A curious equality of integrals involving the prime counting function?

This post discusses the integral, $$I(k)=\int_0^k\pi(x)\pi(k-x)dx$$

where $\pi(x)$ is the prime-counting function. For example, $$I(13)=\int_0^{13}\pi(x)\pi(13-x)dx = 73$$

Using WolframAlpha, the first 50 values for $k=1,2,3,\dots$ are,

$$I(k) = 0, 0, 0, 0, 1, 4, 8, 14, 22, 32, 45, 58, 73, 90, 110, 132, 158, 184, 214, 246, 282, 320, 363, 406, 455, 506, 562, 618, 678, 738, 804, 872, 944, 1018, 1099, 1180, 1269, 1358, 1450, 1544, 1644, 1744, 1852, 1962, 2078, 2196, 2321, 2446, 2581, 2718,\dots$$

While trying to find if the above sequence obeyed a pattern, I noticed a rather unexpected relationship:

Q: For all $n>0$, is it true, $$I(6n+4) - 2\,I(6n+5) + I(6n+6) \overset{\color{red}?}= 0$$

Example, for $n=1,2$, then $$I(10)-2I(11)+I(12)=32-2*45+58 = 0$$ $$I(16)-2I(17)+I(18)=132-2*158+184= 0$$ and so on.

Solution 1:

The answer is yes. Sketch of solution: $$ I(k) = \int_0^k \sum_{p\le x} \sum_{q\le k-x} 1 \,dx = \sum_p \sum_{q\le k-p} \int_p^{k-q} dx = \sum_p \sum_{q\le k-p} (k-(p+q)) = \sum_{m\le k} r(m)(k-m), $$ where $r(m)$ is the number of ways of writing $m$ as the sum of two primes. Then \begin{align} I(6n+6) &{}-2I(6n+5)+I(6n+4) \\ &= \sum_{m \le 6n+4} r(m)\big((6n+6-m)-2(6n+5-m) +(6m+4-m)\big) + r(6n+5) \\&= 0 + r(6n+5); \end{align} and $r(6n+5)=0$ for every $n\ge1$, since the only way the odd integer $6n+5$ can be the sum of two primes is $6n+5=2+(6n+3)$, but $6n+3=3(2n+1)$ is always composite when $n\ge1$.

The same argument gives $I(6n+2)-2I(6n+1)+I(6n) = r(6n+1)$, which is $2$ if $6n-1$ is prime and $0$ otherwise; this is why (as observed by John Omielan) it equals $2$ for $1\le n\le 5$ but $0$ for $n=6$.