Abstract
If \(z\in {\mathbb {C}}\) and \(1\le n\) is a natural number then
where \(d_1=p_1^{r_1}\dots p_m^{r_m }\), \(d_2=q_1^{e_1}\dots q_i^{e_i }\) are the prime decompositions of \(d_1, d_2\). This is one of the identities involving arithmetic functions that we prove using ideas from the paper of Dressler and van de Lune [3].
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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
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
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)
and its associated Dirichlet series is
One has the bound
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
-
(a)
If \(1<\sigma =\mathfrak {R}s, |z|\le 1\) then
$$\begin{aligned} \left( \sum _{n=1}^\infty \frac{ (1-z^{p_1})\cdots (1-z^{p_m})}{n^s}\right) \left( \sum _{n=1}^\infty \frac{ z^{p_1 r_{1}+\cdots +p_m r_{m} }}{n^s}\right) =\zeta (s). \end{aligned}$$ -
(b)
If \(1<\sigma , |z|\le 1\) then
$$\begin{aligned} \sum _{n=1}^\infty \frac{ z^{p_1 r_{1}+\cdots +p_m r_{m} }}{n^s} =\zeta (s)\sum _{n=1}^\infty \frac{ (z^{p_1}-1)\cdots (z^{p_m}-1) z^{(r_1-1)p_1+\cdots +(r_m-1)p_m}}{n^s}. \end{aligned}$$ -
(c)
If \(0<|z|<2^\sigma\) and \(1<\sigma\) then
$$\begin{aligned} \sum _{n=1}^\infty \frac{ z^{\Omega (n)-\omega (n)}(1+z)^{\omega (n)}}{n^s}=\frac{\zeta (s)}{\zeta (2s)}\left( \sum _{n=1}^\infty \frac{ z^{\Omega (n)}}{n^s}\right) . \end{aligned}$$ -
(d)
If \(z\in {\mathbb {C}}\) and \(1<\sigma\) then
$$\begin{aligned} \sum _{n=1}^\infty \frac{(z+2)^{\omega (n)}}{n^s}=\frac{\zeta (s)^2}{\zeta (2s)}\left( \sum _{n=1}^\infty \frac{(-1)^{\Omega (n)-\omega (n)}z^{\omega (n)}}{n^s}\right) . \end{aligned}$$ -
(e)
If \(0<|z|\le 1\) and \(1<\sigma\)then
$$\begin{aligned} \sum _{n=1}^\infty \frac{ z^{\Omega (n)-\omega (n)}(z-1/p_1)\cdots (z-1/p_m)}{n^s}=\frac{\zeta (s)}{\zeta (s+1)}\left( \sum _{n=1}^\infty \frac{ z^{\Omega (n)-\omega (n)}(z-1)^{\omega (n)}}{n^s}\right) . \end{aligned}$$ -
(f)
If \(0<|z|\le 1\) and \(1<\sigma\) then
$$\begin{aligned} \frac{\zeta (s)\zeta (2s)\zeta (3s)}{\zeta (6s)} &= \left( \sum _{n=1}^\infty \frac{(-1)^{\omega _o(n)}(1+z)^{\omega _e(n)}z^{\sum _{i=1}^m\left\lfloor \frac{r_i-1}{2}\right\rfloor }}{n^s}\right) \\ &\qquad \times \left( \sum _{n=1}^\infty \frac{\left\{ (r_1+1)-(r_1-1)z \right\} \cdots \left\{ (r_m+1)-(r_m-1)z\right\} }{n^s}\right). \end{aligned}$$ -
(g)
If \(1<\mathfrak {R}\lambda\) and \(1<\sigma\) then
$$\begin{aligned} \sum _{n=1}^\infty \frac{ 1}{rad(n)^\lambda n^s}=\zeta (s)\left\{ \sum _{n=1}^\infty \frac{ \mu (n)}{n^s}\left( 1-\frac{1}{p_1^\lambda }\right) \cdots \left( 1-\frac{1}{p_m^\lambda }\right) \right\} . \end{aligned}$$ -
(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
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
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
-
(i)
$$\begin{aligned} z^{\Omega (n)-\omega (n)}(1+z)^{\omega (n)}=\sum _{d|n} |\mu (n/d)|z^{\Omega (d)}. \end{aligned}$$
-
(ii)
$$\begin{aligned} (z+2)^{\omega (n)}=\sum _{d|n} 2^{\omega (n/d)}(-1)^{\Omega (d)}(-z)^{\omega (d)}. \end{aligned}$$
-
(iii)
$$\begin{aligned} n(z-1/p_1)\cdots (z-1/p_m)z^{\Omega (n)-\omega (n)}=\sum _{d|n} d\,\, \phi (n/d) z^{\Omega (d)-\omega (d)}(z-1)^{\omega (d)}. \end{aligned}$$
-
(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:
If we set \(g(p)=z\), \(z=1\) in the first equation and \(g(p)=z\) in the second equation one gets:
Using this two formulas in
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
The product of (7) and (8) is equal to \(\zeta (s)\). This proves formula (a) of Theorem 2.
Also
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
Also
The product of this two formulas yields formula (g).
In a similar way using (3) one has that
which can be used to prove that
which is formula (d).
Also
which is formula (e).
Formula (c) follows from (using (9) and (6))
To prove formula (f) observe that after simplification using \(\zeta (s)=\prod _{p}\frac{p^s}{p^s-1}\) one has
Formula (f) follows multiplying
with
Next we prove formula (h). One has the following formal formula
Setting \(g(p)=\epsilon _p p^{11/2}\) with \(|\epsilon _p|\le 1\) in the last formula one gets
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
If one sets
then
which follows using (1). The proof is complete.
References
Apostol, T.: Introduction to Analytic Number Theory. Springer, New York (1976)
Apostol, T.: Modular Functions and Dirichlet Series in Number Theory. Springer, New York (1976)
Dressler, R.E., van de Lune, J.: Some remarks concerning the number theoretic functions \(\omega (n)\) and \(\Omega (n)\). Proc. Am. Math. Soc. 41(2), 403–406 (1973)
Titchmarsh, E.C.: The Theory of the Riemann Zeta-Function, 2nd edn. Oxford Univ. Press, New York (1986). (revised by D. R. Heath-Brown)
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Panzone, P.A. On a paper of Dressler and Van de Lune. Bol. Soc. Mat. Mex. 26, 831–839 (2020). https://doi.org/10.1007/s40590-020-00278-z
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DOI: https://doi.org/10.1007/s40590-020-00278-z