This is the sequence from Example 2.2.6, in which we found a closed formula by recognizing the sequence as the sequence of partial sums of an arithmetic sequence. Indeed, the sequence of first differences is \(1,4,7, 10, 13,\ldots\text{,}\) which itself has differences \(3,3,3,3,\ldots\text{.}\) Thus \(2, 3, 7, 14, 24, 37,\ldots\) is a \(\Delta^2\)-constant sequence.
These are the perfect cubes. The sequence of first differences is \(7, 19, 37, 61, 91, \ldots\text{;}\) the sequence of second differences is \(12, 18, 24, 30,\ldots\text{;}\) the sequence of third differences is constant: \(6,6,6,\ldots\text{.}\) Thus the perfect cubes are a \(\Delta^3\)-constant sequence.
If we take first differences we get \(1,2,4,8,16,\ldots\text{.}\) Wait, what? That's the sequence we started with. So taking second differences will give us the same sequence again. No matter how many times we repeat this we will always have the same sequence, which in particular means no finite number of differences will be constant. Thus this sequence is not \(\Delta^k\)-constant for any \(k\text{.}\)