TITLE: Ruth-Aaron triples, etc
QUESTION [5 upvotes]: A Ruth-Aaron pair is two numbers $(n,n+1)$ such that
their sum of prime factors is equal, counting repeated prime factors.
(The name refers to Hank Aaron's 715 homeruns surpassing Babe Ruth's 714!)
So
$$15 = 3 \cdot 5 \;;\; 3+5 = 8$$
$$16 = 2^4 \;;\; 2+2+2+2 = 2 \cdot 4 = 8$$
The notion can be generalized:
$$417,162 = 2 \cdot 3 \cdot 251 \cdot 277 \;;\; 2+3+251+277 = 533$$
$$417,163 = 17 \cdot 53 \cdot 463 \;;\; 17+53+463 = 533$$
$$417,164 = 2^2 \cdot 11 \cdot 19 \cdot 499 \;;\; 2+2+11+19+499 = 533$$
My question is:

Q. Are there $k$-tuples, for all $k \ge 2$, for which there exists at least
  one $n$ such that $(n, n+1, n+2, \ldots, n+k)$
  has the same sum of prime factors (counting repeated prime factors as above)?


Addendum. I was out of date, relying on a "claimed proof by Erdős" that there
were an infinite number of R-A pairs, which turned out not
to be sound, as Benjamin Dickman explained. So the problem is open even for pairs,
let alone $k$-tuples.
Apparently no Ruth-Aaron quadruple is known.
David Stork wrote Mathematica search code and found there are no such quadruples up to $10^{10}$.

REPLY [7 votes]: Observe that a positive answer to your question would imply that there are infinitely many Ruth-Aaron pairs; however, this infinitude already constitutes an open question.
Consult:

De Koninck, J. M. (2004). Computational Results and Queries in Number Theory. Annales Univ. Sci. Budapest., 23. pp. 149-161. Link (no paywall).

Excerpt (p. 152):

The paper does go on to provide a heuristic argument for a generalized problem in the same direction as your question, though it is in the context of $\beta$ rather than $S$.
Another excerpt (pp. 153-154):

(See the aforelinked citation for the argument.)