Consecutive rising sequence of largest prime factors 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.
 A: As Kevin Buzzard suggests in a comment,
this would be a consequence of one of the
"standard conjectures on primes", namely the 
first
Hardy-Littlewood conjecture
(which is the special case of 
Schinzel's
hypothesis H where all the polynomials are linear).
If $N$ is sufficiently divisible, say $N = {\rm lcm}(1,2,3,\ldots,k)^2$,
then each of $(Nx-i)/i$ for $i=1,2,3,\ldots,k$
is a linear polynomial in $x$ that always takes
integer values not divisible by any prime $\leq k$,
so Hardy-Littlewood predicts the existence of infinitely many $x$such that each of these $(Nx-i)/i$ is prime; then the largest prime factors of
the $k$ integers in $[Nx-k,Nx)$ are in increasing order.
The same argument applied to $(Nx+i)/i$ would likewise produce
infinitely many runs of $k$ integers $Nx+i$ ($i=1,\ldots,k$)
whose largest prime factors are $(Nx+i)/i$ and are thus in decreasing order.
A: I think at this point it is safe to say that the answer to question 1 is, conjecturally yes but nothing has been proved for any $k\ge3$, and the answer to question 2 is, conjecturally half the time (in the limit, as $n\to\infty$) but this too has not been proved. 
