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Questions tagged [mobius-function]

Questions on the Möbius function μ(n), an arithmetic function used in number theory.

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Is it true that this quadruple sum involving $f(x,y,z,p)=x^p+y^p−z^p$, is equal to the sequence $1,0,0,0,...$ only if $p=1$ or $p=2$?

Consider the sum: $$s(n) = \sum _{k=1}^n \left(\sum _{z=1}^n \left(\sum _{y=1}^n \left(\sum _{x=1}^n g(k)\left[\gcd (f(x,y,z,p),n)=k\right]\right)\right)\right)\label{1}\tag{1}$$ where the minimal ...
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Bounding an equivalent expression of Mertens function

Some months ago, I derived the following formula for the Merten's function $M(n)$ using the inclusion-exclusion principle: $$M(n)=1-\pi\left(n\right)+\sum_{p_{i}\leq\frac{n}{p_i}}\left(\pi\left(\frac{...
Juan Moreno's user avatar
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Bounding $S(n) = \sum_{k=1}^n \mu(k) \left( \pi\left(\frac{n}{k}\right) - \pi(\text{gpf}(k)) \right)$ [closed]

I am trying to bound the sum $$S(n) = \sum_{k=1}^n \mu(k) \left( \pi\left(\frac{n}{k}\right) - \pi(\text{gpf}(k)) \right)$$ In other site, I have been given steps for a "proof" that $S(n) = ...
Juan Moreno's user avatar
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Using the binomial formula in the form $(k - k)^n$

While proving a certain property of the number theoretic mobius function, namely that it is invertible in the monoid of multiplicative functions and its inverse is ...
giorgio's user avatar
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lower bound of $\sum_{n=1}^x \frac{\mu(n)}{n}$

Denote by $\mu$ the Mobius function. Poussin showed that $$ \sum_{n=1}^x \frac{\mu(n)}{n} = O(1/\log x), $$ and there are further improvements since. I wonder what is known about lower bound of ...
mathflow's user avatar
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Why $ \prod\limits_{n=1}^{\infty} \biggl (\phi(q^n)^{\mu(n)} \biggr)= 1-q $?

Playing with Euler $\phi $ function (not to be confused with the totient function, here another reference), I found this curious identity (I calculated it for various $q$ with Mathematica and it holds)...
user967210's user avatar
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Comparing two series expressions for $1/\zeta(s)$. What can be said about their complex roots?

The following two expressions involving the inverted Riemann $\zeta(s)$ functions are well known: \begin{align} \frac{1}{\zeta(s)} &= \sum_{n=1}^\infty \frac{\mu(n)}{n^s} \\ -\frac{\zeta'(s)}{\...
Agno's user avatar
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Möbius function of distributive lattice only takes values $\pm 1$ and $0$.

In this Wikipedia article, I found the statement [...] shares some properties with distributive lattices: for example, its Möbius function takes on only values 0, 1, −1. My question is: How it can ...
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Prove that $\sum_{d=1}^{n} M(\lfloor n/d \rfloor) = 1$

In Wikipedia entry for Mertens function it says that From [Lehman, R.S. (1960). "On Liouville's Function". Math. Comput. 14: 311–320.] we have that $$\sum_{d=1}^{n} M(\lfloor n/d \rfloor) = ...
Juan Moreno's user avatar
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Which is the error in this application of Möbius inversion formula?

In Wikipedia the following generalisation of the Möbius inversion formula is given (and proved): Suppose $F(x)$ and $G(x)$ are complex-valued functions defined on the interval $[1, ∞)$ such that $$G(...
Juan Moreno's user avatar
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Riemann Hypothesis follows from the statement $M\left(x\right)=o_x(x^{\frac{1}{2}+\varepsilon})$

Recall that the Mertens function is defined via: $$M(n):=\sum_{n\ge x\ge 1} \mu(x)$$ Where $\mu$ is the Möbius function. Littlewood proved that if $M\left(x\right)=o_x(x^{\frac{1}{2}+\varepsilon})$ ...
linuxbeginner's user avatar
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What did I get wrong in this Mobius function question? [closed]

$f(n):=\sum\limits_{d\mid n}\mu(d)\cdot d^2,$ where $\mu(n)$ is the Möbius function. Compute $f(192).$ First, I found all of the divisors of 192 by trial division by primes in ascending order: $D=\{...
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Sequence notation?

