On consecutive numbers of the same height in the Collatz problem (Q1210572)

The ``Collatz''-problem (or ``\(3x+1\)''- or ``Hasse''- or ``Syracuse''- or ``Kakutani''-problem) is to prove that for every \(n\in\mathbb{N}\) there exists a \(k\) with \(C^{(k)}(n)=1\) where the Collatz function \(C(n)\) takes odd numbers \(n\) to \(3n+1\) and even numbers \(n\) to \(n/2\). The smallest number \(k\) with \(C^{(k)}(n)=1\) (if it exists) is called the \(\text{height}(n)\), the number of divisions by 2 taken in reaching 1 is called the ``total stopping time'' \(\sigma_ \infty(n)\). It is a well- known phenomenon that there are many consecutive numbers all having the same height and the same total stopping time. Found by numerical calculations the author gives an example of 349 such consecutive numbers starting at 9.749.626.154. Using Penning's argument [\textit{P. Penning}, Crux Math. 15, 282-283 (1989)] he shows that if there is one \(k\)-tuple of consecutive numbers all having the same height and the same total stopping time, then there exist infinitely many such \(k\)-tuples. It is conjectured that there are arbitrary long chains of consecutive numbers of this kind. Further the author considers two consecutive numbers with the same number of divison of 2 in the first \(k\) iterations of the Collatz function and studies the density of such pairs.