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

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 $$\Lambda (n)$$.

• Can you give a reference for that divisibility result, or some indication of how the proof goes? It looks very mysterious to me. What does primality have to do with it? Why $p>4n$? – saulspatz Jun 5 at 10:25
• Now I correct the range of p @saulspatz – Pruthviraj Jun 6 at 17:07
• Reference formula math.stackexchange.com/q/3265732/647719 – Pruthviraj Jul 12 at 17:27

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 $$\Lambda(n)=\sum_{p \,\nmid\, f(p,2n)} 1 =\sum_{p, p-1 \,|\, 2n} 1$$
• You mean$\Lambda (n)= \sum_{ p-1 \,|\, 2n}1$ – Pruthviraj Jun 6 at 7:53
• Note that for such arithmetic function we look at the summatory function $\sum_{n \le x}\Lambda(n) = \sum_{p \le x} \lfloor \frac{2x}{p-1} \rfloor$ which is $\sim \sum_{p \le x} \frac{2x}{p} \sim 2x \log \log x$ by the Mertens theorems or the prime number theorem – reuns Jun 6 at 17:27