R is a whole number, a prime to the power of a prime.?

The previous answers assume each digit of R is a prime, but the question says it is a prime to the power of a prime

The possible digits in R are 4=2^2, 8=2^3, 9=3^2. Some possible R values:

4=2^2

8=2^3

9=3^2

67^2=4489

666667^2 = 444444888889

974417^2 = 949488489889

Any prime consisting of 6's followed by a 7 works.

That gives these solutions and probably infinitely many more:

((2*10^n+1)/3)^2, for n = 2, 6, 8, 9, 11, 20, 23, 41, 63, 66, 119, 122, 149, 252, 284, 305, 592, 746, 875, 1204, 1364, 2240, 2403, 5106, 5776, 5813, 12456, 14235, 39606, 55544, 84239.

Some of the large terms correspond to probable primes.

2^5 = 32

3^3 = 27

5^2 = 25

That's the only case where you take the prime^2, since except for 2 and 5, p^2 ends in a 1 or a 9. (This is because p other than 2 or 5 must be odd and end in 1, 3, 7 or 9, so p^2 ends in 1 or 9.)

While searching, I found this interesting number. It's not all primes but I think it's pretty cool:

71^3 = 357911

that 3 5 7 9 11, five consecutive even numbers.

1's not a prime, but

137^3 = 2571353 comes very close to qualifying.

Those were all I could find. I did a computer search at least up to the 50th power for primes up to 113 and didn't find any others. In fact, there weren't any p^n for ANY n <= 50, prime or composite (excluding n=1) where the digits were all prime.

It's a bit surprising that 32 might be the largest such number.

25 is one solution. Twenty five is the same as 5² and both 5 and 2 are prime. Each digit in 25 is also prime.

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