Repeated! Factorials!

Question

Not to be confused with Find the factorial!

Introduction

The factorial of an integer n can be calculated by $$n!=n\times(n-1)\times(n-2)\times(...)\times2\times1$$

This is relatively easy and nothing new. However, factorials can be extended to double factorials, such that $$n!!=n\times(n-2)\times(n-4)\times(...)\times4\times2$$ for even numbers, and $$n!!=n\times(n-2)\times(n-4)\times(...)\times3\times1$$ for odd numbers. But we're not limited to double factorials. For example $$n!!!=n\times(n-3)\times(n-6)\times(...)\times6\times3$$ or $$n!!!=n\times(n-3)\times(n-6)\times(...)\times5\times2$$ or $$n!!!=n\times(n-3)\times(n-6)\times(...)\times4\times1$$ depending on the starting value.

In summary: $${\displaystyle n!^{(k)}={\begin{cases}1&{\text{if }}n=0 \\n&{\text{if }}0<n\leq k\\n\cdot\left({(n-k)!}^{(k)}\right)&{\text{if }}n>k\end{cases}}}$$ where $${\displaystyle n!^{(k)}=n\underbrace{!\dots!}_{k}}$$ Or, in plain English: Subtract the factorial count from the base number repeatedly and multiply all resulting positive integers.

The Challenge

Write a function that will calculate any kind of repeated factorial for any non-negative integer.

Input

Either

A string containing a non-negative base-ten integer, followed by 1 or more exclamation marks. E.g. "6!" or "9!!" or "40!!!!!!!!!!!!!!!!!!!!".

or

The same values represented by two integers: one non-negative base value and one positive value representing the factorial count. This can be done according to any format from the default I/O rules.

Output

The result of said calculation.

Challenge remarks

0! equals 1 by definition. Your code must account for this.
The factorial count is limited by $$ 0 < factorial~count \leq base~value $$outside this range, you are free to output whatever. Aside from 0!, which is the only exception to this rule.

Examples

Input Output 3!!! 3 0! 1 6! 720 9!! 945 10!!!!!!!! 20 40!!!!!!!!!!!!!!!!!!!! 800 420!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! 41697106428257280000000000000000

Try it with an ungolfed Python implementation: Try it online!

General remarks

This is code-golf, so the answer using the fewest bytes in each language wins.
Standard rules, I/O rules and loophole rules apply.
Please include a Try it Online-link to demonstrate your code working.
Please motivate your answer with an explanation of your code.

The examples list 0! but the challenge remarks say that the factorial count will be less than or equal to the base value. — Jonathan Allan
– Jonathan Allan, Commented Aug 5, 2019 at 13:09
@ouflak If it works like 1!, not really. It's more like 1! = 1. 2!! = 2. 3!!! = 3. There's no calculation, because you are at the end of the recursiveness. No 0 in products or else every single factorial would drop down to 0 in the end. — V. Courtois
– V. Courtois, Commented Aug 5, 2019 at 14:16
3!!!!!!! should not be undefined—it should just yield the answer 3. It's the same as 1!!=1 (not undefined). Also your input specification says that there will always be at least one !, so the first example 3 doesn't fit the specification. — Greg Martin
– Greg Martin, Commented Aug 5, 2019 at 18:04
@FabianRöling: But that's not what this is. It's not (3!)! instead it's removing terms from a factorial. It's a misleading name; I came in assuming it was going to be applying the Factorial function repeatedly in a chain and had to read carefully to see what it actually was. Fortunately the question does explain it clearly. A better name might be stride factorial or step factorial or something. — Peter Cordes
– Peter Cordes, Commented Aug 8, 2019 at 5:31

gkgkgkgk · Accepted Answer · 2019-08-05 15:28:25Z

1

JavaScript (Node.js), 35 bytes

(n,i,f=i)=>i-n<1?f:r(n,i-n,f*(i-n))

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answered Aug 5, 2019 at 15:28

gkgkgkgk

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Community · Accepted Answer · 2020-06-17 09:04:33Z

