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 at 10:25
  • $\begingroup$ I prove this result $p|(1+2^{2n}+3^{2n}+...+p^{2n})$ for $p>4n$ without any special formula as a reference,.i rediscovered power sum formula in new way but it's little bits incomplete.but i assume it's true.may be you can use Faulhaber's formula to simplify $\endgroup$ – Pruthviraj Jun 5 at 10:50
  • $\begingroup$ Now I correct the range of p @saulspatz $\endgroup$ – Pruthviraj Jun 6 at 17:07
  • $\begingroup$ Reference formula math.stackexchange.com/q/3265732/647719 $\endgroup$ – 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 $$\Lambda(n)=\sum_{p \,\nmid\, f(p,2n)} 1 =\sum_{p, p-1 \,|\, 2n} 1$$

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

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