Fifteen consecutive integers with exactly four prime factors (Q2759113)

Denote by \(d(n)\) the number of divisors of \(n\) and by \(\Omega(n)\) the number of prime divisors of \(n\), each counted according to multiplicity. \textit{R. Guy} [Unsolved problems in number theory, 2nd ed., Springer, New York (1994; Zbl 0805.11001)] discussed solutions of \(d(n)= d(n+1)= \dots= d(n+k-1)= m\) for given \(m\) and \(k\). Here a similar problem is discussed with \(d(n)\) replaced by \(\Omega(n)\): Given \(m,k>1\), find \(k\) consecutive integers which all have the same \(\Omega\) value, \(m\). If \(m=2\) or \(m=3\), finding examples of sequences of maximum length is easy. The simplest with \(m=2\) is 33, 34, 35 while for \(m=3\) it begins with 211673, with \(k=7\). If \(m>4\), the problem is open. For \(m=4\), the longest sequence begins with NEWLINE\[NEWLINEn= 48899 5430 5677 6531 7569NEWLINE\]NEWLINE and has length equal to 15.