<s>[INST] Show that there is always one integer $t$ with a least prime factor $> 5$ where $x < t \le x+6$ Let $p_k$ be the $k$th prime.
Let $f_2(x) = \lfloor x\rfloor - \left\lfloor\dfrac{x}{2}\right\rfloor$ 
For $k > 1$, let:
$f_{p_k}(x) = f_{p_{k-1}}(\lfloor x\rfloor) - f_{p_{k-1}}\left(\left\lfloor\dfrac{x}{p_k}\right\rfloor\right)$ 
Let $x \ge 1$ be an integer.
Show that $f_5(x+6) - f_5(x) > 0$
I find this problem easy to solve but challenging to answer concisely.
Please let me know if I made a mistake or if my answer could be more concise.
$f_5(x+6) - f_5(x) = \sum\limits_{i|30}\left(\left\lfloor\dfrac{x+6}{i}\right\rfloor-\left\lfloor\dfrac{x}{i}\right\rfloor\right)\mu(i)$ 
where $\mu(i)$ is the möbius function.
Let $r(x,d)$ be the remainder of $x$ when divided by $d$.
Clearly:


*

*$r(x+6,2) = r(x,2)$

*$r(x+6,3) = r(x,3)$

*$r(x+6,6) = r(x,6)$
So that:
$f_5(x+6) - f_5(x) = (x+6 - x) - \left(\dfrac{x+6 - x}{2}\right) - \left(\dfrac{x+6 - x}{3}\right)- \left(\dfrac{x+6 - x - r(x+6,5) + r(x,5)}{5}\right)+ \left(\dfrac{x+6 - x}{6}\right)+ \left(\dfrac{x+6 - x - r(x+6,10) + r(x,10)}{10}\right) + \left(\dfrac{x+6 - x - r(x+6,15) + r(x,15)}{15}\right) - \left(\dfrac{x+6 - x - r(x+6,30) + r(x,30)}{30}\right)$
$= 2 - \left(\dfrac{6 - r(x+6,5) + r(x,5)}{5}\right)+ \left(\dfrac{6 - r(x+6,10) + r(x,10)}{10}\right) + \left(\dfrac{6 - r(x+6,15) + r(x,15)}{15}\right) - \left(\dfrac{6 - r(x+6,30) + r(x,30)}{30}\right)$ 
It follows that if $x \not\equiv 4 \pmod 5$:


*

*$\dfrac{6 - r(x+6,5) + r(x,5)}{5} = 1$
If $x \equiv 4 \pmod 5$:


*

*$\dfrac{6 - r(x+6,5) + r(x,5)}{5} - \dfrac{6 - r(x+6,10) + r(x,10)}{10} = 1$
if $r(x,30) < 24$:


*

*$\dfrac{6 - r(x+6,30) + r(x,30)}{30} = 0$
if $r(x,30) \ge 24$:


*

*$\dfrac{6 - r(x+6,15) + r(x,15)}{15} - \dfrac{6 - r(x+6,30) + r(x,30)}{30} = 0$
So that:
$f_5(x+6) - f_5(x) \ge 2 - 1 + 0 = 1$

Edit 1:
I forgot to mention that $x$ is an integer.  I have updated the question.

Edit 2:
Updated the recurrence relation in order to make it clearer.  One commenter said that the relation wasn't clear.

Edit 3:
Apologies I have no idea why I wrote "kth integer" instead of $k$th prime.  Thanks for your patience.  Now fixed.
 [/INST] Whether or not an integer $n$ has a prime factor $\in\{2,3,5\}$ depends only on $n\bmod{30}$. Among the first $30$ positive integers, we find that $$\tag11,7,11,13,17,19,23,29$$ (and then $31=1+30$) do not have a such a small prime factor. The largest gap between these  is of size $5$  and occurs between $1$ and $7$ as well as between $23$ and $29$. Hence among any six consecutive integers, one is congruent $\bmod{30}$  to a number in $(1)$ and therefore does not have a prime factor $\le 5$. Hence this number is either $\pm1$ or a has smallest prime factor $\ge 7$.
We conclude that the interval $(x,x+6]$ contains an integer with least prime factor $>5$, provided that  $x\ge1$ or $x<-7$.
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