1 Introduction and results

Recall that the arithmetic function \(\omega (n)\) is the number of distinct prime divisors of a positive integer n and \(\Omega (n)\) is the total number of prime divisors of n. In other words if for a natural number \(n\ge 2\) we write \(n=p_1^{r_1}\dots p_m^{r_m }\) with \(p_i\) distinct primes (a notation that we keep throughout this paper where p always denotes a prime) then \(\omega (n)=m\), \(\Omega (n)=r_1+\cdots +r_m\) and \(\Omega (1)=\omega (1)=0\). Write \(\zeta (s)\) for the Riemann–Zeta function.

One knows that \(\omega (n)=O(\log n/ \log \log n)\) which easily implies that \(\sum _{n=1}^\infty z^{\omega (n)}/n^s\) is an entire function of z when \(\mathfrak {R}s =\sigma >1\). Also

$$\begin{aligned} \sum _{n=1}^\infty \frac{z^{\Omega (n)}}{n^s}=\prod _p \left( 1+\frac{z}{p^s}+\frac{z^2}{p^{2s}}+\cdots \right) =\prod _p\frac{1}{1-z/p^s}, \end{aligned}$$

is an analytic function of z if \(1<\sigma\) and \(|z|< 2^\sigma\).

In [3] the following remarkable duality relation was proved.

Theorem 1

(R. Dressler and J. van de Lune) If \(|z|<2^\sigma\)and \(\mathfrak {R}s=\sigma >1\)then

$$\begin{aligned} \left( \sum _{n=1}^\infty \frac{(1- z)^{\omega (n)}}{n^s}\right) \left( \sum _{n=1}^\infty \frac{z^{\Omega (n)}}{n^s}\right) =\zeta (s). \end{aligned}$$

The aim of this note is to obtain similar formulas using the methods of [3].

To state our results we need some definitions: let \(\omega _o(n)\) (\(\omega _e(n)\) respectively) be the number of primes in the decomposition of n with odd (even respectively) exponent. Thus \(\omega _o(2^23^5 5^6)=1,\, \omega _e(2^23^5 5^6)=2\). Note that \(\omega _e(n)+\omega _o(n)=\omega (n)\). \(\left\lfloor x\right\rfloor\) is the floor function and \(\mu\) is the Möbius function. The radical of a number n is defined as \(rad(n)=p_1\cdots p_m\).

Ramanujan’s tau function is defined by (see [2], p. 136)

$$\begin{aligned} z\prod _{n=1}^\infty (1-z^n)^{24}=\sum _{n=1}^{\infty }\tau (n) z^n, \end{aligned}$$

and its associated Dirichlet series is

$$\begin{aligned} \sum _{n=1}^{\infty } \frac{\tau (n)}{n^s}=\prod _p \frac{1}{1-\tau (p)p^{-s}+p^{11-2s}}. \end{aligned}$$
(1)

One has the bound

$$\begin{aligned} |\tau (p)|\le 2 p^{11/2}. \end{aligned}$$
(2)

This result was conjectured by Ramanujan and it was proved by Deligne [2].

The main contribution of this note is the following theorem. Note: in formulas (a)–(g) below it is assumed that in all the sums the term with \(n=1\) is equal to 1.

Theorem 2

  1. (a)

    If \(1<\sigma =\mathfrak {R}s, |z|\le 1\) then

    $$\begin{aligned} \left( \sum _{n=1}^\infty \frac{ (1-z^{p_1})\cdots (1-z^{p_m})}{n^s}\right) \left( \sum _{n=1}^\infty \frac{ z^{p_1 r_{1}+\cdots +p_m r_{m} }}{n^s}\right) =\zeta (s). \end{aligned}$$
  2. (b)

    If \(1<\sigma , |z|\le 1\) then

    $$\begin{aligned} \sum _{n=1}^\infty \frac{ z^{p_1 r_{1}+\cdots +p_m r_{m} }}{n^s} =\zeta (s)\sum _{n=1}^\infty \frac{ (z^{p_1}-1)\cdots (z^{p_m}-1) z^{(r_1-1)p_1+\cdots +(r_m-1)p_m}}{n^s}. \end{aligned}$$
  3. (c)

