Is there a $c > 1$ such that for all $n \ge 1$ the largest integer $\le c^n$ is prime?

Question

Does there exist a real number $c > 1$ such that for every natural number $n > 0$, the number $\lfloor c^n \rfloor$ is prime?

I doubt such a number $c$ is known to exist, since the best similar results I've seen are much weaker. For example, in 1947 William Harold Mills proved that there is a real number $c > 1$ such that for every natural number $n > 0$, the number $\lfloor c^{3^n} \rfloor$ is prime:

• Wikipedia, Mills' constant

and if the Riemann Hypothesis holds, the smallest such $c$ is approximately

$$ c = 1.3063778838630806904686144926\ldots $$

Harder results along these lines are known, e.g. in 2010 Matomäki showed that there exists an uncountable infinity of real numbers $c > 1$ with the property that for every natural $n > 0$, the number $\lfloor c^{2^n} \rfloor$ is prime:

• K. Matomäki, Prime-representing functions, Acta Mathematica Hungarica 128 (4) (2010), 307–314.

However, I haven't seen results like this for $\lfloor c^n \rfloor$. So, I think my question boils down to: is the answer to my original question known to be no, or is it still open?

Answer 1

Lower bounds
The code written below computes lower bounds for the smallest possible such $c$.

• After 15 minutes of computation, my laptop got that $c>41$,
• After 18 hours, it got that $c>59$.

You will find all the details below. Better lower bounds can be get by more computation time.

First steps of the computation
Assume that such $c$ exists. We know that $\lfloor c\rfloor$ is prime, so $c \ge 2$. Then $c^2 \ge 4$, but $\lfloor c^2 \rfloor$ is prime, so $c^2 \ge 5$ and then $c \ge 5^{1/2}$. Now, $\lfloor 5^{1/2} \rfloor = 2$, $\lfloor 5^{2/2} \rfloor = 5$, $\lfloor 5^{3/2} \rfloor = 11$, $\lfloor 5^{4/2} \rfloor = 25$. But $c \ge 5^{1/2}$, so $\lfloor c^4 \rfloor \ge \lfloor 5^{4/2} \rfloor = 25$ and $\lfloor c^4 \rfloor$ is prime, so $\lfloor c^4 \rfloor \ge 29$ (the next prime after $25$). Thus $c \ge 29^{1/4}$. Now $\lfloor 29^{1/4} \rfloor = 2$, $\lfloor 29^{2/4} \rfloor = 5$, $\lfloor 29^{3/4} \rfloor = 12$. But $c \ge 29^{1/4}$, so $\lfloor c^3 \rfloor \ge \lfloor 29^{3/4} \rfloor = 12$ and $\lfloor c^3 \rfloor$ is prime, so $\lfloor c^3 \rfloor \ge 13$ (the next prime after $12$). Thus $c \ge 13^{1/3}$, and so on.

Length records
The computation provides also length records (see all the details below), for example:

• The smallest $x$ such that the sequence $(\lfloor x^n \rfloor)_n$ starts by $27$ prime numbers is $x = 38628691699209543232005302230980383283077723^{1/27} \simeq 41.146$,
• The smallest $x$ such that the sequence $(\lfloor x^n \rfloor)_n$ starts by $35$ prime numbers is $x = 4148748103580890708839735608204733043355735349912591106907117^{1/35} \simeq 53.94$.

Asymptotic: The length record for $y < x$ seems to be approximately $2.5 \dfrac{x}{\ln(x)}$.

Code
Here is the code written in SageMath:

sage: def pp(x,m,N):
....:     if m==0:
....:         return [x,N]
....:     n=1
....:     while is_prime(floor(x^n)):
....:         n+=1
....:     y=(next_prime(floor(x^n)))^(1/n)
....:     if n>N:
....:        print([y.n(),x,n])
....:     if m>0:
....:         return pp(y,m-1,max(N,n))

Computation
Here is the computation proving the lower bounds and the length records mentioned above.

