# Special factorials

Special factorials 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.

This task is an aggregation of lesser-known factorials that nevertheless have some mathematical use.

Special factorials
Superfactorial                 n
sf(n) = ∏ k!
k=1
sf(4) = 1! × 2! × 3! × 4! = 288
Hyperfactorial                 n
H(n) = ∏ kk
k=1
H(4) = 11 × 22 × 33 × 44 = 27,648
Alternating factorial                 n
af(n) = ∑ (-1)n-ii!
i=1
af(3) = -12×1! + -11×2! + -10×3! = 5
Exponential factorial n\$ = n(n-1)(n-2)... 4\$ = 4321 = 262,144

• Write a function/procedure/routine for each of the factorials in the table above.
• Show   sf(n),   H(n),   and   af(n)   where   0 ≤ n ≤ 9.   Only show as many numbers as the data types in your language can handle. Bignums are welcome, but not required.
• Show   0\$,   1\$,   2\$,   3\$,   and   4\$.
• Show the number of digits in   5\$.   (Optional)
• Write a function/procedure/routine to find the inverse factorial (sometimes called reverse factorial). That is, if   5! = 120,   then   rf(120) = 5.   This function is simply undefined for most inputs.
• Use the inverse factorial function to show the inverse factorials of 1, 2, 6, 24, 120, 720, 5040, 40320, 362880, and 3628800.
• Show   rf(119).   The result should be undefined.

Notes
Since the factorial inverse of   1   is both   0   and   1,   your function should return   0   in this case since it is normal to use the first match found in a series.

## 11l

Translation of: Nim
```F sf(n)
V result = BigInt(1)
L(i) 2 .. n
result *= factorial(i)
R result

F hf(n)
V result = BigInt(1)
L(i) 2 .. n
result *= pow(BigInt(i), i)
R result

F af(n)
V result = BigInt(0)
V m = ((n [&] 1) << 1) - 1
L(i) 1 .. n
result += m * factorial(i)
m = -m
R result

F ef(n)
V result = BigInt(1)
L(k) 2 .. n
result = pow(k, result)
R result

F rf(n)
I n == 1
R 0
V result = 1
V p = 1
L p < n
result++
p *= result
I p > n
result = -1
R result

V sfs = (0.<10).map(n -> sf(n))
print(‘First ’sfs.len‘ superfactorials: ’sfs.join(‘ ’))

V hfs = (0.<10).map(n -> hf(n))
print(‘First ’hfs.len‘ hyperfactorials: ’hfs.join(‘ ’))

V afs = (0.<10).map(n -> af(n))
print(‘First ’afs.len‘ alternating factorials: ’afs.join(‘ ’))

V efs = (0.<5).map(n -> ef(n))
print(‘First ’efs.len‘ exponential factorials: ’efs.join(‘ ’))

print("\nNumber of digits of ef(5): "String(ef(5)).len)

print("\nReverse factorials:")
L(n) [1, 2, 6, 24, 119, 120, 720, 5040, 40320, 362880, 3628800]
V r = rf(n)
print(f:‘{n:7}: {I r >= 0 {f:‘{r:2}’} E ‘undefined’}’)```
Output:
```First 10 superfactorials: 1 1 2 12 288 34560 24883200 125411328000 5056584744960000 1834933472251084800000
First 10 hyperfactorials: 1 1 4 108 27648 86400000 4031078400000 3319766398771200000 55696437941726556979200000 21577941222941856209168026828800000
First 10 alternating factorials: 0 1 1 5 19 101 619 4421 35899 326981
First 5 exponential factorials: 1 1 2 9 262144

Number of digits of ef(5): 183231

Reverse factorials:
1:  0
2:  2
6:  3
24:  4
119: undefined
120:  5
720:  6
5040:  7
40320:  8
362880:  9
3628800: 10
```

## ALGOL 68

Works with: ALGOL 68G version Any - tested with release 2.8.3.win32

Uses Algol 68G's LONG LONG INT with default precision, which is long enough for the values needed for the task, except exponential factorial( 5 ). The reverse factorial function uses a UNION(INT,STRING) for the result, so the result can either be an integer or the string "undefined".