    If \(0<|z|<2^\sigma\) and \(1<\sigma\) then

    $$\begin{aligned} \sum _{n=1}^\infty \frac{ z^{\Omega (n)-\omega (n)}(1+z)^{\omega (n)}}{n^s}=\frac{\zeta (s)}{\zeta (2s)}\left( \sum _{n=1}^\infty \frac{ z^{\Omega (n)}}{n^s}\right) . \end{aligned}$$
  4. (d)

    If \(z\in {\mathbb {C}}\) and \(1<\sigma\) then

    $$\begin{aligned} \sum _{n=1}^\infty \frac{(z+2)^{\omega (n)}}{n^s}=\frac{\zeta (s)^2}{\zeta (2s)}\left( \sum _{n=1}^\infty \frac{(-1)^{\Omega (n)-\omega (n)}z^{\omega (n)}}{n^s}\right) . \end{aligned}$$
  5. (e)

    If \(0<|z|\le 1\) and \(1<\sigma\)then

    $$\begin{aligned} \sum _{n=1}^\infty \frac{ z^{\Omega (n)-\omega (n)}(z-1/p_1)\cdots (z-1/p_m)}{n^s}=\frac{\zeta (s)}{\zeta (s+1)}\left( \sum _{n=1}^\infty \frac{ z^{\Omega (n)-\omega (n)}(z-1)^{\omega (n)}}{n^s}\right) . \end{aligned}$$
  6. (f)

    If \(0<|z|\le 1\) and \(1<\sigma\) then

    $$\begin{aligned} \frac{\zeta (s)\zeta (2s)\zeta (3s)}{\zeta (6s)} &= \left( \sum _{n=1}^\infty \frac{(-1)^{\omega _o(n)}(1+z)^{\omega _e(n)}z^{\sum _{i=1}^m\left\lfloor \frac{r_i-1}{2}\right\rfloor }}{n^s}\right) \\ &\qquad \times \left( \sum _{n=1}^\infty \frac{\left\{ (r_1+1)-(r_1-1)z \right\} \cdots \left\{ (r_m+1)-(r_m-1)z\right\} }{n^s}\right). \end{aligned}$$
  7. (g)

    If \(1<\mathfrak {R}\lambda\) and \(1<\sigma\) then

    $$\begin{aligned} \sum _{n=1}^\infty \frac{ 1}{rad(n)^\lambda n^s}=\zeta (s)\left\{ \sum _{n=1}^\infty \frac{ \mu (n)}{n^s}\left( 1-\frac{1}{p_1^\lambda }\right) \cdots \left( 1-\frac{1}{p_m^\lambda }\right) \right\} . \end{aligned}$$
  8. (h)

    Assume that \(\left\{ \epsilon _p\right\}\)is any sequence of complex numbers defined on the set of primes with \(|\epsilon _p|\le 1\)for all p. Write \(n=n_1 n_2\), where \(n_1=p_1\ldots p_t\), \(n_2=p_{t+1}^{r_{t+1}}\ldots p_m^{r_m}\)with \(2\le \min \left\{ r_{t+1},\ldots ,r_m \right\}\)and define

    $$\begin{aligned} a_n&:=\prod _1 \prod _2,\\ \prod _1&:=\prod _{p\in n_1} \left\{ \epsilon _{p} p^{11/2}-\tau (p)\right\} ,\\ \prod _2&:=n_2^{11/2}rad(n_2)^{-11} \prod _{k=t+1}^m \left\{ \left( 1+\epsilon _{p_k}^2\right) p_k ^{11}-\epsilon _{p_k}p_k^{11/2}\tau (p_k)\right\} \epsilon _{p_{k}}^{r_{k}-2}. \\ A(s)&:=\sum _{n=1}^\infty \frac{a_n}{n^s},\\ B(s)&:=\sum _{n=1}^\infty \frac{\mu (n) \epsilon _{p_1}\cdots \epsilon _{p_m} }{n^{s-11/2}}. \end{aligned}$$

    Then if \(13/2< \sigma\)

    $$\begin{aligned} \left( \sum _{n=1}^{\infty } \frac{\tau (n)}{n^s}\right) ^{-1}=A(s)B(s). \end{aligned}$$

    Note: it is understood that the first summand is 1 in any of the last two sums. Also \(\prod _i=1\) if \(n_i=1\).