sage: [x,N]=[2,1]
....: for i in range(1022):
....:     print([x.n(),[i]])
....:     [x,N]=pp(x,2000,N)
....: 
[2.00000000000000, [0]]
[2.23606797749979, 2, 2]
[2.32059578710608, sqrt(5), 4]
[2.36051999685294, 31^(1/4), 6]
[2.36103991787210, 173^(1/6), 7]
[2.36144757900740, 409^(1/7), 8]
[2.36184719247834, 967^(1/8), 9]
[11.4754436180012, 3450844193^(1/9), 10]
[11.4772249460124, 39661481813^(1/10), 11]
[13.1754084688783, 2076849234433^(1/11), 12]
[13.9025228053976, 52134281654579^(1/12), 13]
[17.5369538773784, 14838980942616539^(1/13), 14]
[17.5369538773784, 260230524377962793^(1/14), 15]
[17.5369538773784, 4563650703502319197^(1/15), 16]
[17.5369538773784, 80032531899785490253^(1/16), 17]
[19.5947182358701, [1]]
[23.2708572819963, 172111744128569095516889^(1/17), 18]
[23.2708572819963, 4005187834171404283105501^(1/18), 19]
[23.4071193822604, [2]]
[29.2781355078466, [3]]
[29.2836275624131, 7342984643407766159814138311^(1/19), 20]
[29.2836275624130, 215029227494071397857756115239^(1/20), 22]
[29.2836275624130, 184394194768671251848277906031901^(1/22), 23]
[29.3140651944330, [4]]
[29.6863280163833, [5]]
[29.7199886617202, [6]]
[29.7262158681975, [7]]
[29.7986032159883, [8]]
[31.1080969548337, [9]]
[31.1656716760603, [10]]
[31.3602318150116, [11]]
[31.4936884203380, [12]]
[31.5855793146363, [13]]
[31.5877240204566, [14]]
[31.7708284209810, [15]]
[31.7776275935340, [16]]
[31.9228719777914, [17]]
[31.9561321469378, [18]]
[31.9561321470032, 40248230682190652963808004897577443^(1/23), 24]
[31.9561321470032, 1286177778363154944458206697482326941^(1/24), 25]
[37.0580160390721, [19]]
[37.0584389108062, [20]]
[37.0658517174746, [21]]
[37.1639162040819, [22]]
[37.1669125206849, [23]]
[37.1683288435420, [24]]
[37.1687268547021, [25]]
[37.4113116298099, [26]]
[37.4118531804157, [27]]
[37.4119519979104, 2114750864194724801026123348775078610409^(1/25), 26]
[37.4119519979104, 79116957818792486170093330814381247140659^(1/26), 27]
[37.4133160075168, [28]]
[37.5366773413800, [29]]
[37.5399875044127, [30]]
[37.5479790983487, [31]]
[37.5481223970385, [32]]
[37.7259910967192, [33]]
[37.7273014336926, [34]]
[37.7315275341604, [35]]
[37.7349245483495, [36]]
[37.7783534119867, [37]]
[37.7807097942423, [38]]
[37.7835286660001, [39]]
[37.7863629239001, [40]]
[37.8064776911507, [41]]
[37.8562779668531, [42]]
[37.8623613459024, [43]]
[37.9373990583452, [44]]
[37.9424408570224, [45]]
[37.9426346064883, [46]]
[37.9444485557837, [47]]
[41.1462253191129, 38628691699209543232005302230980383283077723^(1/27), 28]
[41.1521261606709, [48]]
[41.1533074278378, [49]]
............... etc ....................
[53.9436697195599, 4148748103580890708839735608204733043355735349912591106907117^(1/35), 36]
[53.9436702309118, [1015]]
[53.9436784982251, [1016]]
[53.9436794835264, [1017]
[53.9436818954090, [1018]]
[53.9436819071867, [1019]]
[53.9436825661574, [1020]]
[59.0854858349532, [1021]]

We took m=2000 because if m is too big then SageMath crashed (on my laptop), then we iterated $1022$ times, so $2044000$ steps in all. We skipped the intermediate prints for $50 < i < 1015$.

Prints explanation
The computation prints two types of list:

• If it is of size $2$ then it corresponds to [x.n(),[i]] which means that after $2000i$ steps we reached the lower bound x (x.n() is just a numerical approximation).
• If it is of size $3$ then it corresponds to [y.n(),x,n] which means that for all $a < y$, the sequence $(\lfloor a^r \rfloor)_r$ starts by less than $n$ prime numbers, and $n-1$ is realized by $x<y$.

Explicit sequences for the length records

Length 27

sage: x=38628691699209543232005302230980383283077723^(1/27)
sage: x.n()
41.1462253191129
sage: [factor(floor(x^n)) for n in range(1,28)]
[41,
 1693,
 69661,
 2866289,
 117936979,
 4852661521,
 199668704371,
 8215613499241,
 338041484174539,
 13909131075052931,
 572308241207202497,
 23548323844696748659,
 968924638801330588309,
 39867591505359597096497,
 1640400903009876073202443,
 67496305168920571652572271,
 2777218180688008234140354019,
 114272045022925494706426972643,
 4701863312189102352562450647343,
 193463927263003124964953243458331,
 7960310342283989048957005995760097,
 327536722953681474755429159335036199,
 13476899802936024915424299817959748633,
 554523555894713462471785930643956846483,
 22816551175599557614676776265536047443603,
 938814955676288945033539543576440477832567,
 38628691699209543232005302230980383283077723]

