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Find all ordered pairs $(p_{k},p_{k+1})$, where $p_k$ denotes the $k$-th prime, such that for every $m\ \in \mathbb{N}$ there exists $\alpha \in \mathbb{N}$ s.t. $\Omega(\alpha) = m$ so that $p_{k+1} + 1 = \alpha p_{k}$, where $\Omega(x)$ denotes total number of prime numbers in prime factorization of $x$.

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tesgoe
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  • Typically $\Omega$ notation means that there is some constant multiplier that provides a lower bound. Do you have such a constant in mind? Otherwise it seems like the problem is ill-posed and dependent on the unknown constant multiplier. – user2566092 Mar 19 '14 at 21:58
  • Omega means: http://en.wikipedia.org/wiki/Almost_prime – tesgoe Mar 19 '14 at 22:00
  • you should clarify this in your posting, because $\Omega$ has a well-documented different usage in the computer science literature. All you have to do is specify that $\Omega$ counts the number of prime divisors. – user2566092 Mar 19 '14 at 22:12
  • Added this information to the problem description. Thank you for your insight. – tesgoe Mar 19 '14 at 22:17
  • What have you tried so far? – Klangen May 27 '19 at 08:27

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