We remark that the formula in the abstract follows from formula (a) where it is understood there that \((1-z^{p_1})\cdots (1-z^{p_m})\equiv 1\) if \(d_1=1\) and \(z^{q_1 e_{1}+\cdots +q_i e_{i}}\equiv 1\) if \(d_2=1\).

Observe that formula (b) yields, on setting \(z=0\), the well-known relation

$$\begin{aligned} 1=\zeta (s)\sum _{n=1}^\infty \frac{\mu (n)}{n^s}. \end{aligned}$$

Also, the above zeta quotients appearing in formulas (c)–(f) are well-known: if \(\phi (n)\) is the number of numbers less than n and prime to n and the arithmetic function \(\kappa (n)\) is defined by \(\kappa (1)=1\) and \(\kappa (p_1^{r_1}\ldots p_m^{r_m})=r_1\ldots r_m\) then

$$\begin{aligned} \frac{\zeta (s)^2}{\zeta (2s)}&=\sum _{n=1}^\infty \frac{2^{\omega (n)}}{n^s},\\ \frac{\zeta (s)}{\zeta (2s)}&=\sum _{n=1}^\infty \frac{|\mu (n)|}{n^s},\\ \frac{\zeta (s)\zeta (2s)\zeta (3s)}{\zeta (6s)}&=\sum _{n=1}^\infty \frac{\kappa (n)}{n^s},\\ \frac{\zeta (s)}{\zeta (s+1)}&=\sum _{n=1}^\infty \frac{\phi (n)}{n^{s+1}}. \end{aligned}$$

See [1] p. 247, [4] formulas (1,2,7), (1.2.8) and (1.2.12).

This permits, using Dirichlet convolution, to obtain the following corollary.

Corollary 1

If \(z\in {\mathbb {C}}\) in (i), (ii) and (iii) then

  1. (i)
    $$\begin{aligned} z^{\Omega (n)-\omega (n)}(1+z)^{\omega (n)}=\sum _{d|n} |\mu (n/d)|z^{\Omega (d)}. \end{aligned}$$
  2. (ii)
    $$\begin{aligned} (z+2)^{\omega (n)}=\sum _{d|n} 2^{\omega (n/d)}(-1)^{\Omega (d)}(-z)^{\omega (d)}. \end{aligned}$$
  3. (iii)
    $$\begin{aligned} n(z-1/p_1)\cdots (z-1/p_m)z^{\Omega (n)-\omega (n)}=\sum _{d|n} d\,\, \phi (n/d) z^{\Omega (d)-\omega (d)}(z-1)^{\omega (d)}. \end{aligned}$$
  4. (iv)

    If \(\lambda \in {\mathbb {C}}\)then

    $$\begin{aligned} rad(n)^{\lambda }=\sum _{d|n}\mu (d)\left( 1-q_1^\lambda \right) \cdots \left( 1-q_i^\lambda \right) . \end{aligned}$$

    (Note: here \(d=q_1^{e_1}\cdots q_i^{e_i}\) is the prime decomposition of d and if \(d=1\) in the last sum then summand is understood to be 1.)

Proof

Formulas (i), (ii), (iii), (iv) follow from formulas (c), (d), (e), (g) respectively and analytic continuation. \(\square\)

