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clarified and repunctuated
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user44143
user44143

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; there are27: 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,floors; and $x=32340221^{1/5}$, for which the first 98 powers have prime floors.

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; there are 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 9 powers have prime floors.

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.

Source Link
user44143
user44143

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; there are 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 9 powers have prime floors.