sum of the primes with even indices and with odd indices

Question

Referring to https://oeis.org/A077131 and https://oeis.org/A077126 one sees that at a(40) the sum of the primes with odd indices is 7440 and for even indices 7257. Has anyone looked at the possibility of a race between these two, similar to that between primes of the form $4*n+1$ and $4*n+3$? Do the two sets, even indices and odd indices, have the same proportion of the latter two $4*n+1$ and $4*n+3$ as do all the primes as the primes increase?

Answer 1

First, you wrote the sum of even and odd primes in reverse order.

• Sum of the primes with odd indices is 7257
• Sum of the primes with even indices is 7440

As Malcolm stated in the comment;

$$\sum_{n=1}^{N} p_{2n} > \sum_{n=1}^{N} p_{2n-1}$$

If we define a difference $d = \sum_{n=1}^{N} p_{2n} - \sum_{n=1}^{N} p_{2n-1}$, it means you sum the half of difference of primes which gives you $d \sim \frac{p_{2N}}{2}$.

To explain it visually, when calculating d, you are summing (3-2) + (7-5) + (13-11) + ... instead of (3-2) + (5-3) + (7-5) ... ($p_{2N}-p_{2N-1}$)

The latter sum is $p_{2N}-2$ which is asymptotic to $p_{2N}$.

By the corollary of Dirichlet theorem on arithmetic progression, we know that all redisue classes of primes are uniformly distributed.Since the first sum has half as many terms as the second, we can say

$$ \frac{p_{2N}}{2} \sim \sum_{n=1}^{N} p_{2n} - \sum_{n=1}^{N} p_{2n-1}.$$

In your case, N=40 and $p_{2n}=409$. Difference between sum of even and odd indices is 7440 - 7257 = 183. It is close to our prediction $\frac{p_{2N}}{2} \sim 204.5$

For larger N values the difference between $d - \frac{p_{2N}}{2}$ will grow arithmetically, but it will diminish rationally.

In case of N = 50000;

• $p_{2n}=1299709$
• odd sum = 31,130,023,341
• even sum =31,130,675,380
• difference = 652,039
• our prediction = $\frac{p_{2N}}{2} \sim 649,854.5$

But I can't say the same for 4n+1 and 4n+3.