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Let $\pi$ be the prime counting function.

Under what conditions is it proven true that $\pi(n) > \pi(2n) -\pi(n)$, if at all?

- $\mathrm{lcm}(1, 2, 3, \ldots, n)$?
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- Is there a formula to calculate the sum of all proper divisors of a number?

- What does it mean for a “place” to divide?
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- How to find solutions in set of intgers?
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- Is the equality $1^2+\cdots + 24^2 = 70^2$ just a coincidence?
- When can you simplify the modulus? ($10^{5^{102}} \text{ mod } 35$)
- Argument of the Riemann zeta function on Re(s)=1

According to Wikipedia,

$${x\over\ln x-1}\lt\pi(x)\lt{x\over\ln x-1.1}$$

for large $x$, with the lower bound holding for $x\ge5393$ and the upper bound for $x\ge60184$. Thus the inequality $\pi(2n)\lt2\pi(n)$ holds for $n\ge30092$, since the inequality

$${2n\over\ln(2n)-1.1}\lt{2n\over\ln n-1}$$

holds for *all* $n$ (i.e., since $\ln2\gt0.1$). Whether the inequality $\pi(2n)\lt2\pi(n)$ fails for any small $n$ can be “easily” checked.

I’ve written the program Akiva Weinberger suggested above. This is just a straightforward interpretation of the sieve of Eratosthenes, in R.

```
n = 30092
top = 2*n
isPrime = rep(TRUE, top)
isPrime[1] = FALSE
nextprime = 2
while (nextprime < sqrt(top)){
isPrime[seq(2*nextprime, floor(top/nextprime)*nextprime, nextprime)] = FALSE
nextprime = min(which(isPrime[(nextprime+1):top])) + nextprime
}
#isPrime[n] is now TRUE if n is prime and FALSE otherwise
primePi = cumsum(isPrime) #prime counting function, denoted as pi above
f = primePi[seq(2, 2*n, 2)] - 2*primePi[1:n]
which(f>0)
```

The output is the list `[1]`

. That is, $\pi(2k) > 2\pi(k)$ for $k = 1$ and no other $k <= 30092$. As Barry Cipra showed above, we can prove the desired inequality for larger values from of $k$ from known bounds.

If we want to consider the possibility that $\pi(2k) = 2\pi(k)$, we can replace the last line with `which(f>=0)`

. The output here is `[1, 2, 4, 10]`

. And in fact we have $2 = \pi(4) = 2 \pi(2), 4 = \pi(8) = 2 \pi(4), 8 = \pi(20) = 2\pi(10)$.

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