Let us call $k$-tuple $(n+1,n+2,..,n+k)$ a factorazy $k$-tuple if we have $p(n+1)<p(n+2)<\cdots<p(n+k)$ where $p(m)$ denotes biggest prime factor of $m$.
I would like to know:
Is it true that for every $k \in \mathbb N \setminus \{1\}$ there exists at least one factorazy $k$-tuple? If not, what is the maximal value of $k$? For which $k$´s there exist an infinite number of factorazy $k$-tuples? What is known about this topic?
The $k=3$ case is solved in Erdos & Pomerance, On the largest prime factors of $n$ and $n+1$, Aequationes Mathematicae 17 (1978) 311-321, available here. The solution is short enough to give here in its entirety. Let $P(n)$ be the largest prime factor of $n$, let $p$ be an odd prime, and $$k_0=\inf\{\,k:P(p^{2^k}+1)>p\,\}$$ (note that $P(p^{2^{k_0}}+1)\equiv1\bmod{2^{k_0+1}}$, so $k_0<\infty$). Then $$P(p^{2^{k_0}}-1)<P(p^{2^{k_0}})<P(p^{2^{k_0}}+1)$$ and we're done.
Some variants are discussed elsewhere in the paper, and also at B46 in Guy, Unsolved Problems in Number Theory, 3rd edition.
If we assume the prime $k$-tuples conjecture, then there are infinitely many such tuples for any fixed $k$. To see this, consider the $k$ polynomials with integer coefficients: $$ k!x-1, \quad\frac{k!}2x-1, \quad\frac{k!}3x-1, \quad\dots\quad, \frac{k!}kx-1. $$ The product of these polynomials has no common prime factor (check $x=0$), so conjecturally there are infinitely many integers $x$ such that the above expressions are simultaneously prime. But then $$ k!x-1, \quad k!x-2 = 2\bigg( \frac{k!}2x-1 \bigg), \quad k!x-3 = 3\bigg( \frac{k!}3x-1 \bigg), \quad\dots $$ form such a tuple (with, I guess, $n=k!x-k-1$).
Of course the infinitude of primes proves that there are infinitely many such $2$-tuples. I don't know offhand whether the $k=3$ case is known, although it doesn't seem out of the question.