```BEGIN # show the values of some "special factorials"                         #

# super factorial: returns the product of the factorials up to to n      #
PROC sf = ( INT n )LONG LONG INT:
BEGIN
LONG LONG INT f := 1, p := 1;
FOR i TO n DO
f *:= i;
p *:= f
OD;
p
END # sf # ;
# hyper factorial: returns the product of k^k for k up to n              #
PROC hf = ( INT n )LONG LONG INT:
BEGIN
LONG LONG INT p := 1;
FOR i TO n DO
TO i DO p *:= i OD
OD;
p
END # hf # ;
# alternating factorial: returns the sum of (-1)^(i-1)*i! for i up to n  #
PROC af = ( INT n )LONG LONG INT:
BEGIN
LONG LONG INT s := 0;
LONG LONG INT f := 1;
INT sign        := IF ODD n THEN 1 ELSE - 1 FI;
FOR i TO n DO
f   *:= i;
s   +:= sign * f;
sign := - sign
OD;
s
END # hf # ;
# exponential factorial: returns n^(n-1)^(n-2)...                        #
PROC xf = ( INT n )LONG LONG INT:
BEGIN
LONG LONG INT x := 1;
FOR i TO n DO
LONG LONG INT ix := 1;
TO SHORTEN SHORTEN x DO ix *:= i OD;
x := ix
OD;
x
END # hf # ;
# reverse factorial: returns k if n = k! for some integer k              #
#                           "undefined" otherwise                        #
PROC rf = ( LONG LONG INT n )UNION( INT, STRING ):
IF n < 2
THEN 0
ELSE
LONG LONG INT f := 1;
INT           i := 1;
WHILE f < n DO
f *:= i;
i +:= 1
OD;
IF f = n THEN i - 1 ELSE "undefined" FI
FI # rf # ;

print( ( "super factorials 0..9:", newline, "   " ) );
FOR i FROM 0 TO 9 DO print( ( " ", whole( sf( i ), 0 ) ) ) OD;
print( ( newline, "hyper factorials 0..9:", newline, "   " ) );
FOR i FROM 0 TO 9 DO print( ( " ", whole( hf( i ), 0 ) ) ) OD;
print( ( newline, "alternating factorials 0..9:", newline, "   " ) );
FOR i FROM 0 TO 9 DO print( ( " ", whole( af( i ), 0 ) ) ) OD;
print( ( newline, "exponential factorials 0..4:", newline, "   " ) );
FOR i FROM 0 TO 4 DO print( ( " ", whole( xf( i ), 0 ) ) ) OD;

[]INT rf test = ( 1, 2, 6, 24, 119, 120, 720, 5040, 40320, 362880, 3628800 );
print( ( newline, "reverse factorials:", newline ) );
FOR i FROM LWB rf test TO UPB rf test DO
INT tv = rf test[ i ];
print( ( whole( tv, -8 )
, " -> "
, CASE rf( tv )
IN ( INT    iv ): whole( iv, 0 )
, ( STRING sv ): sv
OUT "Unexpected value returned by rf( " + whole( tv, 0 ) + " )"
ESAC
, newline
)
)
OD

END```
Output:
```super factorials 0..9:
1 1 2 12 288 34560 24883200 125411328000 5056584744960000 1834933472251084800000
hyper factorials 0..9:
1 1 4 108 27648 86400000 4031078400000 3319766398771200000 55696437941726556979200000 21577941222941856209168026828800000
alternating factorials 0..9:
0 1 1 5 19 101 619 4421 35899 326981
exponential factorials 0..4:
1 1 2 9 262144
reverse factorials:
1 -> 0
2 -> 2
6 -> 3
24 -> 4
119 -> undefined
120 -> 5
720 -> 6
5040 -> 7
40320 -> 8
362880 -> 9
3628800 -> 10
```