Length 35

sage: x=4148748103580890708839735608204733043355735349912591106907117^(1/35)
sage: x.n()
53.9436697195599
sage: [factor(floor(x^n)) for n in range(1,36)]
[53,
 2909,
 156971,
 8467631,
 456775117,
 24640126081,
 1329178823209,
 71700783437431,
 3867803380382509,
 208643508091551469,
 11254996489620980821,
 607135813330920533459,
 32751133789239563874359,
 1766716344067858106189647,
 95303162952544927897525549,
 5141002345541482903522204573,
 277324532555372631935243953111,
 14959902989298364748262440497297,
 806992065891368174795394951094273,
 43532113468749303653986100712896887,
 2348281951152618500276459340180295249,
 126674945981380601052391522154658777199,
 6833311447762689695342149169976354827237,
 368613895828998383350681753288959741868319,
 19884386250639755929577516344783599870922529,
 1072636764480669473521399910231762010493578667,
 57861963352202618742765105036618265539241629887,
 3121286640396498386560879368924457870106070776561,
 168373655629623512662702993484676929764429005794377,
 9082692868759332058734233410406296531041203742381329,
 489953784276555631429428597665812242911507414311160479,
 26429905116863028778863328677541914579343251077156126963,
 1425726072343366036233932371293234811318245076148559248257,
 76908896357055934782538327507552055223759094236947310955187,
 4148748103580890708839735608204733043355735349912591106907117]

Answer 2

The question is equivalent to that of asking whether the sets $$ E_n := \bigcup_p [\frac{\log p}{n}, \frac{\log(p+1)}{n})$$ for $n=1,2,\dots$ have a non-empty intersection, where $p$ ranges over primes (since $\log c$ would have to lie in all of the $E_n$ in order for the claimed property of $c$ to hold). Probabilistic heuristics suggest that the answer is negative, though a proof seems well out of reach of current technology. Heuristically, it suffices to show that for any $1 < a < b$, the sets $E_n \cap [a,b]$ have empty intersection "almost surely". In particular, it would heuristically suffice to show that as $N \to \infty$, the "probability" that $$ \bigcup_{N \leq n < 2N} E_n \cap [a,b] = \emptyset$$ goes to zero as $N \to \infty$.

Observe that if $\frac{k}{b^{3N}} \in [a,b]$ lies within $\frac{1}{b^{3N}}$ of $E_n$, then $\frac{k}{b^{3N}} = \frac{\log p}{n} + O( \frac{1}{pn})$ for some prime $p \in [a^n,b^n]$. A calculation using the prime number theorem then shows that the discretized set $$ F_n := \{\frac{k}{b^{3N}} \in [a,b]: k \in {\bf N}, \mathrm{dist}(\frac{k}{b^{3N}}, E_n) \leq \frac{1}{b^{3N}} \}$$ has density $O(1/N)$ in $$ F_0 := \{\frac{k}{b^{3N}} \in [a,b]: k \in {\bf N} \}$$ (we allow implied constants to depend on $a,b$). Since $F_0$ is a finite set of cardinality $O( b^{3N} )$, and there is no reason to expect the sets $F_n$ to ``correlate'' with each other, probabilistic heuristics suggest that $\bigcap_{N \leq n < 2N} F_n$ should be empty with "probability" $O(b^{3N}) (O(1/N))^N$, which goes to zero as $N \to \infty$, and hence the "probability" that $\bigcap_{N \leq n < 2N} E_n \cap [a,b]$ is empty will do so also.

Answer 3

I rewrote Sebastien Palcoux's answer in Mathematica as f[{x_, n_}] := If[PrimeQ[Floor[x^n]], {x, n+1}, {NextPrime[x^n]^(1/n), 1}].

Iterating this ten times with NestList[f, {2, 1}, 10] gives, as in his answer:

$$\{2,1\},\{2,2\},\\ \left\{\sqrt{5},1\right\}, \left\{\sqrt{5},2\right\},\left\{\sqrt{5},3\right\}, \left\{\sqrt{5},4\right\},\\\left\{29^{1/4},1\right\}, \left\{29^{1/4},2\right\},\left\{29^{1/4},3\right\},\\ \left\{13^{1/3},1\right\},\dots$$

Iterating this 1,047,399 times confirms that any $c$ would satisfy $c\ge 40$.

Iterating this three million times, with about four minutes of computation time, finds a highest exponent of 27: it produces three values of $x$ with $41\le x\le 42$ where the floors of $x^1,\ldots,x^{27}$ are all prime. In fact $x^{27}$ is a 44-digit prime in all three of those cases.

Other smaller but interesting cases include $x=7691^{1/3}$ and $x=13591^{1/3}$, for which the first 5 powers have prime floors; and $x=32340221^{1/5}$, for which the first 8 powers have prime floors.