Forth (gforth), 50 bytes

: f 1 1 2over / 1+ 0 do 2over i * - 1 max * loop ;

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Code Explanation

: f \ start a new word definition 1 1 \ add placeholder and accumulator to stack 2over / 1+ \ get the number of times to run the loop (num/factorial + 1) 0 do \ start loop from 0 to num/factorial 2over \ copy num and factorial to the top of the stack i * - \ get the current number to multiply by (num - factorial * i) 1 max \ make sure it can't be 0 or negative [set to 1 if it is] * \ multiply accumulator by result loop \ end loop ; \ end the word definition

Xcali · Accepted Answer · 2019-08-05 18:18:15Z

1

Perl 5 `-Mbigint -p`, 45 bytes

$b=s/!//g;$\=1*$_||1;$\*=$_ while($_-=$b)>0}{

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answered Aug 5, 2019 at 18:18

Xcali

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recursive · Accepted Answer · 2019-08-05 18:29:11Z

1

Stax, 6 bytes

å▐█ΩV∟

Run and debug it

It takes input in the form {count} {base}.

answered Aug 5, 2019 at 18:29

recursive

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Giuseppe · Accepted Answer · 2019-08-05 18:49:44Z

Gaia, 6 bytes

…)¦v%Π

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Takes input as n, k, so input of 3 4 would be 3!!!!.

…	 push [0...n-1], or [] if n == 0 )¦	 increment each value (does nothing if []) v	 reverse list %	 take every k'th element Π	 product; product([]) = 1.

Digital Trauma · Accepted Answer · 2019-08-05 20:43:44Z

1

Bash + coreutils, 27

seq -s\* $[$1+!$1] -$2 1|bc

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answered Aug 5, 2019 at 20:43

Digital Trauma

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Night2 · Accepted Answer · 2019-08-06 06:07:35Z

PHP, 47 bytes

Recursive function with tests

function f($n,$c){return$n>0?f($n-$c,$c)*$n:1;}

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PHP, 54 bytes

Loop version (original answer)

for($n=$argv[1],$f=1;$n>0;$f*=$n,$n-=$argv[2]);echo$f;

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First argument is the the base value and second one is the factorial count.

Supports specified valid range, but outputs big numbers in scientific notation. For example the last sample is shown as 4.1697106428257E+31.

ouflak · Accepted Answer · 2019-08-07 06:30:49Z

1

Lua, 150 145 129 bytes

i=io.read()p,c=i.gsub(i,"!","")p=p+0 function f(n,c)r=1 if n~=0 then while(n>0) do r=r*n n=n-c end end return r end print(f(p,c))

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Ported from Charlie's ArnoldC psuedocode. -16 bytes thanks to Charlie

edited Aug 7, 2019 at 6:30

answered Aug 6, 2019 at 16:40

ouflak

2,0892 gold badges18 silver badges28 bronze badges

1

\$\begingroup\$ You can submit only the function as answer, which can also be improved. \$\endgroup\$

Charlie
– Charlie

2019-08-06 17:13:09 +00:00
Commented Aug 6, 2019 at 17:13
\$\begingroup\$ @Charlie, Yeah when I get more comfy with the idea of headers and footers in TIO, I'll experiment a bit more. \$\endgroup\$

ouflak
– ouflak

2019-08-07 06:32:00 +00:00
Commented Aug 7, 2019 at 6:32

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ar4093 · Accepted Answer · 2019-08-07 07:52:06Z

Python 3 (Input as string), 82 bytes

m=lambda n,s:n<1or n*m(n-s,s) f=lambda x:int(m(int(x.split("!")[0]),x.count("!")))

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Note: the outer int() in f is solely for the fact that without it, 0! will output True

Nice approach! True == 1 in Python, so without conversion it is still valid. — Jitse
– Jitse, Commented Aug 8, 2019 at 5:35

Deadcode · Accepted Answer · 2022-08-19 08:53:16Z

Regex `🐇` (RME / PCRE2 _^v10.35+), 25 bytes

^(x+),((?*x+)(?>\1|.*))*$

Attempt This Online! - PCRE2
Try it on replit.com! - RegexMathEngine, in ECMAScript+(?*)+(?>) mode

Takes its input in unary, as two strings of x characters whose lengths represent the numbers, in the order {factorial count, base value}, joined/delimited by a ,. Returns its output as the number of ways the regex can match. (The rabbit emoji indicates this output method.)