## Arturo

```super: \$ => [fold.seed:1 1..& [x y] -> x * factorial y]

hyper: \$ => [fold.seed:1 1..& [x y] -> x * y^y]

alternating: \$[n][
fold 1..n [x y] ->
x + (factorial y) * (neg 1)^n-y
]

exponential: \$ => [fold.seed:1 0..& [x y] -> y^x]

rf: function [n][
if 1 = n -> return 0
b: a: <= 1
while -> n > a [
'b + 1
'a * b
]
if a = n -> return b
return null
]

print "First 10 superfactorials:"
print map 0..9 => super

print "\nFirst 10 hyperfactorials:"
print map 0..9 => hyper

print "\nFirst 10 alternating factorials:"
print map 0..9 => alternating

print "\nFirst 5 exponential factorials:"
print map 0..4 => exponential

prints "\nNumber of digits in \$5: "
print size ~"|exponential 5|"
print ""

[1 2 6 24 120 720 5040 40320 362880 3628800 119]
| loop 'x -> print ~"rf(|x|) = |rf x|"
```
Output:
```First 10 superfactorials:
1 1 2 12 288 34560 24883200 125411328000 5056584744960000 1834933472251084800000

First 10 hyperfactorials:
1 1 4 108 27648 86400000 4031078400000 3319766398771200000 55696437941726556979200000 21577941222941856209168026828800000

First 10 alternating factorials:
0.0 1 1 5 19 101 619 4421 35899 326981

First 5 exponential factorials:
0 1 2 9 262144

Number of digits in \$5: 183231

rf(1) = 0
rf(2) = 2
rf(6) = 3
rf(24) = 4
rf(120) = 5
rf(720) = 6
rf(5040) = 7
rf(40320) = 8
rf(362880) = 9
rf(3628800) = 10
rf(119) = null```

## Bruijn

Implementation for numbers encoded in balanced ternary using Mixfix syntax defined in the Math module:

```:import std/Combinator .
:import std/List .
:import std/Math .

factorial [∏ (+1) → 0 | [0]]

superfactorial [∏ (+1) → 0 | factorial]

hyperfactorial [∏ (+1) → 0 | [0 ** 0]]

alternating-factorial y [[=?0 0 ((factorial 0) - (1 --0))]]

exponential-factorial y [[=?0 0 (0 ** (1 --0))]]

:test ((factorial (+4)) =? (+24)) ([[1]])
:test ((superfactorial (+4)) =? (+288)) ([[1]])
:test ((hyperfactorial (+4)) =? (+27648)) ([[1]])
:test ((alternating-factorial (+3)) =? (+5)) ([[1]])
:test ((exponential-factorial (+4)) =? (+262144)) ([[1]])

invfac y [[[compare-case 1 (2 ++1 0) (-1) 0 (∏ (+0) → --1 | ++‣)]]] (+0)

:test ((invfac (+1)) =? (+0)) ([[1]])
:test ((invfac (+2)) =? (+2)) ([[1]])
:test ((invfac (+6)) =? (+3)) ([[1]])
:test ((invfac (+24)) =? (+4)) ([[1]])
:test ((invfac (+120)) =? (+5)) ([[1]])
:test ((invfac (+720)) =? (+6)) ([[1]])
:test ((invfac (+5040)) =? (+7)) ([[1]])
:test ((invfac (+40320)) =? (+8)) ([[1]])
:test ((invfac (+362880)) =? (+9)) ([[1]])
:test ((invfac (+3628800)) =? (+10)) ([[1]])
:test ((invfac (+119)) =? (-1)) ([[1]])

seq-a [((superfactorial 0) : ((hyperfactorial 0) : {}(alternating-factorial 0)))] <\$> (iterate ++‣ (+0))

seq-b exponential-factorial <\$> (iterate ++‣ (+0))

