A sum involving fractional part function

Question

I was exploring some sum when I came across this sum which I have no idea the value, here is the sum

Let $ N $ be an integer with the prime decomposition $ N = p_1^{k_1} p_2^{k_2} ... p_m^{k_m} $.

Let $ a $ be another integer such that $ 0 < a < N $. Consider the sum

\begin{align} \sum_{1 \leq i \leq m} \left \{ \frac{a}{p_i} \right \} - \sum_{1 \leq i < j \leq m} \left \{ \frac{a}{p_i p_j} \right \} + ... + (-1)^m \left \{ \frac{a}{p_1 p_2 ... p_m} \right \} \end{align} where $ \{ x \} = x - \lfloor x \rfloor $ is the factional part function.

I did some numerical calculations and found that the sum is generally small. For some values of $ a $, the sum may get large, but its absolute value seems to be bounded by $ m - 1 $, where $ m $ is the number of distinct prime factors of $ N $ as above.

My question is, is there currently a known formula/estimate/bound (in terms of $ a $ and $ N $) for the sum above?

(Realize that I did not really phrase the original question that well, so I edited some part of the post)

Answer 1

First observation: the fractional part of the result stays the same if you take the fractional part at the end (only the whole part may change).

Second observation: taking the fractional part of a quotient is the same as taking the remainder of division.

You can rewrite this as

$$\sum \frac{a\mod p_i}{p_i}-\sum \frac{a \mod p_i p_j}{p_i p_j}+\cdots$$

Now observe what happens when you keep adding one more prime to the set of primes $p_i$. With a single prime it's trivial. Adding more, it starts looking like you'll be able to use the Chinese Remainder Theorem once you take them to the common denominator. Can you continue?

Answer 2

For $n \ge 1$ look at $\mu(n)$ the Möbius function, $\text{Lpf}(n), \text{lpf}(n)$ the largest and least prime factor of $n$.

For a given $M = p_m$ and $a$ your sum is

$$-\sum_{n\ge \color{red}{2}} \mu(n) 1_{\text{Lpf}(n) \le M} \left(\frac{a}n-\left\lfloor \frac{a}n \right\rfloor \right)$$

• Since $\displaystyle \frac{\mu(n)1_{\text{Lpf}(n) \le M}}{n}$ is multiplicative we have the Euler product $$\sum_{n\ge 1} \frac{\mu(n)1_{\text{Lpf}(n) \le M}}{n} = \prod_{p} (1+\sum_{k \ge 1} \frac{\mu(p^k)1_{\text{Lpf}(p^k) \le M}}{p^k}) = \prod_{p \le M} \left(1-\frac{1}{p}\right)$$

• Then we look at the other sum

$$\sum_{n \ge 1} \mu(n)1_{\text{Lpf}(n) \le M} \left\lfloor \frac{a}n \right\rfloor = \sum_{n \le a, r \le a/n} \mu(n)1_{\text{Lpf}(n) \le M} = \sum_{r \le a} \sum_{n | r} \mu(n)1_{\text{Lpf}(n) \le M} = \sum_{r \le a} 1_{\text{lpf}(r) > M}$$ With the convention $1_{\text{lpf}(1) > M} = 1$, and I used that $ \sum_{n | r} \mu(n)1_{\text{Lpf}(n) \le M}$ is multiplicative to deduce it is $ = 1_{\text{lpf}(r) > M}$

Overall you get

$$-\sum_{n\ge \color{red}{2}} \mu(n) 1_{\text{Lpf}(n) \le M} \left(\frac{a}n-\left\lfloor \frac{a}n \right\rfloor \right)= a-\lfloor a\rfloor - a\prod_{p \le M} \left(1-\frac{1}{p}\right)+ \sum_{r \le a} 1_{\text{lpf}(r) > M}$$

And I don't see any good reason for it being small in general. The first sum can be approximated with the Mertens theorem, and the second one is trivial when $M \ge a$.