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$\newcommand\bmod{\mathbin\%}$I have been playing around with the Möbius Function and primorials and I am finding results that I am not yet able to understand which I suspect are very elementary.

Here's the current result which is I am working through. Any help is greatly appreciated!

Let $p_k$ be any prime, and write $p_k\#$ for its primorial. Let $x$ be any integer.

It seems based on my calculations that $$\sum_{i \mid p_{k}\#} \left\lfloor{\frac{(x \bmod i) + (p_k \bmod i)}{i}}\right\rfloor\mu(i) \ge -1$$

where $\bmod$ is the remainder so that $5 \bmod 3 = 2$ and $7 \bmod 3 = 1$.

But there exist integers $x$, $y$ such that:

$$\sum_{i \mid p_{k}\#} \left\lfloor{\frac{(x \bmod i) + (y \bmod i)}{i}}\right\rfloor\mu(i) < -1.$$

For example, if $x=13$, $y=23$, $p_k = 5$, then the sum is $-2$.

Is there a well known explanation for this?

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  • $\begingroup$ Perhaps I'm misunderstanding your notation, but in the case $x=13,y=23,p=5$, I get that the sum is $1 + 1 - \lfloor (3 + 3) / 5 \rfloor = 1$. What is the significance of the #? $\endgroup$ Jul 11, 2012 at 16:13
  • $\begingroup$ Right, I see now. $p$# means the primorial of p. $\endgroup$ Jul 11, 2012 at 16:22

1 Answer 1

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$\newcommand\bmod{\mathbin\%}$Note that

$$ \left\lfloor \frac{a \bmod p + b \bmod p}{p} \right\rfloor = \left\lfloor \frac{a+b}{p} \right\rfloor - \left\lfloor \frac{a}{p} \right\rfloor - \left\lfloor \frac{b}{p} \right\rfloor$$

so your expression can be written as $F_{p_k\#}(x+p_k) - F_{p_k\#}(x) - F_{p_k\#}(p_k)$, where

$$ F_n(x) := \sum_{i\mid n} \left\lfloor \frac{x}{i} \right\rfloor \mu(i).$$

Now, the expression $F_n$ can be simplified by elementary number theoretic computations. Indeed, observe that

$$ \left\lfloor \frac{x}{i} \right\rfloor = \sum_{m \leq x: i\mid m} 1$$

and thus after interchanging summations,

$$ F_n(x) = \sum_{m \leq x} \sum_{i\mid n, i\mid m} \mu(i).$$

But by Möbius inversion, $\sum_{i\mid n, i\mid m} \mu(i)$ equals 1 when $n$, $m$ are coprime and 0 otherwise. Thus $F_n(x)$ is nothing more than the number of natural numbers $m$ less than $x$ that are coprime to $n$. In particular, this gives

$$ F_{p_k\#}(x+p_k) \geq F_{p_k\#}(x)$$

and

$$ F_{p_k\#}(p_k) = 1$$

which explains your numerically observed phenomenon.

The quantities here are somewhat reminiscent of the quantity $\pi(x+y)-\pi(x)-\pi(y)$ that occurs in the second Hardy–Littlewood conjecture (which, by the way, is widely believed to be false). Indeed, the quantity $F_{p_k\#}(x)$ (that is, the number of natural numbers up to $x$ that have no prime factor less than or equal to $p_k$) is more commonly denoted $\pi(x,p_k)$ in the analytic number theory literature.

EDIT: I also encountered some related expressions and inequalities in my paper A remark on partial sums involving the Mobius function (and with more advanced versions of these inequalities also in this paper of Granville and Soundararajan: Negative values of truncations to $L(1, \chi)$).

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    $\begingroup$ Thanks, Terry! I really appreciate your analysis! :-) I look forward to reading the articles you cited! Cheers, -Larry $\endgroup$ Jul 12, 2012 at 15:43

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