# return first 10/4 elements of both sequences
main [(take (+10) seq-a) : {}(take (+5) seq-b)]```

## C

```#include <math.h>
#include <stdint.h>
#include <stdio.h>

/* n! = 1 * 2 * 3 * ... * n */
uint64_t factorial(int n) {
uint64_t result = 1;
int i;

for (i = 1; i <= n; i++) {
result *= i;
}

return result;
}

/* if(n!) = n */
int inverse_factorial(uint64_t f) {
int p = 1;
int i = 1;

if (f == 1) {
return 0;
}

while (p < f) {
p *= i;
i++;
}

if (p == f) {
return i - 1;
}
return -1;
}

/* sf(n) = 1! * 2! * 3! * ... . n! */
uint64_t super_factorial(int n) {
uint64_t result = 1;
int i;

for (i = 1; i <= n; i++) {
result *= factorial(i);
}

return result;
}

/* H(n) = 1^1 * 2^2 * 3^3 * ... * n^n */
uint64_t hyper_factorial(int n) {
uint64_t result = 1;
int i;

for (i = 1; i <= n; i++) {
result *= (uint64_t)powl(i, i);
}

return result;
}

/* af(n) = -1^(n-1)*1! + -1^(n-2)*2! + ... + -1^(0)*n! */
uint64_t alternating_factorial(int n) {
uint64_t result = 0;
int i;

for (i = 1; i <= n; i++) {
if ((n - i) % 2 == 0) {
result += factorial(i);
} else {
result -= factorial(i);
}
}

return result;
}

/* n\$ = n ^ (n - 1) ^ ... ^ (2) ^ 1 */
uint64_t exponential_factorial(int n) {
uint64_t result = 0;
int i;

for (i = 1; i <= n; i++) {
result = (uint64_t)powl(i, (long double)result);
}

return result;
}

void test_factorial(int count, uint64_t(*func)(int), char *name) {
int i;

printf("First %d %s:\n", count, name);
for (i = 0; i < count ; i++) {
printf("%llu ", func(i));
}
printf("\n");
}

void test_inverse(uint64_t f) {
int n = inverse_factorial(f);
if (n < 0) {
printf("rf(%llu) = No Solution\n", f);
} else {
printf("rf(%llu) = %d\n", f, n);
}
}

int main() {
int i;

/* cannot display the 10th result correctly */
test_factorial(9, super_factorial, "super factorials");
printf("\n");

/* cannot display the 9th result correctly */
test_factorial(8, super_factorial, "hyper factorials");
printf("\n");

test_factorial(10, alternating_factorial, "alternating factorials");
printf("\n");

test_factorial(5, exponential_factorial, "exponential factorials");
printf("\n");

test_inverse(1);
test_inverse(2);
test_inverse(6);
test_inverse(24);
test_inverse(120);
test_inverse(720);
test_inverse(5040);
test_inverse(40320);
test_inverse(362880);
test_inverse(3628800);
test_inverse(119);

return 0;
}
```
Output:
```First 9 super factorials:
1 1 2 12 288 34560 24883200 125411328000 5056584744960000

First 8 hyper factorials:
1 1 2 12 288 34560 24883200 125411328000

First 10 alternating factorials:
0 1 1 5 19 101 619 4421 35899 326981

First 5 exponential factorials:
0 1 2 9 262144

rf(1) = 0
rf(2) = 2
rf(6) = 3
rf(24) = 4
rf(120) = 5
rf(720) = 6
rf(5040) = 7
rf(40320) = 8
rf(362880) = 9
rf(3628800) = 10
rf(119) = No Solution```

## C++

Translation of: C
```#include <cmath>
#include <cstdint>
#include <iostream>
#include <functional>

/* n! = 1 * 2 * 3 * ... * n */
uint64_t factorial(int n) {
uint64_t result = 1;
for (int i = 1; i <= n; i++) {
result *= i;
}
return result;
}

/* if(n!) = n */
int inverse_factorial(uint64_t f) {
int p = 1;
int i = 1;

if (f == 1) {
return 0;
}

while (p < f) {
p *= i;
i++;
}

if (p == f) {
return i - 1;
}
return -1;
}

/* sf(n) = 1! * 2! * 3! * ... . n! */
uint64_t super_factorial(int n) {
uint64_t result = 1;
for (int i = 1; i <= n; i++) {
result *= factorial(i);
}
return result;
}

/* H(n) = 1^1 * 2^2 * 3^3 * ... * n^n */
uint64_t hyper_factorial(int n) {
uint64_t result = 1;
for (int i = 1; i <= n; i++) {
result *= (uint64_t)powl(i, i);
}
return result;
}

/* af(n) = -1^(n-1)*1! + -1^(n-2)*2! + ... + -1^(0)*n! */
uint64_t alternating_factorial(int n) {
uint64_t result = 0;
for (int i = 1; i <= n; i++) {
if ((n - i) % 2 == 0) {
result += factorial(i);
} else {
result -= factorial(i);
}
}
return result;
}

/* n\$ = n ^ (n - 1) ^ ... ^ (2) ^ 1 */
uint64_t exponential_factorial(int n) {
uint64_t result = 0;
for (int i = 1; i <= n; i++) {
result = (uint64_t)powl(i, (long double)result);
}
return result;
}

void test_factorial(int count, std::function<uint64_t(int)> func, const std::string &name) {
std::cout << "First " << count << ' ' << name << '\n';
for (int i = 0; i < count; i++) {
std::cout << func(i) << ' ';
}
std::cout << '\n';
}

void test_inverse(uint64_t f) {
int n = inverse_factorial(f);
if (n < 0) {
std::cout << "rf(" << f << ") = No Solution\n";
} else {
std::cout << "rf(" << f << ") = " << n << '\n';
}
}

int main() {
/* cannot display the 10th result correctly */
test_factorial(9, super_factorial, "super factorials");
std::cout << '\n';