Cor. Mobius Inversion Formula for multiplicative functions. Let $f,F$ be multiplicative functions such that $F(n)=\sum\limits_{d\mid n}f(d)$. Then $f(n)=\sum\limits_{d\mid n}\mu(\frac nd)F(d)$. Proof. ...
Jason Xu's user avatar
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Why is Möbius function's co-domain $\mathbb{C}$?

I am new to the concept of partially ordered sets. Here's my professor's definition of a Möbius function from her lecture notes: The inverse of zeta function with relative to the convolution product ...
Zek's user avatar
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Generalization of Möbius inversion formula

A generalization of Möbius inversion formula guarantees that if we have $G(n)=\sum_{k=1}^{n} F\left(\frac{n}{k}\right)$, then $F(n)=\sum_{k=1}^{n} \mu(k) G\left(\frac{n}{k}\right)$. If we have $\sqrt{...
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Meaning of $M(n)=O\left(x^{\frac{1}{2}+\epsilon}\right)$

I am trying to fully understand the implications of $M(n)=O\left(n^{\frac{1}{2}+\epsilon}\right)$, where $M(n)$ is Mertens function, being equivalent to Riemann Hypothesis. (i) Is the equivalence ...
Juan Moreno's user avatar
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Möbius inversion formula and $\sum_{k\leq n} \frac{\mu(k)}{k}$

I have tried to apply what is stated at the Generalizations of Möbius inversion formula section of Wikipedia to bound $$\sum_{k\leq n} \frac{\mu(k)}{k}$$ The application seems simple and ...
Juan Moreno's user avatar
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Bounding a partial sum with Möbius inversion formula

I am trying to bound the partial sum $$S(n)=\sum_{k=1}^{\sqrt{n}} \mu(k) \pi\left(\frac{n}{k}\right)$$ Where $\pi(x)$ is the prime counting function, and $\mu(x)$ is the Möbius function. Empirical ...
Juan Moreno's user avatar
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Möbius function on a product of posets is the product of the Möbius function on each poset

I'm trying to prove that $\mu_{P\times Q}((p_1,q_1),(p_2,q_2))=\mu_{P}(p_1,p_2)\mu_{Q}(q_1,q_2)$, and my attempt is by induction. By definition, I get that: $\mu_{P\times Q}((p_1,q_1),(p_2,q_2))=-\...
Grimm Troupe's user avatar
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Freshman's dream and the commutativity of the square root of the Möbius function over the divisors.

Let the infinite matrix $A$ be: $$A(n,k)=\left[ k \mid n \right] \left(\frac{\sqrt{k \, \mu(k)}}{n^s}\right)$$ where $n=1,2,3,4,5,...$ and $k=1,2,3,4,5,...$ Multiply $A$ with its transpose $A^{\mathsf{...
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Möbius function for graphs??

I'm reading this post and I'm getting a little confused. I am trying to find a useful notion of the Mobius function for directed graphs and have had little success in my search. I don't know much ...
joe's user avatar
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Moebius function and expansion in sequence $A280194$

For my research interest, I came accross with OEIS sequence A280194 In particular I am interested how is obtained the value $36203$ in the sequence Expansion of: $\frac{1}{1-\sum_{k>=1}{\mu(k)^2\...
Enzo Creti's user avatar
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Automorphisms on upper half plane

I am supposed to prove that: \begin{align} Aut(\mathbb{H})= \left\{ T:\mathbb{H}\to \mathbb{H}, T(z)=\frac{az+b}{cz+d} \vert a,b,c,d\in\mathbb{R}; ad-bc>0 \right\} \end{align}. where $\mathbb{H}$ ...
MilesDefis's user avatar
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1 answer
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Alternating Dirichlet series involving the Möbius function.

It is well known that: $$\sum_{n=1}^\infty \frac{\mu(n)}{n^s} = \frac{1}{\zeta(s)} \qquad \Re(s) > 1$$ with $\mu(n)$ the Möbius function and $\zeta(s)$ the Riemann Zeta function. Numerical ...
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Proof of the dual Möbius Inversion Formula

The Dual Möbius Inversion Formula: Let $\mathcal D$ be a divisor closed set of natural numbers (i.e., if $d\in \mathcal D$ and $c \mid d$, then $c\in \mathcal D$.) Let $f$ and $g$ be two complex-...
stoic-santiago's user avatar
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1 answer
259 views

Show that $\Phi(n) = \phi(1) +\phi(2) + \dots + \phi(n) =\frac{n^2}{2 \zeta(2)}+O(n \log n)$

