For \(n \in \mathbb{N}\), let \(a_n\) count the number of ternary strings of length \(n\) that contain no consecutive \(1\)s. We find that \(a_n = \left(\frac{1}{2}+\frac{\sqrt{3}}{3}\right)\left(1 + \sqrt{3}\right)^n – \left(\frac{1}{2}-\frac{\sqrt{3}}{3}\right)\left(1 – \sqrt{3}\right)^n\). For a given \(n \geq 0\), we then determine the following for these \(a_n\) ternary strings:
(1)the number of \(0’\)s, \(1’\)s, and \(2’\)s;(2)the number of runs;(3) the number of rises, levels, and descents; and
(4)the sum obtained when these strings are considered as base \(3\) integers.
Following this, we consider the special case for those ternary strings (among the \(a_n\) strings we first considered) that are palindromes, and determine formulas comparable to those in (1) – (4) above for this special case.