/* cannot display the 9th result correctly */
test_factorial(8, hyper_factorial, "hyper factorials");
std::cout << '\n';

test_factorial(10, alternating_factorial, "alternating factorials");
std::cout << '\n';

test_factorial(5, exponential_factorial, "exponential factorials");
std::cout << '\n';

test_inverse(1);
test_inverse(2);
test_inverse(6);
test_inverse(24);
test_inverse(120);
test_inverse(720);
test_inverse(5040);
test_inverse(40320);
test_inverse(362880);
test_inverse(3628800);
test_inverse(119);

return 0;
}
```
Output:
```First 9 super factorials
1 1 2 12 288 34560 24883200 125411328000 5056584744960000

First 8 hyper factorials
1 1 4 108 27648 86400000 4031078400000 3319766398771200000

First 10 alternating factorials
0 1 1 5 19 101 619 4421 35899 326981

First 5 exponential factorials
0 1 2 9 262144

rf(1) = 0
rf(2) = 2
rf(6) = 3
rf(24) = 4
rf(120) = 5
rf(720) = 6
rf(5040) = 7
rf(40320) = 8
rf(362880) = 9
rf(3628800) = 10
rf(119) = No Solution```

## Factor

Works with: Factor version 0.99 2021-02-05
```USING: formatting io kernel math math.factorials math.functions
math.parser math.ranges prettyprint sequences sequences.extras ;
IN: rosetta-code.special-factorials

: sf ( n -- m ) [1..b] [ n! ] map-product ;
: (H) ( n -- m ) [1..b] [ dup ^ ] map-product ;
: H ( n -- m ) [ 1 ] [ (H) ] if-zero ;
:: af ( n -- m ) n [1..b] [| i | -1 n i - ^ i n! * ] map-sum ;
: \$ ( n -- m ) [1..b] [ ] [ swap ^ ] map-reduce ;

: (rf) ( n -- m )
[ 1 1 ] dip [ dup reach > ]
[ [ 1 + [ * ] keep ] dip ] while swapd = swap and ;

: rf ( n -- m ) dup 1 = [ drop 0 ] [ (rf) ] if ;

: .show ( n quot -- )
[ pprint bl ] compose each-integer nl ; inline

"First 10 superfactorials:" print
10 [ sf ] .show nl

"First 10 hyperfactorials:" print
10 [ H ] .show nl

"First 10 alternating factorials:" print
10 [ af ] .show nl

"First 5 exponential factorials:" print
5 [ \$ ] .show nl

"Number of digits in 5\$:" print
5 \$ log10 >integer 1 + . nl

{ 1 2 6 24 120 720 5040 40320 362880 3628800 119 }
[ dup rf "rf(%d) = %u\n" printf ] each nl
```
Output:
```First 10 superfactorials:
1 1 2 12 288 34560 24883200 125411328000 5056584744960000 1834933472251084800000

First 10 hyperfactorials:
1 1 4 108 27648 86400000 4031078400000 3319766398771200000 55696437941726556979200000 21577941222941856209168026828800000

First 10 alternating factorials:
0 1 1 5 19 101 619 4421 35899 326981

First 5 exponential factorials:
0 1 2 9 262144

Number of digits in 5\$:
183231

rf(1) = 0
rf(2) = 2
rf(6) = 3
rf(24) = 4
rf(120) = 5
rf(720) = 6
rf(5040) = 7
rf(40320) = 8
rf(362880) = 9
rf(3628800) = 10
rf(119) = f
```

## Fermat

```Function Sf(n) = Prod<k=1, n>[k!].

Function H(n) = Prod<k=1, n>[k^k].

Function Af(n) = Sigma<i=1,n>[(-1)^(n-i)i!].

Function Ef(n) = if n < 2 then 1 else n^Ef(n-1) fi.

Function Rf(n) =
for r = 1 to n do
rr:=r!;
if rr=n then Return(r) fi;
if rr>n then Return(-1) fi;
od.

for n=0 to 9 do !!(Sf(n), H(n), Af(n)) od;
!!' ';
for n=0 to 4 do !!Ef(n) od;
!!' ';
for n=1 to 10 do !!Rf(n!) od;
!!Rf(119)
```