2 Proof

Recall that if \(2\le n\) is an integer then we write \(n=p_1^{r_1}\cdots p_m^{r_m}\) with \(p_i\) different primes. We record the following formal formulas:

$$\begin{aligned} \prod _{p}\left( 1+\frac{g(p)}{p^s-z}\right)= & {} \prod _{p}\left( 1+\frac{g(p)}{p^s}+\frac{z g(p)}{p^{2s}}+\frac{z^2 g(p)}{p^{3s}}+\cdots \right) \nonumber \\= & {} \sum _{n=1}^\infty \frac{ z^{\Omega (n)-\omega (n)}g(p_1)\cdots g(p_m)}{n^s}, \end{aligned}$$
(3)
$$\begin{aligned} \prod _{p}\frac{p^s}{p^s-g(p)}= & {} \prod _{p}\left( 1+\frac{g(p)}{p^s}+\frac{g(p)^2}{p^{2s}}+\frac{g(p)^3}{p^{3s}}+\cdots \right) \nonumber \\= & {} \sum _{n=1}^\infty \frac{ g(p_1)^{r_{1}}\cdots g(p_m)^{r_{m}}}{n^s}. \end{aligned}$$
(4)

If we set \(g(p)=z\), \(z=1\) in the first equation and \(g(p)=z\) in the second equation one gets:

$$\begin{aligned}&\prod _{p}\left( 1+\frac{z}{p^s-1}\right) =\sum _{n=1}^\infty \frac{ z^{\omega (n)}}{n^s}, \end{aligned}$$
(5)
$$\begin{aligned}&\prod _{p}\frac{p^s}{p^s-z}=\sum _{n=1}^\infty \frac{ z^{\Omega (n)}}{n^s}. \end{aligned}$$
(6)

Using this two formulas in

$$\begin{aligned} \zeta (s)=\prod _{p}\frac{p^s}{p^s-1}=\prod _{p}\left( \frac{p^s}{p^s-z}\right) \left( 1+\frac{1-z}{p^s-1}\right) \end{aligned}$$

yields Theorem 1.

The values of z and \(\sigma\) depend on each formula in such way that the products or sums involved are absolutely convergent. For example assume that \(|z|\le 1\) and \(1<\sigma\). Then if \(g(p)=1-z^p\), \(z=1\) in (3) and \(g(p)=z^p\) in (4) one has

$$\begin{aligned}&\prod _{p}\left( 1+\frac{1-z^p}{p^s-1}\right) =\sum _{n=1}^\infty \frac{ (1-z^{p_1})\cdots (1-z^{p_m})}{n^s}, \end{aligned}$$
(7)
$$\begin{aligned}&\prod _{p}\frac{p^s}{p^s-z^p}=\sum _{n=1}^\infty \frac{ z^{p_1 r_{1}+\cdots +p_m r_{m} }}{n^s}. \end{aligned}$$
(8)

The product of (7) and (8) is equal to \(\zeta (s)\). This proves formula (a) of Theorem 2.

Also

$$\begin{aligned} \begin{aligned} \prod _{p}\left( \frac{p^s-1}{p^s-z^p}\right)&=\prod _{p}\left( \left\{ 1-\frac{1}{p^s}\right\} \left\{ 1+\frac{z^p}{p^s}+\frac{z^{2p}}{p^{2s}}+\cdots \right\} \right) \\&=\sum _{n=1}^\infty \frac{ (z^{p_1}-1)\cdots (z^{p_m}-1) z^{(r_1-1)p_1+\cdots +(r_m-1)p_m}}{n^s}. \end{aligned} \end{aligned}$$

Formula (b) follows using (8) and observing that \(\prod _{p}\frac{p^s}{p^s-z^p}\left( \frac{p^s-1}{p^s-z^p}\right) ^{-1}=\zeta (s)\).

If \(g(p)=1/p^\lambda\), \(z=1\) in (3) then

$$\begin{aligned} \begin{aligned} \prod _{p}\left( 1+\frac{1/p^\lambda }{p^s-1}\right) =\sum _{n=1}^\infty \frac{ 1}{rad(n)^\lambda n^s}. \end{aligned} \end{aligned}$$

Also

$$\begin{aligned} \begin{aligned} \prod _{p}\left( \frac{p^s-1+1/p^\lambda }{p^s}\right) ^{-1}&=\prod _{p}\left( 1-\frac{1}{p^s}\left\{ 1-\frac{1}{p^\lambda }\right\} \right) ^{-1}\\ {}&=\left\{ \sum _{n=1}^\infty \frac{ \mu (n)}{n^s}\left( 1-\frac{1}{p_1^\lambda }\right) \cdots \left( 1-\frac{1}{p_m^\lambda }\right) \right\} ^{-1}. \end{aligned} \end{aligned}$$

The product of this two formulas yields formula (g).