Set $\Phi(n) = \phi(1) + \phi(2) + \cdots + \phi(n)$. I want to understand why $$ \Phi(n) = \frac{n^2}{2\zeta(2)} + O(n\log n). $$ I am going to show the proof that I have, and I would be very ...
Oswald Diaz's user avatar
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Invariance of the Möbius function under order isomorphism

Let $(X,\leq)$ and $(Y,\leq)$ be locally finite partially ordered sets with Möbius functions $\mu_X$ respectively $\mu_Y$, and let $\varphi:X\to Y$ be an order isomorphism. I want to show that \begin{...
Bart's user avatar
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Summing a series with number-theoretic coefficients

Is there a simpler form to $f(x)=\sum_{r=1}^{\infty}\left(\sum_{d|r}d\mu(d)\right)x^r/r$ where $\mu(d)$ is the Mobius function and $\sum_{d|r}$ is a sum over the divisors $d$ of $r$? I searched the ...
Valentine Michael Smith's user avatar
4 votes
1 answer
186 views

Calculate $ \sum_{k=1}^{n} k\cdot \mu(k) $

Problem: Given $n$, Calculate $ \sum_{k=1}^{n} k\cdot \mu(k) $ This is the oeis series. My Thoughts: I need a sublinear algo, possibly something of the order of $n^{3/4}$ or $n^{2/3}$ time. Any ...
sibillalazzerini's user avatar
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Two poset properties: are they related?

A bunch of infinite posets $P$ with $\hat 0$ have the following property For every $x\in P$, the principal filter $\{ y\in P : y\ge x\}$ is isomorphic as a poset to $P$ itself. Examples include ${\...
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What is the Dirichlet serie of The function $A$ (OEIS A001414) which gives the sum of prime factors (with repetition) of a number $n$?

The function $A$ (OEIS A001414) which gives the sum of prime factors (with repetition) of a number $n$ and defined by $$ A(n)=\sum \limits_{p^{\alpha}\parallel n}\alpha p $$ is this serie calculated ...
Es-said En-naoui's user avatar
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0 answers
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New $ \overset {s}{\underset{k=x}{ \lower 3pt { \LARGE\Xi}}} $ operator and Möbius function

At the beginning I will explain some concepts and at the end I will ask the specific question. Let's consider some operator $ \overset {s}{\underset{k=x}{ \lower 3pt { \LARGE\Xi}}} $ on function $f ...
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Characterization of Möbius-monotonicity

We say that an arithmetic function $f:\mathbb{Z}^+\to\mathbb{C}$ is Möbius-monotone if $\forall n\geq1:(\mu*f)(n)\geq0$ (where $*$ denotes de Dirichlet convolution), i.e. if there exists a non-...
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Prove the sum of nth complex primitve roots of unity is $\mu(n)$ [duplicate]

I am trying to prove the following statement, Let ω be a complex nth root of unity. We say that ω is a primitive nth root if $\omega^m\neq1$ for any positive $m < n$. Prove that for all $n\in\...
Mathsbot69's user avatar
-1 votes
1 answer
99 views

Product version of Möbius Inversion formula

I am currently learning about Möbius Inversion. In Wikipedia , the sum version is proved using convolutions which I can follow along with. However, it also lists a product version of this formula: $g(...
Meow's user avatar
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2 answers
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Definition of the Möbius function from Vinogradov's book

Ivan Vinogradov in the book "Elements of Higher Mathematics" (page 357) gives the following definition of the Möbius function: The Möbius function is a multiplicative function defined by ...
Aragorn's user avatar
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Proving that $\sum_{k\le x}\frac{k}{\varphi(k)}=Ax+O(\log x)$

I'm hoping to show that $\sum_{k\le x}\frac{k}{\varphi(k)}=Ax+O(\log x)$, for some positive constant $A$ but I keep getting stuck. I know that $\sum_{k\le x}\frac{1}{\varphi(k)} = O(\log x)$ and $\...
James2390's user avatar
1 vote
1 answer
343 views

Prove this generalization of Möbius inversion formula

Generalizations list this one as a generalization of Möbius inversion formula- Suppose $F(x)$ and $G(x)$ are complex-valued functions defined on the interval $[1, ∞)$ such that $$G(x)=\sum_{1\le n\le ...
Sayan Dutta's user avatar
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3 votes
6 answers
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Möbius inversion and Möbius function as sum of cosines

Let $\mu (n)$ be the Möbius function. I want to prove the following formula: $$\mu (n)=\sum_{\substack{1\leq k \leq n\\ (k,n)=1}}\cos \frac{2k\pi}{n}.$$ Let $F(n)$ be the right hand side, then by ...
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Perron's formula application - zeros of $L(s,\chi )$

