What is relation between $\Lambda (n)$ and $\pi (n)$

Question

Definition

Let's $\Lambda$ be the function, defined as $$\Lambda (n)= \sum_{p \nmid f(p,2n)} 1 $$

And $f(p,2n) = \sum_{i=1}^{p} i^{2n}$

Such that $p$ is prime and $n\in\mathbb{N}.$

We can prove

$$p\mid \sum_{i=1}^{p} i^{2n}$$

For $p>2n+1$

So $$\Lambda (n) \leq \pi (2n+1)$$

Example

$\Lambda (1) = 2$

Question

what is formula for $\Lambda (n)$.

Answer 1

Let $g_p$ be a generator of $(\Bbb{Z}/p\Bbb{Z})^\times$ then

$$f(p,k)=\sum_{i =1}^p i^k\equiv \sum_{i \in (\Bbb{Z}/p\Bbb{Z})^\times}i^k \equiv \sum_{l=0}^{p-2} (g_p^l)^k \equiv \cases{\frac{g_p^{(p-1)k}-1}{g_p^k-1} \equiv 0\bmod p \text{ if } p-1 \nmid k \\ \sum_{l=0}^{p-2} 1 \equiv -1 \bmod p\text{ otherwise}}$$ Thus your function is $$\Lambda(n)=\sum_{p \,\nmid\, f(p,2n)} 1 =\sum_{p, p-1 \,|\, 2n} 1$$