Fix $0 < c < 1/3$. Show that there exists a strictly increasing sequence $a_k > 0$, a sequence $b_k \geq 0$, and an infinite set $K \subseteq \mathbb{Z}_{>0}$ such that \begin{align}\label{1}\tag{1} \begin{array}{ll} b_k = 2c a_k - 100 \log a_k > 0 & \text{ if } k \in K \\ b_k = 0 & \text{ otherwise } \end{array} \end{align} \begin{align}\label{2}\tag{2} c a_{k} + 100 \log a_k > 2k + 2\sum_{i=1}^{k-1} b_i \quad \text{ if } k \in K \end{align} \begin{align} \label{3}\tag{3} a_k > a_{k-1} + \log a_{k-1} + b_k \quad \text{ for all large } k \\ \label{4}\tag{4} a_k < (1-2c)^{-1} \left( a_{k-1} - \sum_{i=1}^{k-1} b_i \right)\quad \text{ for all large } k \end{align}
To get an idea of what's going on and to suggest that this isn't totally impossible, consider the simpler problem where $b_k = 0$ for all $k$. In other words, suppose $K$ is allowed to be empty.
Then it is clear that $a_k = k^2$ satisfies (1),(2),(3),(4).
It seems intuitive to me that if we choose $b_k = 2c a_k - 100 \log a_k$ on a rapidly growing sequence $k = k_j$, and if we adjust $a_k$ a little bit when $k=k_j$, then we can still get (1),(2),(3),(4) to hold. But I am having a lot of trouble making this rigorous.
Edit: Cross-Posted at Mathstack Exchange. It's been here for a while without progress. https://math.stackexchange.com/questions/2456086/proving-the-existence-of-a-sequence-with-recursive-growth-constraints
