], [1, 4, 4], [1, 5, 5], [1, 4, 6], [1, 5, 7], [1, 6, 12, 8], [1, 6, 13, 9], [1, 5, 13, 10], [1, 6, 15, 11], [1, 6, 13, 12], [1, 7, 15, 13], [1, 6, 15, 14], [1, 5, 15, 15], [1, 6, 20, 30, 16], [1, 7, 23, 33, 17], [1, 7, 21, 32, 18], [1, 8, 24, 35, 19], [1, 8, 22, 34, 20].

Is there a good reason why these numbers should be positive? I know this would follow if it could be shown that the sequence $${m(x)+1\choose k}D_{m(x)-j+1-k}(x)$$ is unimodal for $k\in\{0,m(x)-j\}$, but I don't see how this could be proved.

  • $\begingroup$ What is $\Omega(n)$? $\endgroup$ – Wolfgang Jan 25 '15 at 18:14
  • $\begingroup$ $\Omega(n)$ is the total number of prime factors of $n$, counting multiplicities. $\endgroup$ – Kevin Smith Jan 25 '15 at 18:17

The answer is no, on both counts. Certainly I was hasty to post this question based on the observation that the zeros have negative real parts and the coefficients are positive in the observed ranges. Here's a simple explanation of why:

First note that $D_k(x)$ may be written as

$$D_k(x)=\sum_{n\leq x}\prod_{p|n}{\alpha_p+k-1\choose \alpha_p}$$

where $\alpha_p$ is the exponent of $p$ in the canonical factorisation of $n$. $D_k(x)$ is thus a polynomial in $k$ of degree $m(x)$, so

$$\sum_{k=0}^{m+1}{m+1\choose k}(-1)^kD_{m+1-k-j}(x)=0.$$ Therefore $$c(j,x)=-\sum_{k=m-j+1}^{m+1}{m+1 \choose k} (-1)^{m-k-j}D_{m+1-k-j}(x).$$ For example then, $$c(1,x)=m(x)+1-D_{-1}(x)$$ where $D_{-1}(x)=M(x)$ is Mertens' function. It follows from known results that $M(x)$ is $\Omega_{\pm}(x^{1/2})$, so $c(1,x)$ will eventually be negative of square root order. As for the zeros, by virtue of the fact that the coefficients are real and $c(0,x)=1$, we have $$c(1,x)=-\sum_{\rho:P(\rho,x)=0}\frac{1}{\rho}=-\sum_{\rho:P(\rho,x)=0}\frac{\Re{\rho}}{|\rho|^2},$$ so the same argument applies.

On the other hand, from the linear relation above, it is easy to see that $c(m(x),x)=x$, so it is certainly not so obvious that $c(m(x)-1,x)$, $c(m(x)-2,x)$, etc, will also change sign eventually. These questions are tied up with the Riemann hypothesis.


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