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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 I was out of date, relying on a "claimed proof by Erdős" that there were an infinite number of R-A pairs, which turned out not to be sound, as Benjamin Dickman explained.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 $10^{10}$.

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    $\begingroup$ Did Erdos ever meet either Ruth or Aaron? $\endgroup$
    – Will Jagy
    Commented Dec 24, 2014 at 0:43
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    $\begingroup$ @WillJagy Actually, Erdos is known (in a piece of mathematical folklore) as having met Aaron and signed a baseball together, as reported by Pomerance (mentioned in my answer below). See here: "Carl Pomerance (now at Dartmouth College), who had a long and fruitful collaboration with Paul, reports having a baseball autographed by both of them, occasioned by their both having received honorary degrees at Emory University in 1995" in connection with the Ruth-Aaron numbers asked about. $\endgroup$
    – Benjamin Dickman
    Commented Dec 24, 2014 at 0:59
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    $\begingroup$ JO'R: Apparently only two triples are known. Besides what you list in your post, the other known one is: 6913943284, 6913943285, 6913943286 (source). That link checks up to at least 532,320,600,592 (I had to "stop" the page because it was very slow to load!) so I think you would need to look pretty far to find a quadruple. $\endgroup$
    – Benjamin Dickman
    Commented Dec 24, 2014 at 1:20
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    $\begingroup$ Apparently, here's a picture of Erdos and Aaron: www2.math.ou.edu/~jalbert/courses/hank_erdos.png $\endgroup$
    – verret
    Commented Dec 24, 2014 at 7:38
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    $\begingroup$ It seems that, contra Hoffman, Erdos never claimed to prove the infinitude of Ruth--Aaron pairs. After the original Ruth--Aaron paper appeared in the Journal of Recreational Mathematics, Pomerance learned of Erdos's interest from a letter (not a phone call). In the letter, Erdos says he can prove that the Ruth--Aaron numbers have density $0$. Far from claiming a proof that there are infinitely many Ruth--Aaron numbers, Erdos says that problem "seems hopeless" ! The original letter is available online --- see the very first scan at cah.utexas.edu/collections/math_erdos.php $\endgroup$
    – so-called friend Don
    Commented Dec 24, 2014 at 17:41

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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):

enter image description here

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):

enter image description here

(See the aforelinked citation for the argument.)

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    $\begingroup$ Ah, I was out of date, due to "an erroneous claimed proof by Erdős" that there were an infinite number of Ruth-Aaron pairs. Thanks for the correction and update! $\endgroup$
    – Joseph O'Rourke
    Commented Dec 24, 2014 at 0:13
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    $\begingroup$ (Note that I was "counting repeated prime factors," which increases the solutions, but still leaves matters quite open.) $\endgroup$
    – Joseph O'Rourke
    Commented Dec 24, 2014 at 0:16

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