Dividing Divisive Divisors

Question

Given a positive integer \$n\$ you can always find a tuple \$(k_1,k_2,...,k_m)\$ of integers \$k_i \geqslant 2\$ such that \$k_1 \cdot k_2 \cdot ... \cdot k_m = n\$ and $$k_1 | k_2 \text{ , } k_2 | k_3 \text{ , } \ldots \text{ , }k_{m-1}|k_m.$$ Here \$a|b\$ means \$b\$ is a multiple of \$a\$, say "a divides b". If \$n>1\$ all entries \$k_i\$ must be at least \$2\$. For \$n=1\$ we have no such factor and therefore we get an empty tuple.

In case you're curious where this comes from: This decomposition is known as invariant factor decomposition in number theory and it is used in the classification of finitely generated Abelian groups.

Challenge

Given \$n\$ output all such tuples \$(k_1,k_2,...,k_m)\$ for the given \$n\$ exactly once, in whatever order you like. The standard sequence output formats are allowed.

Examples

  1: () (empty tuple)
  2: (2)
  3: (3)
  4: (2,2), (4)
  5: (5)
  6: (6)
  7: (7)
  8: (2,2,2), (2,4), (8)
  9: (3,3), (9)
 10: (10)
 11: (11)
 12: (2,6), (12)
108: (2,54), (3,3,12), (3,6,6), (3,36), (6,18), (108)

Related: http://oeis.org/A000688, List all multiplicative partitions of n

Answer 1

Haskell, 66 62 60 bytes

f n=[n|n>1]:[k:l:m|k<-[2..n],l:m<-f$div n k,mod(gcd n l)k<1]

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Answer 2

05AB1E, 13 bytes

Òœ€.œP€`êʒüÖP

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Ò                      # prime factorization of the input
 œ€.œ                  # all partitions
     P                 # product of each sublist
      €`               # flatten
        ê              # sorted uniquified
         ʒ             # filter by:
          üÖ           #  pairwise divisible-by (yields list of 0s or 1s)
            P          #  product (will be 1 iff the list is all 1s)

Answer 3

Jelly, 17 bytes

ÆfŒ!Œb€ẎP€€QḍƝẠ$Ƈ

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Answer 4

JavaScript (V8),  73  70 bytes

Prints the tuples in descending order \$(k_m,k_{m-1},...,k_1)\$.

f=(n,d=2,a=[])=>n>1?d>n||f(n,d+1,a,d%a[0]||f(n/d,d,[d,...a])):print(a)

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Commented

f = (             // f is a recursive function taking:
  n,              //   n   = input
  d = 2,          //   d   = current divisor
  a = []          //   a[] = list of divisors
) =>              //
  n > 1 ?         // if n is greater than 1:
    d > n ||      //   unless d is greater than n,
    f(            //   do a recursive call with:
      n,          //     -> n unchanged
      d + 1,      //     -> d + 1
      a,          //     -> a[] unchanged
      d % a[0] || //     unless the previous divisor does not divide the current one,
      f(          //     do another recursive call with:
        n / d,    //       -> n / d
        d,        //       -> d unchanged
        [d, ...a] //       -> d preprended to a[]
      )           //     end of inner recursive call
    )             //   end of outer recursive call
  :               // else:
    print(a)      //   this is a valid list of divisors: print it

Answer 5

05AB1E, 17 15 14 bytes

ѦIиæʒPQ}êʒüÖP

Very slow for larger test cases.

-1 byte thanks to @Grimy.

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Explanation:

Ñ               # Get all divisors of the (implicit) input-integer
 ¦              # Remove the first value (the 1)
  Iи            # Repeat this list (flattened) the input amount of times
                #  i.e. with input 4 we now have [2,4,2,4,2,4,2,4]
    æ           # Take the powerset of this list
     ʒ  }       # Filter it by:
      PQ        #  Where the product is equal to the (implicit) input
         ê      # Then sort and uniquify the filtered lists
          ʒ     # And filter it further by:
           ü    #  Loop over each overlapping pair of values
            Ö   #   And check if the first value is divisible by the second value
             P  #  Check if this is truthy for all pairs

                # (after which the result is output implicitly)

Answer 6

Wolfram Language (Mathematica), 78 76 72 71 67 bytes

If[#>(p=1##2),Join@@If[i∣##,##~#0~i,{}]~Table~{i,2,#/p},{{##2}}]&

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Recursive search tree.

---

Brute force solution, 64 bytes:

Union@Cases[Range@#~Tuples~#,{a__,__}/;1a==#&&a>=2&&1∣a:>{a}]&

Trivial modification of my Mathematica solution to List all multiplicative partitions of n.

Since this needs to check \$n^n\$ tuples, try a more efficient version using the same logic.

Answer 7

JavaScript, 115 bytes

f=(n,a=[],i=1)=>{for(;i++<n;)n%i||(a=a.concat(f(n/i).filter(e=>!(e[0]%i)).map(e=>[i].concat(e))));return n>1?a:[a]}

I will write an explanation later

Answer 8

Japt, 22 bytes

â Åï c à f@¥XשXäv eÃâ

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â Åï c à f@¥XשXäv eÃâ     :Implicit input of integer U
â                          :Divisors
  Å                        :Slice off the first element, removing the 1
   ï                       :Cartesian product
     c                     :Flatten
       à                   :Combinations
         f                 :Filter by
          @                :Passing each sub-array X through the following function
           ¥               :  Test U for equality with
            X×             :  X reduced by multiplication
              ©            :  Logical AND with
               Xä          :  Consecutive pairs of X
                 v         :  Reduced by divisibility
                   e       :  All truthy?
                    Ã      :End filter
                     â     :Deduplicate