This post discusses the integral,

I(k)=∫k0π(x)π(k−x)dxwhere π(x) is the prime-counting function. For example,

I(13)=∫130π(x)π(13−x)dx=73Using WolframAlpha, the first 50 values for k=1,2,3,… 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,…

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)−2I(6n+5)+I(6n+6)?=0Example, 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.

**Answer**

The answer is yes. Sketch of solution:

I(k)=∫k0∑p≤x∑q≤k−x1dx=∑p∑q≤k−p∫k−qpdx=∑p∑q≤k−p(k−(p+q))=∑m≤kr(m)(k−m),

where r(m) is the number of ways of writing m as the sum of two primes. Then

I(6n+6)−2I(6n+5)+I(6n+4)=∑m≤6n+4r(m)((6n+6−m)−2(6n+5−m)+(6m+4−m))+r(6n+5)=0+r(6n+5);

and r(6n+5)=0 for every n≥1, 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≥1.

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≤n≤5 but 0 for n=6.

**Attribution***Source : Link , Question Author : Tito Piezas III , Answer Author : Greg Martin*