I hope this is okay for the site, I asked on math exchange with no answer.

Consider the function $f(n)$ defined on the natural integers which returns the largest prime factor of $n$, and is $0$ for $1$. For example, $f(9)=3$,$f(15)=5$.

A beautiful riddle says that there are infinitely many $n$ so that $f(n)<f(n+1)<f(n+2)$.

I have 2 questions:

1.Given $k$, are there infinitely many $n$ so that $f(n)<f(n+1)...<f(n+k)$?

2.How often is $f(n)<f(n+1)$?

I'd guess is true because we can take large primes that are near each other, and with chinese reminder theorm make $n+i$ be divisible by $p_i$ and hope that after we divide all the other prime factors are small.

Proof of the riddle:

Lemma 1, if $f(a)<c, f(b)<c$, then $f(ab)<c$.

For any odd prime $q$ we find such different $n$, as there are infinitely many different odd primes we're done.

Choose an odd prime $q$, try $n+1=q$. Obviously $f(n)<f(n+1)$.

If $f(n+1)<f(n+2)$ we're done. Otherwise, choose $n+1$=$q^2$. Now $f(n)=f(q^2-1)=f((q−1)(q+1))$, so by lemma 1, by setting $a=q-1$, $b=q+1$, $c=q$ we get $f(n)<f(n+1)$. Again if $f(n+1)<f(n+2)$ we're done, otherwise choose $n+1=q^4$ and keep going like that. Assume by contradiction this goes on forever, then $q=f(q^{2^k})>f(q^{2^k}+1)$ for all $k$, but $gcd(q^{2^k}+1,q^{2^m}+1) = 1$ for all different $m,n$, and so eventually they contain primes larger than $q$, contradiction.