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Is the set $S:=\{n\in\mathbb{N} \mid \text{$n$, $n+1$ and $n+2$ have the same number of divisors}\}$ infinite?

Example: $33\in S$.

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    $\begingroup$ The answer is surely yes, even for triples $(3p,2q,5r)$ with prime $p,q,r$. Another question is how to prove. $\endgroup$ Jul 14, 2016 at 6:01
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    $\begingroup$ Please use LaTeX for mathematical notation. $\endgroup$
    – Todd Trimble
    Jul 14, 2016 at 7:55
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    $\begingroup$ @Fedor: why are you sure? $\endgroup$
    – user35593
    Jul 14, 2016 at 10:30
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    $\begingroup$ Heath-Brown proved that there are infinitely many $n$ for which $n$ and $n+1$ have the same number of divisors. It is conjectured that your statement is true. See Richard Guy's Unsolved Problems in Number Theory Problem B18. Indeed, there it is conjectured that there are infinitely many $n$ such that $n, n+1, n+2$ are all the product of two primes (this is stronger than your conjecture and weaker than Fedor's). $\endgroup$
    – Tony Huynh
    Jul 14, 2016 at 10:35
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    $\begingroup$ It is a very special case of widely believed (I think, cause of probabilistic heuristics) en.m.wikipedia.org/wiki/Dickson%27s_conjecture $\endgroup$ Jul 14, 2016 at 11:51

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I will convert my comment into an answer, since I suspect it is still the state of the art.

The version of your question for two consecutive integers was proved in

Heath-Brown, D. R. (1984). The divisor function at consecutive integers. Mathematika, 31(01), 141-149. ISO 690

See this paper of Hildebrand (the Heath-Brown paper is behind a paywall) for an improvement of Heath-Brown's result.

As Fedor Petrov mentions, your conjecture is very likely true, but is still an open problem (as far as I know). Indeed, your conjecture appears explicitly in Richard Guy's Unsolved Problems in Number Theory (Section 18B). This section also contains a lot of other related conjectures and results. For example, Erdős conjectured that for any $k$, there exist $k$ consecutive integers with the same number of divisors.

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