Threshing floor sequence

We have $a_n^2 \; \ge \; a_n\lfloor a_n \rfloor \; = \; 49^n + 2n + 1 \; > \; 49^n$ and hence $a_n \ge 7^n$ , so that $\lfloor a_n \rfloor \ge 7^n$ . Moreover $\lfloor a_n\rfloor^2 \; \le \; a_n\lfloor a_n\rfloor \; = \; 49^n + 2n + 1 \; < \; 49^n + 2\times 7^n + 1 \; = \; (7^n + 1)^2$ so that $\lfloor a_n \rfloor < 7^n + 1$ . Thus $\lfloor a_n \rfloor = 7^n$ , so that $a_n \; = \; 7^n + \frac{2n+1}{7^n}$ This means that $\begin{aligned} \sum_{n=1}^N a_n & = \; \sum_{n=1}^N 7^n + \sum_{n=1}^N \frac{2n+1}{7^n} \; = \;\tfrac76(7^N-1) + \frac{5 \times 7^N - (3N+5)}{9\times 7^N} \\ \sum_{n=1}^N \tfrac12a_n & = \; \tfrac{7}{12}(7^N-1) + \tfrac{5}{18} - \frac{3N+5}{18 \times 7^N} \end{aligned}$ If $N$ is odd, then $7^N \equiv 7 \pmod{12}$ , and hence $7(7^N-1) \equiv 6 \pmod{12}$ . Thus $\tfrac{7}{12}(7^N-1) \; =\; X_N + \tfrac12$ where $X_N$ is an integer. Thus $\sum_{n=1}^N \tfrac12a_n \; = \; X_N + \tfrac79 - \frac{3N+5}{18 \times 7^N} \; = \; X_N + \tfrac79 - \varepsilon_N$ It is clear that $0 < \varepsilon_{2017} < \tfrac79$ (in fact $\varepsilon_{2017} \approx 1.086 \times 10^{-1702}$ ) so that $S \; = \; \left\lfloor \sum_{n=1}^{2017}\tfrac12a_n\right\rfloor \; = \; X_{2017} \; = \; \tfrac{7}{12}(7^{2017}-1) - \tfrac12$ and hence $2S+1 \; = \; \tfrac76(7^{2017}-1)$