Late starters

Let $a_j$ be the $j$ th term of the sequence, and $S_j$ the sum of the first $j$ terms of the sequence. We are told that $\begin{aligned} a_{j+1} + a_j & = \; S_{j-1} + n \\ S_{j+1} & = \; 2S_{j-1} + n \\ S_{j+1} + n & = \; 2\big(S_{j-1} + n\big) \end{aligned}$ for all $j \ge 2$ . Thus we deduce that $S_{2j+1}+n = 2^j(S_1+n)$ and $S_{2j}+n = 2^{j-1}(S_2+n)$ for all $j \ge 1$ , and hence $S_{2j+1} \; = \; 2^ja_1 + n(2^j-1) \hspace{2cm} S_{2j} \; = \; 2^{j-1}(a_1+a_2) + n(2^{j-1}-1)$ for all $j\ge 1$ . Thus we deduce that $\begin{array}{rclcl} a_{2j} & \;=\; & 2^{j-1}a_2 & \hspace{2cm} & j \ge 1 \\ a_{2j+1} & = & 2^{j-1}(a_1-a_2+n) & & j \ge 1 \end{array}$ Since the sequence $(a_j)_j$ is strictly monotonic increasing, we deduce that $a_2 > a_1$ and that $0 < a_{2j+1} - a_{2j} = 2^{j-1}[a_1-2a_2+n] \hspace{2cm} 0 < a_{2j+2}-a_{2j+1} = 2^{j-1}[3a_2-a_1-n]$ Thus the conditions of the problem are satisfied provided that $a_2 > a_1 \hspace{2cm} 2a_2-a_1 < n < 3a_2-a_1$ We are assuming that $a_j \in \mathbb{N}$ is a positive integer for all $j \ge 1$ . Then $n \ge 1 + 2a_2-a_1 = 1 + a_2 + (a_2-a_1) \ge 2 + a_2 \ge 3 + a_1 \ge 4$ and so the smallest possible value of $n$ is $4$ . But then $3 \ge 2a_2-a_1 \ge a_2 + 1$ and hence $a_2 \le 2$ . Since $a_2 > a_1 \ge 1$ , we deduce that $a_2=2$ , so that $a_1=1$ . Thus we must have $a_1-1$ , $a_2-2$ , $n=4$ , and hence $a_{2j} \; =\; 2^j \hspace{2cm} j \ge 1 \hspace{5cm} a_{2j+1} \; = \; \left\{\begin{array}{lll} 3 \times 2^{j-1} & \hspace{1cm} & j \ge 1 \\ 1 & & j = 0 \end{array}\right.$

Suppose that $N$ is a positive integer, and that there are $2N+1$ rounds. Then the total number of points that can be scored is $S_{2N+1} = 2^N + 4(2^N-1) = 5 \times 2^N-4$ . The maximum score of the player ranked second is obtained if only two players score, and the winner obtains as little more than half the possible total as possible, with the player ranked second scoring the rest. The largest possible score for the second player is then $\tfrac12(5 \times 2^N - 4) - 1 = 5 \times 2^{N-1} - 3$ . We need to show that it is possible for one player to score $5 \times 2^{N-1} - 3$ .

Adding up subcollections of the numbers $a_1,a_2,a_4,\ldots,a_{2N}$ can give any integer between $0$ and $2^{N+1}-1$ . Adding up subcollections of the numbers $a_3,a_5,\cdots,a_{2N+1}$ can given any integer of the form $3u$ where $0 \le u \le 2^N-1$ . Thus the possible scores for any player are numbers of the form $3u + v$ where $0 \le u \le 2^N-1$ and $0 \le v \le 2^{N+1}-1$ . With $u = 2^{N-1}-1$ and $v = 2^N$ we deduce that a player can score $5 \times 2^{N-1} - 3$ .

In this case we have $N = 1009$ , and so the maximum possible score of a player who comes second is $5 \times 2^{1008} - 3$ , making the answer $5 + 2 + 1008 + 3 = \boxed{1018}$ .