Is there a result in the spirit of Bertrand-Chebyshev which talks about the existence of prime powers between n and 2n (or 3n or something like that) for n large?
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3 Answers
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It follows from the prime number theorem that for fixed $k$, provided $n$ is sufficiently large, there is a prime $p$ such that $p^k$ is between $n$ and $2n$. Does this answer your question?
answered Apr 27, 2011 at 15:48
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$\begingroup$ Thank you very much for your quick answer. Yes, it answers the question, and it goes to show the power of the PNT. $\endgroup$– CheboluCommented Apr 27, 2011 at 16:01
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1$\begingroup$ I wonder if there is an elementary proof of this without using PNT, like Erdős' proof of Bertrand's Postulate? $\endgroup$ Commented Apr 27, 2011 at 17:24
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Actually there is a power of 2. It goes to show the power of binary arithmetic ... : write 2n in binary and write zeroes after the initial one.
answered Apr 27, 2011 at 18:03
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2$\begingroup$ In other words, if $2^{k - 1}$ is the largest power of 2 less than n, then $n \leq 2^k < 2n$? $\endgroup$ Commented Apr 27, 2011 at 18:14
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For fixed $k$, the existence of a prime power $p^k$ between $n$ and $2n$ is (asymptotically) equivalent to the existence of a prime $p$ between $m$ and $\sqrt[k]{2} m$ where $m = \sqrt[k]{n}$. Bachraoui gives an elementary proof here that there exists a prime between $m$ and $\frac{3}{2} m$ for sufficiently large $m$.
I remember reading on MO that it is known that similar elementary proofs exist for showing the existence of primes between $m$ and $(1 + \epsilon) m$ for any $\epsilon > 0$ and, furthermore, that the existence of these proofs itself depends on the PNT. However, I can't track down where I read this.
answered Apr 27, 2011 at 18:06
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3$\begingroup$ Possibly it is this question you have in mind: mathoverflow.net/questions/53498/nontrivial-circular-arguments $\endgroup$– user9072Commented Apr 27, 2011 at 19:00
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