After applying Perron's formula I have a complex integral involving something like \[ \sum _{n=1}^\infty \frac {\mu (n)}{n^s}e\left (\frac {an}{q}\right ).\] As usual ("usual" meaning e.g. ...
tomos's user avatar
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1 vote
2 answers
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$\varphi(x,n)=\sum_{d|n}\mu(d)\left\lfloor\frac xd\right\rfloor$ and $\sum_{d|n}\varphi\left(\frac xd,\frac nd\right)=\lfloor x\rfloor$

If $x$ is real, $x\ge 1$, let $\varphi(x,n)$ denote the number of positive integers less than or equal to $x$ that are relatively prime to $n$. Prove that $$\varphi(x,n)=\sum_{d|n}\mu(d)\left\lfloor\...
Sayan Dutta's user avatar
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1 answer
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Need help with finding Dirichlet generating function

I am unable to find the following generating function: $$\tag{1}{\sum _{n=1}^{\infty } \frac{2^{-k_n} \mu \big(\frac{n}{2^{k_n}}\big)}{n^s}}$$ $\mu$ is Möbius function, $k_n$ is the highest integer ...
azerbajdzan's user avatar
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2 votes
1 answer
580 views

Product form of Mobius Inversion formula: $g(n)=\prod_{d|n}f(d)^{a(n/d)}\iff f(n)=\prod_{d|n} g(d)^{b(n/d)}$

Product form of the Möbius inversion formula: If $f(n)>0$ for all $n$ and if $a(n)$ is real, $a(1)\neq 0$, prove that $$g(n)=\prod_{d|n}f(d)^{a(n/d)}\iff f(n)=\prod_{d|n} g(d)^{b(n/d)}$$ where $b=a^...
Sayan Dutta's user avatar
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7 votes
3 answers
354 views

Prove that $\sum_{d|n}\mu(d)\log^m d=0$

Prove that $$\sum_{d|n}\mu(d)\log^m d=0$$ if $m\ge 1$ and $n$ has more than $m$ distinct prime factors. I tried using Induction and kind of succeeded in the sense that if we write down the case $m=2$ ...
Sayan Dutta's user avatar
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7 votes
1 answer
202 views

Count pairwise coprime triples such that the maximum number of the triple is not greater than N

Problem Statement: Given N you are to count the number of pairwise coprime triples which satisfy $1≤a,b,c≤N$. Example: For example N=3, valid triples are (1,1,1),(1,1,2),(1,2,1),(2,1,1),(1,1,3),(1,3,1)...
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2 answers
65 views

The image of a Möbius transformation

I have the following problem it seems easy but I couldn't find the answer! For the following $m(z)$ $=$ $\frac {z - i}{iz-1 }$ show that $m$($\mathbb{H}$) $=$ $\mathbb{D}$ I tried to show that |$m(z)...
Ola alfares's user avatar
1 vote
0 answers
114 views

Calculate values of Möbius function

Let $(P,\leq)$ be a partial order with $P = \lbrace 0,1,2,3,4 \rbrace$ such that $0 \leq 1 \leq 4$,$0 \leq 2 \leq 4$,$0 \leq 3 \leq 4$. I want to calculate the values of the mobius function $\mu : P \...
Orb's user avatar
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2 votes
0 answers
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Using Mobius Inversion Formula

We are given the following: $f(n)=\prod_{d|n}g(d)$ and asked to show: $g(n)=\prod_{d|n} f(d)^{\mu(\frac{n}{d})}$ The hint given says to use logarithms Here's what I tried doing: $log(f(n))=\prod_{d|n}...
user8083's user avatar
  • 199
5 votes
2 answers
158 views

$\lim_{s \to 1^+} 1/\zeta(s) = 0$ obvious or not?

I read the statement that $$\frac{1}{\zeta(s)} = \sum_{n=1}^\infty \frac{\mu(n)}{n^s} \textrm{ for } \Re(s) > 1 \qquad (*)$$ In fact I can guess what the proof is: just expand both $\zeta$ and the ...
Vincent's user avatar
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4 votes
1 answer
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Analytic continuation of this function to $|z|>1$?

Consider the following function: $$\sum_{n=1}^\infty \sum_{s | n} s \mu(s) z^{n^2},$$ where $\mu(s)$ is the Mobius function. This converges for $|z|<1$. Does there exist an analytic continuation to ...
nathan benjamin's user avatar

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