Climb a step to a prime

Question

The title of Numberphile's newest video, 13532385396179, is a fixed point of the following function f on the positive integers:

Let n be a positive integer. Write the prime factorization in the usual way, e.g. 60 = 2² · 3 · 5, in which the primes are written in increasing order, and exponents of 1 are omitted. Then bring exponents down to the line and omit all multiplication signs, obtaining a number f (n). [...] for example, f (60) = f (2² · 3 · 5) = 2235.

(The above definition is taken from Problem 5 of Five $1,000 Problems - John H. Conway)

Note that f (13532385396179) = f (13 · 53² · 3853 · 96179) = 13532385396179.

Task

Take a positive composite integer n as input, and output f(n).

Another example

48 = 2⁴ · 3, so f (48) = 243.

Testcases

More testcases are available here.

   4 -> 22
   6 -> 23
   8 -> 23
  48 -> 243
  52 -> 2213
  60 -> 2235
 999 -> 3337
9999 -> 3211101

Answer 1

Python, 166 162 159 bytes

You guys are much better. This is what I used! (the algorithm that solved it calls this)

from primefac import*
def c(n):
 x=factorint(n)
 a=''
 for i in range(len(x)):
  l=min(x.keys())
  a+=str(l)
  if x[l]>1:a+=str(x[l])
  x.pop(l)
 return int(a)

Answer 2

Brachylog, 8 bytes

ḋoọc;1xc

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Explanation

Example input: 60

ḋ          Prime decomposition: [5,3,2,2]
 o         Order: [2,2,3,5]
  ọ        Occurences: [[2,2],[3,1],[5,1]]
   c       Concatenate: [2,2,3,1,5,1]
    ;1x    Execute 1s: [2,2,3,5]
       c   Concatenate: 2235

You can use ℕ₂ˢ (select all integers greater than or equal to 2) instead of ;1x, which is probably more readable and more in the spirit of Brachylog.

Answer 3

Jelly, 6 bytes

ÆFFḟ1V

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Explanation

ÆF      Get prime factorisation of input as prime-exponent pairs.
  F     Flatten.
   ḟ1   Remove 1s.
     V  Effectively flattens the list into a single integer.

Answer 4

Mathematica, 43 36 Bytes

Row@Flatten@FactorInteger@#/. 1->""&

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Answer 5

CJam, 8 bytes

limF:~1-

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Explanation

li  e# Read input and convert to integer.
mF  e# Get prime factorisation as prime-exponent pairs.
:~  e# Flatten.
1-  e# Remove 1s.
    e# Implicitly print a flattened representation of the list.

Answer 6

05AB1E, 10 bytes

Òγʒ¬?gDië?

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Ò          # Push list of prime factors with duplicates
 γ         # Break into chunks of consecutive elements
  ʒ        # For each
   ¬?      #   Print the first element
     gD    #   Push the length of this chunk twice
       ië  #   If not 1
         ? #     Print the length

Answer 7

05AB1E, 12 11 bytes

Òγvy¬sgD≠×J

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Explanation

Ò            # calculate prime factors with duplicates
 γ           # group consecutive equal elements
  vy         # for each group
    ¬        # get the head without popping
     sg      # push the length of the group
       D≠×   # repeat the length (length != 1) times
          J  # join

Answer 8

Pyth, 12 bytes

smjk_>hddr8P

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alternative, 12 bytes

smjk<_AdGr8P

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explanation

smjk_>hddr8P
           PQ  # prime factorization (already in correct order) of the implicit input: [3, 3, 11, 101]
         r8    # length encode: [[2, 3], [1, 11], [1, 101]]
 m             # map over the length encoded list (lambda variable: d)
     >hdd      # take the d[0] last elements of d (so only the last for d[0]==1 and all else)
    _          # reverse that list
  jk           # join into a string
s              # conatenate the list of strings

Answer 9

Pyth, 11 bytes

jksm_-d1r8P

Try here

Answer 10

Python 2, 99 bytes

n=input()
r=''
p=2
while~-n:
 e=0
 while n%p<1:e+=1;n/=p
 r+=str(p)*(e>0)+str(e)*(e>1);p+=1
print r

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If inputs are restricted to be below 2147483659, both str(...) may be replaced by `...` saving 6 bytes (this program will be very slow for numbers affected anyway!).

