On the pi function and its arguments in geometric progressions

[Tomasz Ordowski]

Hello SeqFans! 

Find all numbers n such that 

pi(2n)^2 = pi(n)*pi(4n).

Are there only finitely many n such that 

pi(2n)^2 >= pi(n)*pi(4n) ? 

If so, find the largest such n = N. 

Conjecture: pi(2n)^2 < pi(n)*pi(4n) for n > N. 

Best, 

Tom Ordo 

[Jack Brennen]

It appears that the largest such n might be 51342.

There are no other such numbers between 51343 and 10^8.

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[M F Hasler]

On Mon, Jun 15, 2026 at 7:46 AM Tomasz Ordowski <tomaszo...@gmail.com> wrote:

Hello SeqFans! 

Find all numbers n such that 

pi(2n)^2 = pi(n)*pi(4n).

For me, n=867 gives the last  p2^2-p1*p4 =  0, 

with  [p2=pi(2*n), p1=pi(n), p4=pi(4*n)]   =  [270, 150, 486]

 

Are there only finitely many n such that 

pi(2n)^2 >= pi(n)*pi(4n) ? 

On Mon, Jun 15, 2026 at 10:44 AM 'Jack Brennen' via SeqFan <seq...@googlegroups.com> wrote:

It appears that the largest such n might be 51342.

I agree.

 

If so, find the largest such n = N. 

Conjecture: pi(2n)^2 < pi(n)*pi(4n) for n > N. 

Well, that's exactly the definition of N, right?

- Maximilian

[Tomasz Ordowski]

Right, thanks!

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