1, 4, 8, 9, 16, 25, 27, 32

Take the arithmetic progression $101k, 102k, \ldots, 199k$ , which has 99 terms. We wish to find a k such that $102k, 102k, \ldots, 198k$ are powers of positive integers, while 101k and 199k aren't. To be specific, we let S denote the set of primes < 198 and excluding 101, then pick k of the form:

$k =\prod_{p \in S} p^{m_p} = 2^{m_2} \times 3^{m_3} \times 5^{m_5} \times\ldots$

for suitable non-negative integers $m_2, m_3, m_5, m_7, \ldots$ .

Pick any distinct primes $q_{102}, q_{103}, \ldots, q_{198}$ ; for every prime $p\in S$ solve the following congruences:

$\begin{aligned} m_p + \nu(102, p) &\equiv 0 \pmod {q_{102}}, \\ m_p + \nu(103, p) &\equiv 0 \pmod {q_{103}},\\ &\vdots \\ m_p + \nu(198, p) &\equiv 0 \pmod {q_{198}}. \end{aligned}$

Here, $\nu(n, p)$ is the largest j such that $p^j | n$ . Note that since all prime factors of $102, \ldots, 198$ lie in S, we have:

$102 = \prod_{p\in S} p^{\nu(102, p)}, \ 103 = \prod_{p\in S} p^{\nu(103, p)}, \ \ldots, \ 198 = \prod_{p\in S} p^{\nu(198, p)}.$

By Chinese Remainder Theorem, there's a solution for $m_p$ in the above congruences. Since $k =\prod_{p \in S} p^{m_p}$ , we see that:

$102k = \prod_{p\in S} p^{\nu(102, p)} p^{m_p} = \prod_{p\in S} p^{m_p + \nu(102, p)}$

is a $q_{102}$ -th power, and likewise for each of $i = 103, \ldots, 198$ , the multiple $ik$ is a $q_i$ -th power. Thus, we have a sequence which satisfies (1).

This sequence clearly satisfies (2) as well since 101 and 199 are prime and k is not a multiple of 101 or 199.