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Let $r=a/b$ be a rational number in lowest terms, larger than $1$, and not an integer (so $b > 1$).

Q. Does the sequence $$ \lfloor r \rfloor, \lfloor r^2 \rfloor, \lfloor r^3 \rfloor, \ldots, \lfloor r^n \rfloor, \ldots$$ always contain an infinite number of primes?

For example, for $r=13/4$, $$(r,r^2,r^3,r^4,r^5, \ldots) = (3.25, 10.56, 34.33, 111.57, 362.59, \ldots)$$ and so the floors are $$(3, 10, 34, 111, 362, \ldots)$$ The next prime after $3$ occurs at $$\lfloor r^{34} \rfloor = 253532870351270971$$ and I only find four primes up to $\lfloor r^{1000} \rfloor$.

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For transcendental r, the answer can be yes: en.wikipedia.org/wiki/Mills%27_constant –  Terry Tao Jan 3 '14 at 3:13
Heuristics is, of course, in favor of the conjecture, but only barely. It seems very hard. I wonder if one can prove that there are only finitely many $k$ such that $[r^{k^2}]$ is prime (for some specific $r$). –  Lev Borisov Jan 3 '14 at 3:28
I wonder if anything is different if one takes $r$ to be real but not an integer or if $r$ is rational but one rounds to the nearest integer. In case of both modifications one can arrange for the closest integer to $r^n$ to always be even, for example $r=1+\sqrt{2}$ has $r^n$ differs from the closest even integer by $\frac{-1}{r^n}$. –  Aaron Meyerowitz Jan 3 '14 at 9:03
@GilKalai: Piatetski-Shapiro showed that if $1<r<c$ for a constant $c$ quite close to $1$ then $[n^r]$ takes on infinitely many primes (he gave an asymptotic formula for the number of such primes). The permissible value of $c$ has been increased subsequently by many authors. –  Lucia Jan 3 '14 at 14:58
Gil, I think that Lev is referring to a Borel-Cantelli type heuristic argument. We imagine that the probability that "$[r^k]$ is prime" is $1/\log(r^k)=1/k\log(r)$. Since the series $$\sum_{k=1}^{\infty}\frac{1}{k\log(r)}$$ diverges, there are ("almost surely") infinitely many instances in which $[r^k]$ is a prime. –  Victor Protsak Jan 4 '14 at 0:21

2 Answers 2

The question is much too hard. Forman and Shapiro proved that $[r^n]$ is composite infinitely often for $r=3/2$ and for $r=4/3$. Dubickas and his students have found a few more results along these lines.

EDIT: Here is a link to the thesis of Novikas, written under the supervision of Dubickas, reviewing and extending the work of Forman and Shapiro.

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Thanks, Gerry! I believe this is the F.-S. paper to which you refer: Forman, W., and H. N. Shapiro. "An arithmetic property of certain rational powers." Communications on Pure and Applied Mathematics 20.3 (1967): 561-573. (Journal link) –  Joseph O'Rourke Jan 3 '14 at 11:55

I recently asked a related question (inspired by this one!) . I accepted an answer but also provided an answer of my own. That answer illustrates the fact that for real $r$ one can arrange to have $\lfloor r^n \rfloor$ all even. One can pick the successive terms of $\lfloor r \rfloor, \lfloor r^2 \rfloor, \lfloor r^3 \rfloor, \ldots, \lfloor r^n \rfloor, \ldots$ fairly freely. By this I mean being slightly vague (or uncommitted) about the value of $r$ and specifying it by providing the sequence of floor values one term at a time. It is easy to preserve freedom of choice enough to have about $\lfloor r \rfloor$ consistent choices at each stage (the smallest of which should be rejected in order to preserve the freedom for the following stage).

Let me relate it to other similar problems and implicitly suggest that for almost all non-integer rationals greater than $1$ there will be infinitely many prime floors. That we might not be able to prove it true for even one rational and that there may not be even one counter-example (at least that we can discover).

Given a positive real $r$ and a real "base" $b \gt 1$ define the base $b$ expansion of $r$ to be the integer sequence $r_b=(x_0,x_1,x_2,\cdots)$ with $x_k=\lfloor rb^k \rfloor$. So this particular integer sequence has $\lim\frac{x_{k+1}}{x_k}=b.$

  • Does the sequence $\pi_{10}=3,31,314,3141,\cdots$ contain infinitely many primes? One would expect so and even that there are about $\frac{\ln{N}}{\ln{10}}$ up to $n=N$ (in an appropriate sense), but one also does not expect to see a proof.

  • What about the same problem with $b=3$ or $b=4$? Same mutatis-mutandis.

