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


$\Lambda (1) = 2$


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

  • 1
    $\begingroup$ 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 $p>4n$? $\endgroup$ – saulspatz Jun 5 '19 at 10:25
  • $\begingroup$ Now I correct the range of p @saulspatz $\endgroup$ – Pruthviraj Jun 6 '19 at 17:07
  • $\begingroup$ Reference formula math.stackexchange.com/q/3265732/647719 $\endgroup$ – Pruthviraj Jul 12 '19 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 $$\Lambda(n)=\sum_{p \,\nmid\, f(p,2n)} 1 =\sum_{p, p-1 \,|\, 2n} 1$$

  • $\begingroup$ You mean$\Lambda (n)= \sum_{ p-1 \,|\, 2n}1$ $\endgroup$ – Pruthviraj Jun 6 '19 at 7:53
  • $\begingroup$ 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 $\endgroup$ – reuns Jun 6 '19 at 17:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.