Mertens Sums requiring Fewer Values of the M\"obius Function

Abstract

We discuss cetain identities
involving $\mu(n)$ and $M(x) = \sum _{n \leq x}\mu (n)$,
the functions of M\"obius and Mertens.
These allow calculation of $M(N^d)$,
for $d=1,2,3,\ldots\ $, as a sum of
$O_d \left( N^d(\log N)^{2d - 2}\right)$ terms, each a product of the
form $\mu(n_1) \cdots \mu(n_r)$ with $r\leq d$ and
$n_1, \ldots , n_r\leq N$.
We prove a more general identity in which
$M(N^d)$ is replaced by $M(g,K)=\sum_{n\leq K}\mu(n)g(n)$, where
$g(n)$ is an arbitrary totally multiplicative function, while
each $n_j$ has its own range of summation, $1,\ldots , N_j$.
This is not new, except perhaps in that
$N_1,\ldots , N_d$ are arbitrary, but our
proof (inspired by an identity of E.~Meissel, 1854) is new.
We are mainly interested in the case $d=2$, $K=N^2$, $N_1=N_2=N$,
where the identity has the form
$M(g, N^2) = 2 M(g,N) - {\bf m}^{\rm T} A {\bf m}$,
with $A$ being the $N\times N$~matrix of elements
$a_{mn}=\sum _{k \leq N^2 /(mn)}\,g(k)$, while
${\bf m}=(\mu (1)g(1),\ldots ,\mu (N)g(N))^{\rm T}$.
Our results in Sections~2 and~3 of the paper
assume that $g(n)$ equals $1$ for all $n$.
The Perron-Frobenius theorem applies in this case:
we find that $A$ has one large positive eigenvalue,
approximately~$(\pi^2 /6)N^2$,
with eigenvector approximately ${\bf f} = (1,1/2,1/3,\ldots ,1/N)^{\rm T}$,
and that, for large $N$, the
second-largest eigenvalue lies in $(-0.58 N, -0.49 N)$.
Section~2 includes estimates for the traces
of~$A$ and~$A^2$ (though, for ${\rm Tr}(A^2)$, we omit part of the proof).
In Section~3 we discuss ways to approximate ${\bf m}^{\rm T} A {\bf m}$,
using the spectral decomposition of $A$, or (alternatively) Perron's formula:
the latter
approach leads to a contour integral involving the Riemann zeta-function.
We also discuss using the identity
$A = N^{2\,} {\bf f}^{\,} \!{\bf f}^T -
\textstyle{1\over 2} {\bf u} {\bf u}^T + Z$,
where ${\bf u} = (1,\ldots ,1)^{\rm T}$ and $Z$ is the $N\times N$ matrix
of elements $z_{mn} = - \psi(N^2 / (mn))$,
with $\psi(x)=x - \lfloor x\rfloor - \textstyle{1\over 2}$.