^(x+), # \1 = input factorial count; tail = input base value ( (?*x+) # Assert tail > 0; multiply the number of possible matches by tail (?> # Atomic group – lock in this match once it's finished \1 # tail -= \1 | # or .* # tail = 0 ) )* # Iterate the above as many times as possible, minimum zero $ # Assert tail == 0

Regex `🐇` (RME / PCRE2 _^v10.35+), 26 bytes

^((?=(x*).*,\2)(?*x+)\2)*,

Attempt This Online! - PCRE2
Try it on replit.com! - RegexMathEngine, in ECMAScript+(?*) mode

Takes the two numbers in the order {base value, factorial count}. Although this is 1 byte longer, it does not need an atomic group to be this length, using an atomic lookahead instead.

^ # tail = input base value ( (?= # Atomic lookahead (x*) # \2 = maximum value ≤ tail for which the following matches: .*, # tail = input factorial count \2 # Assert tail ≥ \2 ) (?*x+) # Assert tail > 0; multiply the number of possible matches by tail \2 # tail -= \2 )* # Iterate the above as many times as possible, minimum zero , # Assert tail == 0

The Thonnu · Accepted Answer · 2023-08-09 15:40:40Z

Thunno 2, 5 bytes

RrẈhp

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Explanation

RrẈhp # Implicit input Rr # Push the range [n..1] Ẉ # Uninterleave into k pieces hp # Product of first group # Implicit output

Toby Speight · Accepted Answer · 2025-08-21 17:21:48Z

C (gcc), 26 bytes

f(b,c){b>c?b*=f(b-c,c):0;}

Simple recursive implementation:

int f(int base, int count) { return base > count ? base * f(base-count, count) : b; }

There are equivalent expressions of the same length, e.g.:

f(b,c){b*=b>c?f(b-c,c):1;}

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Stack Exchange Network

Repeated! Factorials!

Introduction

The Challenge

Input

Output

Challenge remarks

Examples

General remarks

42 Answers 42

JavaScript (Node.js), 35 bytes

Forth (gforth), 50 bytes

Code Explanation

Perl 5 `-Mbigint -p`, 45 bytes

Stax, 6 bytes

Gaia, 6 bytes

Bash + coreutils, 27

PHP, 47 bytes

PHP, 54 bytes

Lua, 150 145 129 bytes

Python 3 (Input as string), 82 bytes

Regex `🐇` (RME / PCRE2 _^v10.35+), 25 bytes

Regex `🐇` (RME / PCRE2 _^v10.35+), 26 bytes

Thunno 2, 5 bytes

Explanation

C (gcc), 26 bytes

Linked

Hot Network Questions

Repeated! Factorials!

Introduction

The Challenge

Input

Output

Challenge remarks

Examples

General remarks

42 Answers 42

JavaScript (Node.js), 35 bytes

Forth (gforth), 50 bytes

Code Explanation

Perl 5 -Mbigint -p, 45 bytes

Stax, 6 bytes

Gaia, 6 bytes

Bash + coreutils, 27

PHP, 47 bytes

PHP, 54 bytes

Lua, 150 145 129 bytes

Python 3 (Input as string), 82 bytes

Regex 🐇 (RME / PCRE2 v10.35+), 25 bytes

Regex 🐇 (RME / PCRE2 v10.35+), 26 bytes

Thunno 2, 5 bytes

Explanation

C (gcc), 26 bytes

Linked

Related

Hot Network Questions

Perl 5 `-Mbigint -p`, 45 bytes

Regex `🐇` (RME / PCRE2 _^v10.35+), 25 bytes

Regex `🐇` (RME / PCRE2 _^v10.35+), 26 bytes