In a similar way using (3) one has that

$$\begin{aligned} \prod _{p}\left( 1+\frac{z}{p^s-u}\right) =\sum _{n=1}^\infty \frac{ u^{\Omega (n)-\omega (n)}z^{\omega (n)}}{n^s}, \end{aligned}$$
(9)

which can be used to prove that

$$\begin{aligned} \frac{\zeta (s)^2}{\zeta (2s)}&=\prod _{p}\frac{p^s+1}{p^s-1}=\prod _{p}\frac{p^s+z+1}{p^s-1}\frac{p^s+1}{p^s+z+1}\\&=\prod _{p}\left( 1+\frac{z+2}{p^s-1}\right) \left( 1+\frac{z}{p^s+1}\right) ^{-1}\\&=\left( \sum _{n=1}^\infty \frac{ (z+2)^{\omega (n)}}{n^s}\right) \left( \sum _{n=1}^\infty \frac{ (-1)^{\Omega (n)-\omega (n)}z^{\omega (n)}}{n^s}\right) ^{-1}, \end{aligned}$$

which is formula (d).

Also

$$\begin{aligned} \frac{\zeta (s)}{\zeta (s+1)}&= {} \prod _{p}\frac{p^s-1/p}{p^s-1}=\prod _{p}\left( \frac{p^s-1}{p^s-z}\right) ^{-1}\left( \frac{p^s-1/p}{p^s-z}\right) \\&= {} \left( \sum _{n=1}^\infty \frac{ z^{\Omega (n)-\omega (n)}(z-1)^{\omega (n)}}{n^s}\right) ^{-1}\\&\quad\times \left( \sum _{n=1}^\infty \frac{ z^{\Omega (n)-\omega (n)}(z-1/p_1)\cdots (z-1/p_m)}{n^s}\right) , \end{aligned}$$

which is formula (e).

Formula (c) follows from (using (9) and (6))

$$\begin{aligned} \frac{\zeta (s)}{\zeta (2s)}&=\prod _{p}\frac{p^s+1}{p^s}=\prod _{p}\left( 1+\frac{1+z}{p^s-z}\right) \left( \frac{p^s}{p^s-z}\right) ^{-1}\\&=\left( \sum _{n=1}^\infty \frac{ z^{\Omega (n)-\omega (n)}(1+z)^{\omega (n)}}{n^s}\right) \left( \sum _{n=1}^\infty \frac{ z^{\Omega (n)}}{n^s}\right) ^{-1}. \end{aligned}$$

To prove formula (f) observe that after simplification using \(\zeta (s)=\prod _{p}\frac{p^s}{p^s-1}\) one has

$$\begin{aligned} \frac{\zeta (s)\zeta (2s)\zeta (3s)}{\zeta (6s)}=\prod _{p}\frac{p^{2s}-p^s+1}{(p^s-1)^2}. \end{aligned}$$

Formula (f) follows multiplying

$$\begin{aligned}&\prod _{p}\frac{p^{2s}-p^s+1}{p^{2s}-z}=\prod _{p}\left( 1-\frac{1}{p^s}+\frac{1}{p^{2s}}\right) \left( 1+\frac{z}{p^{2s}}+\frac{z^2}{p^{4s}}+\cdots \right) \\&\quad =\prod _{p}\left( 1-\frac{1}{p^{s}}-\frac{z}{p^{3s}}-\frac{z^2}{p^{5s}}-\cdots +\frac{(1+z)}{p^{2s}}+\frac{z(1+z)}{p^{4s}}+\frac{z^2(1+z)}{p^{6s}}+\cdots \right) \\&\quad =\sum _{n=1}^\infty \frac{(-1)^{\omega _o(n)}(1+z)^{\omega _e(n)}z^{\sum _{i=1}^m\left\lfloor \frac{r_i-1}{2}\right\rfloor }}{n^s}, \end{aligned}$$

