Multifactorial

From Rosetta Code
Multifactorial is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

The factorial of a number, written as is defined as

A generalization of this is the multifactorials where:

Where the products are for positive integers.

If we define the degree of the multifactorial as the difference in successive terms that are multiplied together for a multifactorial (The number of exclamation marks) then the task is to

  1. Write a function that given n and the degree, calculates the multifactorial.
  2. Use the function to generate and display here a table of the first 1..10 members of the first five degrees of multifactorial.

Note: The wikipedia entry on multifactorials gives a different formula. This task uses the Wolfram mathworld definition.

C++

<lang cpp> /*Generate multifactorials to 9

 Nigel_Galloway
 November 14th., 2012.
  • /

int main(void) {

 for (int g = 1; g < 10; g++) {
   int v[12];
   for (int n = 1; n < 11; n++) {
     v[n] = (g < n)? v[n-g]*n : n;
     std::cout << v[n] << " ";
   }
   std::cout << std::endl;
 }
 return 0;

} </lang>

Output:
1 2 6 24 120 720 5040 40320 362880 3628800
1 2 3 8 15 48 105 384 945 3840
1 2 3 4 10 18 28 80 162 280
1 2 3 4 5 12 21 32 45 120
1 2 3 4 5 6 14 24 36 50
1 2 3 4 5 6 7 16 27 40
1 2 3 4 5 6 7 8 18 30
1 2 3 4 5 6 7 8 9 20
1 2 3 4 5 6 7 8 9 10

Haskell

This example does not show the output mentioned in the task description on this page (or a page linked to from here). Please ensure that it meets all task requirements and remove this message.
Note that phrases in task descriptions such as "print and display" and "print and show" for example, indicate that (reasonable length) output be a part of a language's solution.


<lang haskell>mulfac k = 1:s where s = [1 .. k] ++ zipWith (*) s [k+1..]

-- for single n mulfac1 k n = product [n, n-k .. 1]

main = mapM_ (print . take 10 . tail . mulfac) [1..5]</lang>

Perl 6

<lang perl6>sub mfact($n, :$degree = 1) {

   [*] $n, $n - $degree ...^ * <= 0;

}

for 1 .. 5 -> $degree {

   say "$degree: ", map &mfact.assuming(:$degree), 1 .. 10;

}</lang>

Output:
1: 1 2 6 24 120 720 5040 40320 362880 3628800
2: 1 2 3 8 15 48 105 384 945 3840
3: 1 2 3 4 10 18 28 80 162 280
4: 1 2 3 4 5 12 21 32 45 120
5: 1 2 3 4 5 6 14 24 36 50

Python

Python: Iterative

<lang python>>>> from functools import reduce >>> from operator import mul >>> def mfac(n, m): return reduce(mul, range(n, 0, -m))

>>> for m in range(1, 11): print("%2i: %r" % (m, [mfac(n, m) for n in range(1, 11)]))

1: [1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800]
2: [1, 2, 3, 8, 15, 48, 105, 384, 945, 3840]
3: [1, 2, 3, 4, 10, 18, 28, 80, 162, 280]
4: [1, 2, 3, 4, 5, 12, 21, 32, 45, 120]
5: [1, 2, 3, 4, 5, 6, 14, 24, 36, 50]
6: [1, 2, 3, 4, 5, 6, 7, 16, 27, 40]
7: [1, 2, 3, 4, 5, 6, 7, 8, 18, 30]
8: [1, 2, 3, 4, 5, 6, 7, 8, 9, 20]
9: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

10: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] >>> </lang>

Python: Recursive

<lang python>>>> def mfac2(n, m): return n if n <= (m + 1) else n * mfac2(n - m, m)

>>> for m in range(1, 6): print("%2i: %r" % (m, [mfac2(n, m) for n in range(1, 11)]))

1: [1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800]
2: [1, 2, 3, 8, 15, 48, 105, 384, 945, 3840]
3: [1, 2, 3, 4, 10, 18, 28, 80, 162, 280]
4: [1, 2, 3, 4, 5, 12, 21, 32, 45, 120]
5: [1, 2, 3, 4, 5, 6, 14, 24, 36, 50]

>>> </lang>

Tcl

Works with: Tcl version 8.6

<lang tcl>package require Tcl 8.6

proc mfact {n m} {

   set mm [expr {-$m}]
   for {set r $n} {[incr n $mm] > 1} {set r [expr {$r * $n}]} {}
   return $r

}

foreach n {1 2 3 4 5 6 7 8 9 10} {

   puts $n:[join [lmap m {1 2 3 4 5 6 7 8 9 10} {mfact $m $n}] ,]

}</lang>

Output:
1:1,2,6,24,120,720,5040,40320,362880,3628800
2:1,2,3,8,15,48,105,384,945,3840
3:1,2,3,4,10,18,28,80,162,280
4:1,2,3,4,5,12,21,32,45,120
5:1,2,3,4,5,6,14,24,36,50
6:1,2,3,4,5,6,7,16,27,40
7:1,2,3,4,5,6,7,8,18,30
8:1,2,3,4,5,6,7,8,9,20
9:1,2,3,4,5,6,7,8,9,10
10:1,2,3,4,5,6,7,8,9,10