A curious equality of integrals involving the prime counting function?

This post discusses the integral,
I(k)=k0π(x)π(kx)dx

where π(x) is the prime-counting function. For example,
I(13)=130π(x)π(13x)dx=73

Using 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)?=0

Example, for n=1,2, then
I(10)2I(11)+I(12)=32245+58=0
I(16)2I(17)+I(18)=1322158+184=0
and so on.

Answer

The answer is yes. Sketch of solution:
I(k)=k0pxqkx1dx=pqkpkqpdx=pqkp(k(p+q))=mkr(m)(km),
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)=m6n+4r(m)((6n+6m)2(6n+5m)+(6m+4m))+r(6n+5)=0+r(6n+5);
and r(6n+5)=0 for every n1, 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 n1.

The same argument gives I(6n+2)2I(6n+1)+I(6n)=r(6n+1), which is 2 if 6n1 is prime and 0 otherwise; this is why (as observed by John Omielan) it equals 2 for 1n5 but 0 for n=6.

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

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