Amazing prime numbers!

If the $n$ th prime number is $p_n$ , then $p_{36} + \sum_{j=68}^{1461}p_j \; = \; 8303321 \; = \; \frac12\sum_{j=1}^{2019}p_j$ so the split is possible.

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Now for detail...

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Proposition Any positive integer $N \ge 7$ can be written as the sum of distinct positive integers, none of which are greater than $\mathrm{max}(11,N-7)$ .

Simple checking shows that this works for all $7 \le N \le 27$ . Now suppose that $N \ge 28$ , and that this result holds for all integers $n$ between $7$ and $N-1$ . If we set $M \; = \; \left\lfloor \frac{N-6}{2}\right\rfloor$ then $M > 10$ and so Bertrand's (no longer a) Postulate tells us that there exists a prime number $p$ such that $M+1 \le p \le 2M-1$ . But then $p \le 2M-1 \le 2\left(\tfrac{N-6}{2}\right) - 1 = N-7$ , so that $N-p \ge 7$ . By induction, we deduce that $N-p$ can be written as a sum of distinct primes, none of which are greater than $\mathrm{max}(11,N-p-7)$ .

Note that $p \ge M+1 > 11$ and $2p > 2M \ge N-6 > N-7$ , so that $p > N-p-7$ . Thus $p > \mathrm{max}(11,N-p-7)$ , and hence we deduce that $N$ can be written as a sum of distinct integers, none of which are greater than $p$ . Since $p \le N-7 \le \mathrm{max}(11,N-7)$ , we deduce that none of the primes that sum to $N$ are greater than $\mathrm{max}(11,N-7)$ , and so the desired result follows by induction.

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Suppose that $N \ge 3$ is odd, and let $p_1,p_2,...,p_N$ be the first $N$ primes. This list contains $2$ and an even number of odd primes, so we deduce that $2X \; =\; \sum_{n=1}^Np_n$ is an even integer (so that $X \in \mathbb{N}$ ). Find an integer $K$ such that $\sum_{n=K}^N p_n \; > \; X \; \ge \; \sum_{n=K+1}^Np_n$ If $N=3$ then since $2+3=5$ , we can split the first $N=3$ primes into two sets with equal sums. If $N=5$ we can split the first $N=5$ prime numbers as $3+11 = 2 + 5 + 7$ . Otherwise $N \ge 7$ , so that $2\sum_{n=1}^Kp_n \; \le \; 2X = \sum_{n=1}^Np_n \ge 2 + 3 +5 + 7 + 11+ 13 + 17 = 58$ so that $\sum_{n=1}^Kp_n \ge 29$ which implies that $K \ge 6$ . We can therefore suppose that $K \ge 6$ (so that $p_K \ge 13$ ). There are now various cases to consider:

• If $\sum_{n=K+1}^N p_n = X$ , then we have a suitable split of the first $N$ primes.
• If $Z = X - \sum_{n=K+1}^N p_n$ is such that $Z \ge 7$ , then we know that $Z$ can be written as a sum of distinct primes, none of which exceed $Z < p_K$ . This means that we can write $X$ as the sum of a collection of a collection of distinct primes, all less than $p_K$ , and the primes $p_{K+1}$ to $p_N$ .
• if $Z = 2,3,5$ , then $Z$ is prme and $Z \le 5 < p_K < p_{K+1}$ and $X = Z + \sum_{n=K+1}^Np_n$ .
• If instead $Z = 1,4,6$ then $X - \sum_{n=K+2}^Np_n \; =\; Z + p_{K+1} \; \ge \; 7$ since $K \ge 6$ , and so $X - \sum_{n=K+2}^Np_n$ can be written as a sum of distinct primes, none greater than $\mathrm{max}\left(11,X - \sum_{n=K+2}^Np_n-7\right) < p_{K+1}$ (since $p_{K+1} > 11$ ), and so no prime is greater than $p_K$ . Thus we have written $X$ as a sum of a subset of $p_1,p_2,...,p_N$