Prime Powers of Primes

Question

For the purpose of this challenge, a Prime Power of a Prime (PPP) is defined as a number that can be defined as a prime number to the power of a prime number. For example, 9 is a PPP because it can be represented as 3^2. 81 on the other hand is not a PPP because it can only be represented as 3^4, and 4 is not prime. The first few PPPs are: 4, 8, 9, 25, 27, 32, 49, 121, 125, 128, 169, 243, 289, 343... This is OEIS sequence A053810

Your Task:

Write a program or function that for an input integer n returns/outputs the nth PPP, either 1-indexed or 0-indexed, whichever you prefer.

Input:

An integer between 0 and 1,000, received through any reasonable method.

Output:

The PPP at the index indicated by the input.

Test Cases:

These are 1-indexed, and so, if your program takes 0-indexed input, the same output should be arrived at for the stated input - 1.

3  -> 9
6  -> 32
9  -> 125

Scoring:

This code-golf,lowest score in bytes wins!

Answer 1

05AB1E (legacy),  9  7 bytes

Saved 2 bytes thank to @KevinCruijssen

µNÓ0Kp»

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µ           # while counter_variable != input:
 N          #   push iteration counter                       e.g. 125
  Ó         #   get prime exponents                          -> [0, 0, 3]
   0K       #   filter out zeros                             -> [3]
     p      #   is prime?                                    -> [1]
      »     #   join with newlines: we use » instead of J
            #   so that [0,1] is not interpreted as truthy   -> 1
            #   implicit: if 1, increment counter_variable

Answer 2

Husk, 10 bytes

!fȯṗ§*ELpN

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Explanation

!fȯṗ§*ELpN  Implicit input.
 f       N  Filter the natural numbers by this function:
  ȯṗ§*ELp    Argument is a number, say 27.
        p    Prime factors: [3,3,3]
       L     Length: 3
      E      Are all elements equal: 1
    §*       Multiply last two: 3
  ȯṗ         Is it prime? Yes, so 27 is kept.
!           Index into remaining numbers with input.

Answer 3

Jelly, 12 11 10 bytes

1 byte thanks to Dennis.

ÆN€*þ`FṢị@

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Answer 4

Haskell, 95 85 80 bytes

-10 bytes thanks to @Lynn
-5 bytes thanks to @WillNess

0-based

(!!)[x|x<-[2..],p<-[[i|i<-[2..x],all((>)2.gcd i)[2..i-1]]],or[y^e==x|e<-p,y<-p]]

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Explanation

(!!)                    -- point-free expression, partially evaluate index-access operator
[x|x<-[2..]             -- consider all integers x>=2
,p<-                    -- let p be the list of all primes <=x
[[                      -- list of a list so p ends up as a list
i|i<-[2..x],            -- consider all i<=x to be potentially prime
all((>)2.gcd i)[2..i-1] -- if the gcd of i with all smaller integers is
                        -- smaller than 2 then this i is actually prime
 ]],or                  -- if any of the following list entries is true
[y^e==x|                -- the condition y^e==x holds for x with ...
e<-p,y<-p]              -- y and e being prime, i.e. x is a PPP,
]                       -- then add this x to the output sequence / list

Answer 5

Actually, 14 bytes

Based on Mr. Xcoder's Pyth solution. Golfing suggestions welcome. Try it online!

;ur♂P;∙⌠iⁿ⌡MSE

Ungolfing

                Implicit input n
;ur             Duplicate and push [0..n]
   ♂P           Push the 0th to nth primes
     ;∙         Push Cartesian square of the primes
       ⌠iⁿ⌡M    Reduce each list in the Cartesian square by exponentiation
            SE  Sort the list and get the nth index (0-indexed)

Answer 6

Mathematica, 48 bytes

Sort[Join@@Array[(p=Prime)@#^p@#2&,{#,#}]][[#]]&   

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but Martin Ender had a better idea and saved 6 bytes

Mathematica, 42 bytes

Sort[Power@@@Prime@Range@#~Tuples~2][[#]]&   

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Answer 7

Python 2, 163 157 137 136 bytes

• Saved six bytes by using input() rather than defining a function.
• Saved four bytes thanks to Felipe Nardi Batista; merging two loops.
• Saved sixteen bytes thanks to ASCII-only.
• Saved a byte thanks to ArBo.

p=input();r=i=0;e=lambda p:all(p%d for d in range(2,p))
while~-i<p:
 r+=1
 for x in range(r*r):y=x%r;x/=r;i+=x**y==r>e(x)>0<e(y)
print r

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Answer 8

R + numbers, 57 bytes

function(n,x=numbers::Primes(2*n))sort(outer(x,x,"^"))[n]

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outer is such a handy function.

Fairly certain this will always work. Will make a formal argument when I have the time.

Answer 9

Pyth, 15 bytes

e.f/^FR^fP_TSZ2

Try it here! or Verify more test cases.

Explanation

e.f/^FR^fP_TSZ2  - Full program. Q means input.

 .f              - First Q inputs with truthy results. Uses the variable Z.
        fP_TSZ   - Filter the range [1, Z] for primes.
       ^      2  - Cartesian square. Basically the Cartesian product with itself.
    ^FR          - Reduce each list by exponentiation.
  /              - Count the occurrences of Z in ^.
e                - Last element.

