In this post when $n>1$ we assume that $n=\prod_{k=1}^{\omega(n)}p_{n}^{e_{k}}$ its factorization in prime powers, where $\omega (n)$ is the number of distinct primes. It is well known the definition of Möbius function and its role related with Dirichlet product of aritmetical functions, provide us the inversion formula. In this post I (copy) define another modified Möbius function distinct to Peter Brown function, (see [1] for Peter's definition). Then I define a modified Möbius function as $m(1)=1$ and for $n>1$, $m(n)=(-i)^{\omega (n)}$ if $e_{1}=e_{2}=\cdots =e_{\omega(n)}=1$, and zero in other case (compare with Peter's definition). In [1] Peter leave to the reader the proof of the following theorem, that I prove with my definition, after I ask to me about another modified number theoretical functions using another definition (as Peter used).
Theorem (Peter G. Brown) i) The function $m$ is multiplicative but not completly myltiplicative. ii) Has Dirichlet inverse as $m^{-1}(n)=i^{\Omega (n)}$, where $\Omega (n)$ is the number of distinct prime factors counted with multiplicity, and $m^{-1}(1)=1$.
Proof. i) Let $a, b >1$ with $gcd(a,b)=1$,
if some $a$ or $b$ are divisible by a square then $square|ab$ too, thus $m(a)m(b)=0=m(ab)$;
if neither $a$ nor $b$ are divisible by a square then $m(a)m(b)=(-i)^{\omega (a)+\omega(b)}=m(ab)$, since are coprime.
Thus $m$ is multiplicative, but since $0=m(12)$ is distinct to $m(6)m(2)=(-i)^{2}(-i)=(-i)^{3}$, $m$ isn't completly multiplicative.
ii) Following page 30 and 31 from [2], $m^{-1}(1)=1\neq 0$ implies there exists an unique arithmetical function, the so called inverse with respect Dirichlet product, in this product since if both $f$ and $g\star f$ are multiplicative functions then $g$ is also multiplicative. Thus I compute for a prime $p$ and an integer $k\geq 1$
$$m^{-1}(p^{k})=\frac{-1}{m(1)}\sum_{d|p^{k}, d<p^{k}}m(\frac{p^{k}}{d})m^{-1}(d)$$
and this equal to $-m(p)m^{-1}(p^{k-1})$, since rest summands are equal to zero, computing previous values we obtain a closed form, that we can prove by induction, $m^{-1}(p^{k})=i^{k}$. Thus since $m^{-1}$ is multiplicative $m^{-1}(n)=\prod_{j=1}^{\omega (n)}m^{-1}(p_{j}^{e_{j}})=\prod_{j=1}^{\omega (n)}i^{e_{j}}=i^{e_{1}+\cdots +e_{\omega(n)}}$, and as Peter said this is equal to $i^{\Omega (n)}$. This finish the proof of the theorem.
Now, here I don't follow Peter's definition of modified function, for a relation $g(n)=\sum_{d|n}f(d)\mu (n/d)$, where $\mu$ is the Möbius function, I define the modified function $g_{m}$ as $g_{m}(n)=\sum_{d|n}f(d)m (n/d)$ (as a Dirichlet product display, it is $g_{m}=f\star m$, compare with Peter's definition). I compute the modified Euler's totient functions as $\phi_{m}(n)=\sum_{d|n}d m(n/d)$, equal to $\prod _{k=1}^{\omega(n)}p_{k}^{e_{k}}(1-i\cdot p_{k}^{-1})$, and for the so called identity function, its modified, following Peter, is $I_{m}(n)=\sum_{d|n}1\cdot m(d)=(1-i)^{\omega (n)}$ if $n>1$, and $I_{m}(1)=1$.
Question. From, respectively, $$(\sigma^{-1})_{m}(n)=\sum_{d|n}(d\mu (d))\cdot m(\frac{n}{d})$$ and $$\Lambda_{m}(n)=\sum_{d|n}(\log d)\cdot m(\frac{n}{d})$$ can you find in a closed form $\sigma_{m}$ and $\Lambda_{m}$, the modified functions for sum of divisor function and Mangoldt function?
Basically my definitions and results are a copy of Peter's first steps in his first paper of modified Möbius function. I am interested in other definitions because could have more arithmetical charge. Thanks in advance.
References (there is an open access here to [1] if you want read it www.austms.org.au/gazette):
[1] Brown, P. G., Modifying Möbius, The Australian Mathematical Society Gazette, Vol 37, No. 4, 2010, pp. 244-249.
[2] Tom M. Apostol, Introduction to Analytic Number Theory, Springer (1979).