Answer 11

Ohm, 11 bytes

o:_]D2<?O;J

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Explanation

o:_]D2<?O;J
o           # Push prime factors with powers from input (Format [[prime,power],...]
 :          # For each...
  _          # Push current element
   ]         # flatten
    D        # Duplicate power
     2<? ;   # Is the power smaller than 2?
        O     # Delete top of stacks
          J  # Join

Answer 12

PHP, 88 bytes

for($i=2;1<$a=&$argn;)$a%$i?$i++:$a/=$i+!++$r[$i];foreach($r as$k=>$v)echo$k,$v<2?"":$v;

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Answer 13

Japt, 19 bytes

k ó¥ ®¯1 pZlÃc fÉ q

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Explanation

 k ó¥  ®   ¯  1 pZlà c fÉ  q
Uk ó== mZ{Zs0,1 pZl} c f-1 q  // Ungolfed
                              // Implicit: U = input number
Uk                            // Break U into its prime factors.
   ó==                        // Group into runs of equal items.
       mZ{         }          // Map each item Z in this to
          Zs0,1               //   Z.slice(0, 1) (the array of the first item),
                pZl           //   with Z.length added at the end.
                              // This returns an array of prime-exponent pairs (Jelly's ÆF).
                     c        // Flatten.
                       f-1    // Filter to the items X where X - 1 is truthy (removes '1's).
                           q  // Join the resulting array into a single string.
                              // Implicit: output result of last expression

Answer 14

C#, 206 100 bytes

n=>{var r="";for(int d=1,c;++d<=n;){c=0;while(n%d<1){c++;n/=d;}r+=c>0?d+(c>1?c+"":""):"";}return r;}

Full/Formatted version:

using System;

class P
{
    static void Main()
    {
        Func<int, string> func = n =>
        {
            var r = "";
            for (int d = 1, c; ++d <= n;)
            {
                c = 0;
                while (n % d < 1)
                {
                    c++;
                    n /= d;
                }

                r += c > 0 ? d + (c > 1 ? c + "" : "") : "";
            }

            return r;
        };

        Console.WriteLine(func(4));
        Console.WriteLine(func(6));
        Console.WriteLine(func(8));
        Console.WriteLine(func(48));
        Console.WriteLine(func(52));
        Console.WriteLine(func(60));
        Console.WriteLine(func(999));
        Console.WriteLine(func(9999));

        Console.ReadLine();
    }
}

Answer 15

Javascript - 91 bytes

(x,r='',i=1,n)=>{while(x>i++){for(n=0;x%i<1;n++)x/=i;r+=(n>0?i+'':'')+(n>1?n:'')}return r}

Explanation

(x,r='',i=1,n)=>(          // input x is the number to process, r, i, n are default values only
    while(x>i++){          // iterate over i until x
        for(n=0;x%i<1;n++) // iterate over n until i is not a factor of x
            x/=i;          // factor i out of x
        r+=(n>0?i+'':'')   // append i to r if n > 0
            +(n>1?n:'')    // append n to r if n > 1
                           // i+'' prevents adding i and n before appending to r
    }
    return r               // return r by comma-operator and arrow function syntax
)

Answer 16

Java 8, 103 chars

Pretty straightforward solution.

n->{String r="";int d=2,c;while(n>1){c=0;while(n%d<1){c++;n/=d;}if(c>0)r+=d;if(c>1)r+=c;d++;}return r;}

Ungolfed:

private static Function<Integer, String> f = n->{
    String result = "";
    int divisor = 2, count;
    while (n>1) {
        count = 0;
        while (n % divisor < 1) {
            count++;
            n /= divisor;
        }
        if (count > 0) result += divisor;
        if (count > 1) result += count;
        divisor++;
    }
    return result;
};

Answer 17

Octave, 69 bytes

@(a)printf('%d',(f=[[~,c]=hist(b=factor(a),d=unique(b));d](:))(f~=1))

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Ended up being quite long, but this will generate the desired output.

Essentially we use the histogram function to count the number of occurrences of the unique values in the prime factorisation of the input value.

• The result of the factor() function gives the prime factors in ascending order
• we then find unique() values in that array
• hist() returns the number of occurrences

Once we have the two arrays (one for unique factors, one for counts), we concatenate the arrays vertically (one on top of the other), and then flatten. This interleaves the factors with counts.

Finally we display the result as a string ensuring to skip any 1's in the final array. The only time 1's can appear is if the count was 1 because 1 will never be a prime factor. This elimination is done before converting to a string so it won't affect things like the number 10.

Answer 18

Ruby, 45 + 7 bytes

Requires the flag -rprime.

->n{(n.prime_division.flatten-[1]).join.to_i}

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Answer 19

Pyth - 16 bytes

V{PQpNK/PQNItKpK

Try it

Another solution:

sm`d-.nm(d/PQd){PQ1

Answer 20

R, 72 bytes

x=rle(pracma::factors(scan()));x$l[x$l<2]='';paste0(x$v,x$l,collapse='')

Requires the pracma package, which is not installed on TIO.