Answer 10

Javascript 137 133 bytes

P=n=>{for(p=[i=2];j=++i<n*9;j^i&&p.push(i))
for(;j*j<=i;)j=i%++j?j:i
x=[]
for(i of p)
for(j of p)
x[i**j]=1
return Object.keys(x)[n]}

console.log(P(1000))
console.log(P(800))
console.log(P(9))
console.log(P(5))

**normal algorithem(100ms result) P =n => {

  for(p=[i=2];f=++i<=n*10;!f||p.push(i))
    for(j=0;f&&(x=p[j++])*x<=i;)
      f=i%x
  x=[]
  T=0
  for(i of p)
  for(j of p)
  {
    l= i**j
    if(++T>n &&x.length<l )
    break
    x[l] = 1
  }
  return Object.keys(x)[n]
}

Answer 11

APL (Dyalog Extended), 15 bytes

{⍵⌷∧∊∘.*⍨¯2⍭⍳⍵}

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Explanation

{⍵⌷∧∊∘.*⍨¯2⍭⍳⍵}

              ⍵   Right argument. Our input.
{              }  Wraps the function in dfn syntax which allows us to use ⍵.
             ⍳     Range [1..⍵].
          ¯2⍭     Get the n-th prime for each n in the range.
      ∘.*⍨        Get the prime powers of each prime.
     ∊            Flatten the list.
    ∧             In Extended, this is monadic sort ascending.
 ⍵⌷               Get the input-th index of the list of prime powers of primes.

Answer 12

Perl 6, 50 bytes

{(sort [X**] (^7028,^24)>>.grep(&is-prime))[$_-1]}

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  (^7028,^24)            # create 2 ranges from 0
     >>.grep(&is-prime)  # grep for primes in both
 [X**] ...               # calc each exponential pair (2^2, 2^3, 2^5...)
(sort ... )[$_-1]        # sort and get value at index n-1

The reasons for the 24 and 7028 are that the largest value (n=1000) is 49378729, which is 7027^2, and the largest prime power of 2 which fits under that is 23. So covering 2..7027 ^ 2..23 includes all the items in the first 1000 (and a lot of spares).

Answer 13

Pyth - 13 bytes

e.f&P_lPZ!t{P

Test Suite.

Answer 14

PARI/GP, 48 bytes

f(n)=[x|x<-[1..4^n],isprime(isprimepower(x))][n]

If you do not count the f(n)= part, that is 43 bytes.

---

Another approach without the set notation which does not check so many unnecessary cases:

f(n)=c=0;i=1;while(c<n,i++;isprime(isprimepower(i))&&c++);i

Answer 15

Java 8, 211 bytes

import java.util.*;n->{List l=new Stack();for(int a=2,b;a<132;a++)for(b=2;b<132;b++)if(p(a)*p(b)>0)l.add(Math.pow(a,b));Collections.sort(l);return l.get(n);}int p(int n){for(int i=2;i<n;n=n%i++<1?0:n);return n;}

Very inefficient method.. It basically calculates all PPP's from 2² through 999⁹⁹⁹ 132¹³² and stores it in a List, then sorts that List, and then gets the n'th item from that List.

EDIT: Instead of using 999⁹⁹⁹ which results in a List of 28,225 items, I now use 132¹³² which results in a List of just 1,024 items. This improves the performance quite a bit, and is perfectly acceptable since the challenge states we should support an input from index 0 through 1,000. (Changing 1e3 to 132 doesn't affect the byte-count, though.)

Explanation:

Try it here.

import java.util.*;           // Required import for List, Stack and Collections

n->{                          // Method with integer as parameter and Object as return-type
  List l=new Stack();         //  List to store the PPPs in
  for(int a=2,b;a<132;a++)    //  Loop (1) from 2 to 1,000 (exclusive)
    for(b=2;b<132;b++)        //   Inner loop (2) from 2 to 1,000 (exclusive)
      if(p(a)*p(b)>0)         //    If both `a` and `b` are primes:
        l.add(Math.pow(a,b)); //     Add the power of those two to the List
                              //   End of loop (2) (implicit / single-line body)
                              //  End of loop (1) (implicit / single-line body)
  Collections.sort(l);        //  Sort the filled List
  return l.get(n);            //  Return the `n`'th item of the sorted List of PPPs
}                             // End of method

int p(int n){                 // Separated method with integer as parameter and return-type
  for(int i=2;                //  Index integer (starting at 2)
      i<n;                    //  Loop from 2 to `n` (exclusive)
    n=n%i++<1?                //   If `n` is divisible by `i`:
       0                      //    Change `n` to 0
      :                       //   Else:
       n                      //    Leave `n` the same
  );                          //  End of loop
  return n;                   //  Return `n` (which is now 0 if it wasn't a prime)
}                             // End of separated method

Answer 16

J, 21 bytes

{[:/:~@,[:^/~p:@i.@>:

Zero-indexed anonymous function.

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Trying to get back into the swing of things but I seem to have forgotten all tricks to make good monadic chains.

Short explanation

Constructs a table of prime powers from the 0th prime to the prime at the index of the input plus 1 (to account for 0). Flattens this list and sorts it and then indexes into it. I realize now that this might give incorrect results for some values since the table might not be large enough -- in that case I'd edit in a hardcoded value like 1e4 which should suffice. I can't prove it one way or the other (it passes for the test cases given), so do let me know if this is an issue.

Also 21 bytes

3 :'y{/:~,^/~p:i.>:y'