Considering the $\textit{Divisor Summatory Function}$, $D(n)$, defined as $$ D(n) = \sum_{k=1}^{n}d(k) , $$ where $$ d(n) = \sum_{k|n}^{n}1. $$

One can observe the following pattern in the values of $D(n)$,

$$ \lbrace{D(n)\rbrace}=\lbrace \overbrace{1,3,5,\;}^{3,odd}\overbrace{8,10,14,16,20,}^{5,even}\overbrace{23,27,29,35,37,41,45,}^{7,odd}\cdots \rbrace $$

where groups of odd elements alternate with groups of even elements and where the $n^{th}$ group has $2n-1$ elements, (we can see that the pattern pesists). Now, based on this (but it is not necessary that this pattern of $D(n)$ is verified for all $n$ we can only assume that such a similar pattern exists), and considering that any number can be written as $$ \begin{align*} n=p_{1}^{\alpha_{1}}\cdot p_{2}^{\alpha_{2}}\cdot p_{3}^{\alpha_{3}} \cdots p_{n}^{\alpha_{n}} \end{align*} $$ where $p_{i}$ are prime numbers, one can define the following arithmetical functions $$ a(n)= \begin{cases} 1, & \text{if all } D(\alpha_1), D(\alpha_2), \ldots, D(\alpha_n) \text{ are even} , \\\\ \\\\ 0, & \text{if one or more of the } D(\alpha_{i}) \text{ is odd}, \end{cases} $$ and $$b(n)\;=\; \begin{cases} (-1)^{n_1+n_2+\cdots+n_i}, &\text{if } \alpha_i = n_i^2, \\\\ \\\\ 0, & \text{if } \alpha_i \text{ is not of the form } n_i^2, \end{cases} $$ ($b(n)=$A197774) then we can define two Dirichlet series $A(s)=\sum_{k=1}^{\infty}\frac{a(k)}{n^{s}}$ with $a(1)=1$ and $B(s)=\sum_{k=1}^{\infty}\frac{b(k)}{n^{s}}$ with $b(1)=1$.

Both of the Dirichlet series have, respectively, the following Euler's products $$ A(s) = \prod_{p\in \mathbb{P}}\left(1+\frac{1}{p^{4s}}+\cdots+\frac{1}{p^{8s}}+\frac{1}{p^{16s}}+\cdots+\frac{1}{p^{24s}}+\cdots\right) $$ and $$ B(s) = \prod_{p\in \mathbb{P}}\left(1 - \frac{1}{p^{s}}+\frac{1}{p^{4s}}-\frac{1}{p^{9s}}+\frac{1}{p^{16s}}-\frac{1}{p^{25s}}+\frac{1}{p^{36s}}-\cdots\right) $$ First we can see that $A(s)$ is absolutely convergent for $s>\frac{1}{4}$. Secondly we can observe that $B(s)$ is related to $\vartheta_{4}(0,x)=1-x+x^{4}-x^{9}+x^{16}\cdots$ (the Jacobi Theta function) by $$ \begin{equation*} B(s) = \prod_{p\in \mathbb{P}}\left(\frac{1}{2} \vartheta_{4}(0,p^{-s})+1 \right) \end{equation*} $$ and thirdly $$ \begin{equation*} \zeta(s) = \frac{A(s)}{B(s)} \end{equation*} $$ We can think of $b(n)$ as a generalization of $\mu(n)$, the Möbius function and we can assume that if $B(s)$ converges for $\Re{s}>\frac{1}{2}$ then $\zeta(s)$ has no zeros on the right of $\frac{1}{2}$, just like the Mertens function, where $$ \begin{equation*} \frac{1}{\zeta(s)} = s\int_{1}^{\infty}\frac{M(x)}{x^{s+1}}dx \end{equation*} $$ similarly for $\zeta(s)$ we have $$ \begin{equation*} \zeta(s) = \frac{A(s)}{ s\int_{1}^{\infty}\frac{B(x)}{x^{s+1}}dx} \end{equation*} $$ where $M(x)$ is the Mertens function $$ \begin{equation*} M(x)=\sum_{1\leq n \leq x}\mu(x) \end{equation*} $$ and $B(x)$ is $$ \begin{equation*} B(x)=\sum_{1 \leq n \leq x}b(x) \end{equation*} $$ My question is: Were these Dirichlet series, $A(s)$ and $B(s)$, studied before and related to the $\zeta(s)$-function the way I did? Or... is this something new?


  • 9
    $\begingroup$ The pattern "3 odd, 5 even, 7 odd etc." is a simple consequence of the fact that $d(n)$ is odd if and only if $n$ is a square. $\endgroup$
    – GH from MO
    Nov 28 '11 at 16:55
  • 2
    $\begingroup$ Posted a while ago on MSE math.stackexchange.com/questions/73354/… $\endgroup$ Nov 28 '11 at 17:32
  • $\begingroup$ @Eric: Thanks, I linked my response below there. $\endgroup$
    – GH from MO
    Nov 28 '11 at 17:38
  • $\begingroup$ A.Neves, Can I get your email id to contact you by email? $\endgroup$ Apr 16 '21 at 17:22

If $B(s)$ is any Dirichlet series which factors as an Euler product $\prod_{p\in\mathbb{P}}B_p(s)$ satisfying $$ B_p(s)=1-\frac{1}{p^s}+O\left(\frac{1}{p^{2\sigma}}\right),\qquad s=\sigma+it, $$ then $B(s)=A(s)/\zeta(s)$, where $A(s)$ is given by an Euler product which is absolutely and locally uniformly convergent for $\Re(s)>1/2$. As a result, the poles of $B(s)$ are precisely the zeros of $\zeta(s)$ in $\Re(s)>1/2$, hence if the Dirichlet series $B(s)$ converges here, then the Riemann Hypothesis follows.

In fact it is not hard to relate directly the partial sums of the coefficients $b(n)$ of $B(s)$ to those of $\mu(n)$ and vice versa, since $b$ is a convolution $b=a\ast\mu$ where $a(n)$ is a sparse sequence, namely the coefficients of $A(s)$. In short, the exact shape of $b(n)$ is irrelevant as long as the above conditions on $B_p(s)$ are met, which in turn can be translated to: (1) the coefficients are multiplicative, (2) $b(p)=-1$ for $p$ prime, (3) $b(p^k)$ is not too large for $k>1$. Your example is one of many.


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