  • what about $r_{10}=\lfloor 10^n r \rfloor$ for other reals $r?$ Well, we can pick the decimal expansion to avoid primes but otherwise we might predict infinitely many. The case $r=1/9$ is the question "Are there infinitely many prime repunits?" We know that we can only get a prime for $n$ prime, but might guess that something like $\frac{\ln{\ln{N}}}{\ln{10}}$ is the correct growth rate. We could switch this to questions about Mersenne primes by using the non-terminating binary expansion of $c=1$ (or $c=1/2$).

  • What about $\pi_{\pi}=\lfloor \pi^n \pi \rfloor$ in place of $\lfloor 3^n \pi \rfloor$? Either is a sequence of integers each roughly triple the one before and nothing much should be different.

  • The main point of my answer to my question is that we can build up a feasible integer sequence $\lfloor r^n \rfloor$ or $\lfloor c r^n \rfloor$ term by term somewhat like a decimal expansion. Eventually typically having about $r$ choices for the next term. At times the next choice can be forced by previous ones but that does not persist. So we can avoid primes if we build $r$ by startiing with the expansion. There are specific reals $r$ with special behavior (see the rest of this answer) but no reason to expect that any non-integer rationals are among them.

  • In special cases like $r=\sqrt[3]{7}$ we know that every third term of $r_r$ is non-prime, but that still suggests infinitely many prime floors, just at a slower growth rate.

  • There are certain algebraic integers (Pisot-Vijarayaghavan numbers) $r$ where the distance from $r^n$ to the nearest integer goes to $0$ exponentially fast. Then there is a recurrence relation for $\lfloor r^n \rfloor$ and things can sometimes be cooked to make these all even (for example). With the right definitions ($\lceil rb^n \rceil$ for $r=\frac{1}{\sqrt{5}}, b=\frac{1+\sqrt{5}}2$ ) we get the question "are there infinitely many (odd) prime Fibonacci numbers?" As with repunits, only prime $n$ could give a prime here and the situation should be about the same.

  • Suppose that we look at possible sequences $\lceil r^n \rceil.$ Then every initial segment of the sequence of Fermat numbers $3,5,9,17\cdots$ is feasible and the nested intervals are specifying a real in $\cap (2,\sqrt[k]{2^k+1}].$ These intervals have empty intersection but $\cap [2,\sqrt[k]{2^k+1}]=\{{2\}}$. In this non-standard expansion only the terms with $n=2^j$ could possibly be prime. Hueristics suggest that we know all the Fermat Primes (just $0 \le j \le 4$) but altering the heuristic arguments can (evidently) support a case that a positive proportion of choices $n=2^j$ give Fermat primes.

In conclusion: Give a real constant $r$ and a base $b \in \mathbf{R}$, the integer sequence $\lfloor r b^n\rfloor$ gives the base $b$ expansion of $r$. This is a sequence of integers each roughly $b$ times the previous and generally the previous choices give an interval of length $b$ for possible values of the next term. If we start with the expansion we can avoid primes, but a "random" constant $r\ $ should yield a predictable infinite number of primes subject to adjustments we feel that we understand.

Not that much is different if we use $b=r$. There are nameable reals with exceptional expansions $r_r$ (Integers and Pisot numbers), $r=\sqrt[j]{k}$ a root of an integer does certain special things when $j \mid n$. We have a lot of latitude to construct $r_r$, the "base $r$" expansion of $r$ to get an integer sequence enjoying certain properties, but there is no reason to think that this would lead to a rational number.

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Could you clarify your last sentence: Are you saying that there is no reason to suspect that there are rationals all of whose power-floors are even or have finitely many primes? The latter would support the hypothesis that the rational power-floors always include an infinite number of primes. –  Joseph O'Rourke Jan 8 '14 at 11:56
Let me clarify the last sentence by deleting it and saying something less muddy up front. I think that with probability $1$ any particular non-integer (positive, greater than $1$ ) rational gives infinitely many prime floors (with a predictable logarithmic growth rate). Also that it would be very impressive to prove this for even one rational and that this might be true but essentially unprovable for all rationals (or else there is either an easy counter-example.) –  Aaron Meyerowitz Jan 8 '14 at 16:17
It's not clear to me in what sense one can talk about rational numbers having a certain property "with probability 1". –  Victor Protsak Jan 8 '14 at 18:09
@Victor Protsak: I too have said things like that and am now trying to figure out what I could have meant, but keep running up against the rationals having measure zero and the fact that cardinality-based stuff will be sensitive to the particular ordering chosen. –  Vandermonde Jun 16 at 1:09
@Victor Protsak. In a comment above you said (17 months ago) that for a fixed real $r>1$ it happens "almost surely" that $[r^k]$ is prime infinitely often. Of course integer $r$ is excluded. Is there any reason to take away the "almost surely" if we add the restriction that $r$ must be rational? But "almost surely" means "with probability $1$." By the way, I'd rather change that to $\lfloor r^k \rfloor$ on the chance that $[x]$ means nearest integer and the example $x=1+\sqrt(2).$ –  Aaron Meyerowitz Jun 16 at 5:25

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