I am unsure how to start/proceed with the following problem:
Let $p_1,\cdots , p_m$ be the first $m$ odd primes and let $P(x,m)$ be the number of odd integers $\leq x$ and not divisible by any of these primes. Let $\{ u \} = [u + \frac{1}{2} ]$ be the integer nearest to $u$. Show that $P(x,m) = \sum_a \{\frac{x}{2a} \} - \sum _b \{\frac{x}{2x}\}$, where $a$ and $b$ run over all products of an even and an odd number of primes among $p_1,\cdots , p_m,$ respectively.
I have proved that $$\prod_{p\leq x} \left( 1 - \frac{1}{p} \right) < \frac{1}{\log x}$$ for $x \geq 2$ and $$\pi (x) << \frac{x}{\log \log x}$$
Is there a way to proceed/conclude this problem with this information, or should we be using a different approach?
Simply double counting and principle of inclusion and exclusion is enough,
Firstly, from the right side of the equality $\sum_a \{\frac{x}{2a} \} - \sum _b \{\frac{x}{2b}\}$ .......................................................(1)
$a$ has an even number of distinct prime factors$\implies \mu (a)=1$
$b$ has an odd number of distinct prime factors,$\implies \mu (b)=-1$
the sum can be rewritten into $\sum_a \mu(n) \{\frac{x}{2n}\}$....................................................................................(2)
Start off by experimentation, to count number of odd integers $≤x$ and not divisible by any of these primes.
we know even numbers are not welcomed, we start off having $\lceil \frac{x}{2}\rceil$, which is equal to $ \{ \frac{x}{2}\}$ by intuition.
This is a term in $(1)$, where $a=1$, having zero prime factor.
Then we are to prove with $a=p_{a_1}p_{a_2}p_{a_3}...p_{a_j}$, which has $j$ distinct prime factors, the following lemma
Lemma: $\{\frac{x}{2a}\}$ counts the number of odd $k\le x$, which is divisble by $a$.
Proof: For $k\le x$, $\{a,2a,3a,...,\lfloor\frac{x}{a}\rfloor \cdot a\}$ are all factors of a.
Since we need only odd factors, we get $\{a,3a,5a,...,ga\}$
Where $g\cdot a$ is the last factor with odd $g$.
When $\lfloor\frac{x}{a}\rfloor$ is odd, then $g=\lfloor\frac{x}{a}\rfloor $ and there will be $\frac{(\lfloor\frac{x}{a}\rfloor)+1}{2}$ number of such multiples of $a$.
When $\lfloor\frac{x}{a}\rfloor -1$ is odd, then $g=\lfloor\frac{x}{a}\rfloor -1$, and there will be $\frac{(\lfloor\frac{x}{a}\rfloor -1)+1}{2}$ number of such multiples of $a$.
Both of them are equal to $\{\frac{x}{2a}\}$, this is proved below.
WLOG, if $\frac{(\lfloor\frac{x}{a}\rfloor)+1}{2}$ is not the closest integer to $\frac{x}{2a}$
then $|\frac{(\lfloor\frac{x}{a}\rfloor)+1}{2}-\frac{x}{2a}|\gt1$,
$\lfloor\frac{x}{a}\rfloor=\frac{x}{a}+e$ where $|e|\lt 1$,
it becomes $|\frac{x}{2a}+\frac{e}{2}+\frac{1}{2}-\frac{x}{2a}|\gt1 \implies |\frac{e}{2}+\frac{1}{2}|\gt1\implies|\frac{e}{2}|+|\frac{1}{2}|\gt|\frac{e}{2}+\frac{1}{2}|\implies1=|\frac{1}{2}|+|\frac{1}{2}|\gt|\frac{e}{2}|+|\frac{1}{2}|\gt1$,
Contradiction.
Then consider sets $A_{p_1},A_{p_2},...,A_{p_m}$, where the set $A_{p_i}$ represents set of number that is odd multiple of $p_i$, and is $\le x$.
Let P be the set of numbers of odd integers $≤x$ and not divisible by any of the stated primes.
By principle of inclusion and exclusion,
$|P|= \{ \frac{x}{2}\} -\big[ \{|A_{p_1}|+|A_{p_2}|+...+|A_{p_m}|\bigr] + \big[|A_{p_1}\cap A_{p_2}| +...\big]-...$
For terms like $(-1)^j |A_{p_{a_1}}\cap A_{p_{a_2}}...\cap A_{p_{a_j}}|$ and $a=p_{a_1}p_{a_2}p_{a_3}...p_{a_j}$, this term is equivalent to term $\mu(a)\{\frac{x}{2a}\} $ in (2).
So proof completed.