\((0, 1, 3, 7, 15, 31, \ldots )\text{.}\) Try adding 1 to each term and we get powers of 2. You might guess this because each term is a little more than twice the previous term (the powers of 2 are exactly twice the previous term). Closed formula: \(f_n = 2^{n} - 1\text{.}\)
in-context