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Let $n\in N$.

Let $\mu (n)$ be the classical Mobius function. In other words, it vanishes at square-full numbers, equals $+1$ if $n$ has an even number of distinct prime factors, and equals $-1$ is $n$ has an odd number of distinct prime factors.

Let $p_i$ be the $i$-th prime.

Let $p_{max} (n)$ be the largest prime factor of $n$.

Let $p_{nextmax} (n)$ be the first prime larger than $p_{max} (n)$.

Let $f(x)$ and $g(x)$ be integer functions of real variables. In other words, $f,g:R \to Z$.

Of interest is the following identity:

$$f(x)=\sum_{n p_{nextmax} (n) \leq x} \mu (n) g\left( \frac{x}{n} \right) $$

We notice that this sum runs over $n$, and roughly $n \leq \sqrt{x}$, not $n\leq x$, because of the condition $n p_{nextmax} (n) \leq x$. Thus, an easier variant of this question considers the identity:

$$f(x)=\sum_{n\leq \sqrt{x}} \mu (n) g\left( \frac{x}{n} \right) $$

Or, even more generally, the identity:

$$f(x)=\sum_{n\leq y\leq x} \mu (n) g\left( \frac{x}{n} \right) $$

Can the above identities be Mobius inverted? If so, what would the result be?

Thank you.

anonymous
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  • what's the inspiration behind this problem? (not that there has to be one) – jg mr chapb Jul 27 '17 at 18:37
  • @Dis-integrating Hi. Yes, there is the inspiration behind this. It's RH. I expect this identity to be Mobius invertable. One should use the general Mobius inversion that is defined in abstract form over rings. I'm not quite fluent at abstract Mobius theory, so I decided to ask the question directly. The answer would save lots of time. Besides, I expect the answer to be rather trivial, because one simply has to prove that all conditions are met to use an abstract form of Mobius theory. I like your nick-name, Dis-integrating! – anonymous Jul 27 '17 at 18:45
  • what does the $m$ stand for in $p_m(n)$? – roger Jul 27 '17 at 21:19
  • @roger Hi. Subscript $m$ in $p_m (n)$ stands for "maximal". – anonymous Jul 27 '17 at 21:21
  • ok then it should be $n= \prod_{i=1}^{p_m(n)} p_i^{e_i(n)}$ then no? – roger Jul 27 '17 at 21:23
  • also in the identity with $f$ and $g$ does $x/n$ have to be an integer? I think it should be for it to be Mobius invertible. – roger Jul 27 '17 at 21:25
  • @roger No, index $i$ counts primes $p_i$. So for $i=1$ one has $p_1 =2$ in the prime factorization. For $i=2$ one has $p_2 = 3$, for $i=3$ a prime factor $p_3 = 5$ enters the prime factorization product. Eventually, there's the maximal prime $p_m (n)$ such that $e_m (n) \ne 0$. – anonymous Jul 27 '17 at 21:25
  • yes you're right sorry. – roger Jul 27 '17 at 21:27
  • @roger Ratio $x/n$ need not be an integer in general. This is true for classical Mobius inversion as well. Functions $f$ and $g$ can have support in, say, [0,$+\infty$>. – anonymous Jul 27 '17 at 21:28
  • The notation is poor. Both "m" and "m+1" are used as names, rather than values. I think it would be better to have $v(n) = {\max m\big|p_m | m}$ where $p_m$ is the $m$-th prime. Then $p_{v(n)}$ is the largest prime that divides $n$ and $p_{v(n)+1}$ is the next prime. – marty cohen Jul 27 '17 at 21:32
  • Here's a pdf document link that explains the general Mobius inversion with examples: http://www.unirioja.es/cu/jvarona/downloads/actas-sjtn2007-BenitoNavasVarona.pdf . – anonymous Jul 27 '17 at 21:32
  • @marty cohen Hi. True. But, index $m$ does label the $m$-th prime, so it's useful both ways: as a label as well as the actual number. Prime $p_m$ is the $m$-th prime indeed. It's also "maximal". I tried to keep it as simple as possible. – anonymous Jul 27 '17 at 21:34
  • I think this adds to the confusion. For example,what is the difference between $p_m$ and $p_m(n)$? What is $p_{2m}(n)$? – marty cohen Jul 27 '17 at 21:51
  • @marty cohen True. Luckily, $p_{2m} (n)$ isn't used here... Since $p_m \equiv p_m (n)$, i found this useful, so I used just $m$, because $n$ is the only variable here. I'll change this if it really amounts to confusion. Thanks. – anonymous Jul 27 '17 at 21:54
  • @marty cohen Done! – anonymous Jul 27 '17 at 22:03
  • is the "identity" $f(x) = \sum_{(...)} \mu(n) g(x/n)$ just a definition for $f(x)$? or do you have specific functions $f,g$ in mind? – Harry Richman Jul 27 '17 at 22:25
  • @Harry Richman Hi. I have specific functions in mind. Unfortunately, I cannot disclose what these specific functions are. However, the result should not depend on functions $f$ and $g$, simply because if the result holds for any particular pair of functions $F$ and $G$ with some support, then it holds for arbitrary functions $f$ and $g$ with the same support. See, for instance, http://www.unirioja.es/cu/jvarona/downloads/actas-sjtn2007-BenitoNavasVarona.pdf . – anonymous Jul 27 '17 at 22:37

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