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A Generalized Prime Number Theorem?

Conjecture

Let $n$ and $k$ be positive integers with $n - 50 > k^2 > 0$ and $n$ sufficiently large. Then for the odd primes we have, when $p$ is the biggest odd prime $\le n$, $ 3^k + 5^k + 7^k + 11^k + ... + p^k \sim \frac{n^{k+1}}{(k+1) ( \log(n) - \log(k) ) } $ I wonder if you guys have seen it before ?

How to prove it ?

Any useful references for $k > 1$ ?

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    I slightly edited it.2012-09-04

2 Answers 2

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For $k, you can prove the above using partial summation along with the prime number theorem, but it is provably false for $k\sim x^{u}$ when we assume RH. For a complete solution, take a look at this past answer. Specifically, I prove that $\sum_{p\leq x}p^{k}=\text{li}\left(x^{k+1}\right)+O\left(x^{k+1}E(x)\right),$ where $\text{li}(x)=\int_2^x \frac{1}{\log t}dt$ is the logarithmic integral, and where $E(x)$ is any positive increasing function which bounds the error term $\pi(x)-\text{li}(x)$. This gives the asymptotic $\sum_{p\leq x}p^{k}\sim \frac{x^{k+1}}{(k+1)\log x}$ uniformely for all $k, and if you use the Walfisz bound it can be taken slightly further to $e^{c(\log x)^{3/5}}$ with some doubly logarithmic terms in the exponent.

If you assume the Riemann Hypothesis, this shows that your above expressions is not correct for $k\approx x^u$ where $0. On RH we have that $E(x)\ll x^{\frac{1}{2}+\epsilon}$, and so the asymptotic holds for all $k. Then for $k=x^u$ we have $\log x-\log k=(1-u)\log x$, which yields the asymptotic $(1-u)\frac{x^{k+1}}{(k+1)},$ which is off by the constant factor $(1-u)$.

There may be a simpler way to prove that your expression is incorrect for $k\sim x^u$ that does not require RH. I think you need to be clever though.

Hope that helps,

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    Hmm i wonder why i got the factor (1-u) then. Is it possible that the factor (1-u) fits better for small x and fails for large x ? Could i have made roundoff errors or to brute estimates ? I need some time to think about this. Im a bit confused :)Thanks.2012-09-04