with

$$\begin{aligned} \prod _{p}\frac{p^{2s}-z}{(p^s-1)^2}&=\prod _{p}\left( 1-\frac{z}{p^{2s}}\right) \left( 1+\frac{2}{p^{s}}+\frac{3}{p^{2s}}+\cdots \right) \\&=\prod _{p} \left( 1+\frac{2}{p^{s}}+\frac{3-z}{p^{2s}}+\frac{4-2z}{p^{3s}}+\cdots \right) \\&=\sum _{n=1}^\infty \frac{\left\{ (r_1+1)-(r_1-1)z \right\} \cdots \left\{ (r_m+1)-(r_m-1)z\right\} }{n^s}. \end{aligned}$$

Next we prove formula (h). One has the following formal formula

$$\begin{aligned} \prod _{p}&\frac{p^s-\tau (p)+p^{11}/p^s}{p^s-g(p)} =\prod _{p}\left( 1-\frac{\tau (p)}{p^s}+\frac{p^{11}}{p^{2s}}\right) \left( 1+\frac{g(p)}{p^s}+\frac{g(p)^2}{p^{2s}}+\cdots \right) \\&=\prod _{p}\left\{ 1+\frac{g(p)-\tau (p)}{p^s}+\sum _{j=2}^\infty \frac{g(p)^{j-2} \left\{ g(p)^2-g(p)\tau (p)+p^{11}\right\} }{p^{js}}\right\} \\&=\sum _{n=1}^\infty \frac{\prod _{k=1; r_k=1}^{m} \left\{ g(p_k)-\tau (p_k)\right\} \prod _{k=1; r_k\ge 2}^{m} \left\{ g(p_k)^2-g(p_k)\tau (p_k)+p_k^{11}\right\} g(p_k)^{r_k-2}}{n^s}. \end{aligned}$$

Setting \(g(p)=\epsilon _p p^{11/2}\) with \(|\epsilon _p|\le 1\) in the last formula one gets

$$\begin{aligned} A(s):=\prod _{p}&\frac{p^s-\tau (p)+p^{11}/p^s}{p^s-\epsilon _p p^{11/2}} =\sum _{n=1}^\infty \frac{a_n}{n^s}, \end{aligned}$$

where \(a_n\) is defined as follows. To ease the notation we write \(n=n_1 n_2\), where \(n_i\) are positive integers such that \(n_1=p_1\ldots p_t\), \(n_2=p_{t+1}^{r_{t+1}}\ldots p_m^{r_m}\) with \(2\le \min \left\{ r_{t+1},\ldots ,r_m \right\}\). Then

$$\begin{aligned} a_n&:=\prod _1 \prod _2,\\ \prod _1&:=\prod _{p\in n_1} \left\{ \epsilon _{p} p^{11/2}-\tau (p)\right\} ,\\ \prod _2&:=\prod _{k=t+1}^m \left\{ (1+\epsilon _{p_k}^2) p_k ^{11}-\epsilon _{p_k}p_k^{11/2}\tau (p_k)\right\} \epsilon _{p_{k}}^{r_{k}-2} p_{k}^{\frac{11}{2}(r_{k}-2)}\\ &\,=n_2^{11/2}rad(n_2)^{-11}\prod _{k=t+1}^m \left\{ (1+\epsilon _{p_k}^2) p_k ^{11}-\epsilon _{p_k}p_k^{11/2}\tau (p_k)\right\} \epsilon _{p_{k}}^{r_{k}-2}. \end{aligned}$$

If one sets

$$\begin{aligned} B(s):=\prod _{p}\frac{p^s-\epsilon _p p^{11/2}}{p^s}=\sum _{n=1}^\infty \frac{\mu (n) \epsilon _{p_1}\cdots \epsilon _{p_m} }{n^{s-11/2}}, \end{aligned}$$

then

$$\begin{aligned} A(s)B(s)=\left( \sum _{n=1}^{\infty } \frac{\tau (n)}{n^s}\right) ^{-1}, \end{aligned}$$

which follows using (1). The proof is complete.