Multifactorial

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 $n!$ is defined as $n!=n(n-1)(n-2)...(2)(1)$

A generalization of this is the multifactorials where:

$n!=n(n-1)(n-2)...(2)(1)$

$n!!=n(n-2)(n-4)...$

$n!!!=n(n-3)(n-6)...$

$n!!!!=n(n-4)(n-8)...$

$n!!!!!=n(n-5)(n-10)...$

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.

Perl 6

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

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

}

for 1 .. 5 -> $degree {

   say "$degree: ", map { mfact($_, :$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>