## FreeBASIC

Only goes up to H(7) due to overflow. Using a library with big int support is possible, but would only add bloat without being illustrative.

```function factorial(n as uinteger) as ulongint
if n<2 then return 1 else return n*factorial(n-1)
end function

function sf(n as uinteger) as ulongint
dim as ulongint p=1
for k as uinteger = 1 to n
p*=factorial(k)
next k
return p
end function

function H( n as uinteger ) as ulongint
dim as ulongint p=1
for k as uinteger = 1 to n
p*=k^k
next k
return p
end function

function af( n as uinteger ) as longint
dim as longint s=0
for i as uinteger = 1 to n
s += (-1)^(n-i)*factorial(i)
next i
return s
end function

function ef( n as uinteger ) as ulongint
if n<2 then return 1 else return n^ef(n-1)
end function

function rf( n as ulongint ) as integer
dim as uinteger r=0,rr
while true
rr=factorial(r)
if rr>n then return -1
if rr=n then return r
r+=1
wend
end function

for n as uinteger = 0 to 7
print sf(n), H(n), af(n)
next n
print
for n as uinteger = 0 to 4
print ef(n);" ";
next n
print : print
for n as uinteger =0 to 9
print rf(factorial(n));" ";
next n
print rf(119)```

## Go

Translation of: Wren
```package main

import (
"fmt"
"math/big"
)

func sf(n int) *big.Int {
if n < 2 {
return big.NewInt(1)
}
sfact := big.NewInt(1)
fact := big.NewInt(1)
for i := 2; i <= n; i++ {
fact.Mul(fact, big.NewInt(int64(i)))
sfact.Mul(sfact, fact)
}
return sfact
}

func H(n int) *big.Int {
if n < 2 {
return big.NewInt(1)
}
hfact := big.NewInt(1)
for i := 2; i <= n; i++ {
bi := big.NewInt(int64(i))
hfact.Mul(hfact, bi.Exp(bi, bi, nil))
}
return hfact
}

func af(n int) *big.Int {
if n < 1 {
return new(big.Int)
}
afact := new(big.Int)
fact := big.NewInt(1)
sign := new(big.Int)
if n%2 == 0 {
sign.SetInt64(-1)
} else {
sign.SetInt64(1)
}
t := new(big.Int)
for i := 1; i <= n; i++ {
fact.Mul(fact, big.NewInt(int64(i)))
sign.Neg(sign)
}
return afact
}

func ef(n int) *big.Int {
if n < 1 {
return big.NewInt(1)
}
t := big.NewInt(int64(n))
return t.Exp(t, ef(n-1), nil)
}

func rf(n *big.Int) int {
i := 0
fact := big.NewInt(1)
for {
if fact.Cmp(n) == 0 {
return i
}
if fact.Cmp(n) > 0 {
return -1
}
i++
fact.Mul(fact, big.NewInt(int64(i)))
}
}

func main() {
fmt.Println("First 10 superfactorials:")
for i := 0; i < 10; i++ {
fmt.Println(sf(i))
}

fmt.Println("\nFirst 10 hyperfactorials:")
for i := 0; i < 10; i++ {
fmt.Println(H(i))
}

fmt.Println("\nFirst 10 alternating factorials:")
for i := 0; i < 10; i++ {
fmt.Print(af(i), " ")
}

fmt.Println("\n\nFirst 5 exponential factorials:")
for i := 0; i <= 4; i++ {
fmt.Print(ef(i), " ")
}

fmt.Println("\n\nThe number of digits in 5\$ is", len(ef(5).String()))

fmt.Println("\nReverse factorials:")
facts := []int64{1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800, 119}
for _, fact := range facts {
bfact := big.NewInt(fact)
rfact := rf(bfact)
srfact := fmt.Sprintf("%d", rfact)
if rfact == -1 {
srfact = "none"
}
fmt.Printf("%4s <- rf(%d)\n", srfact, fact)
}
}
```
Output:
```First 10 superfactorials:
1
1
2
12
288
34560
24883200
125411328000
5056584744960000
1834933472251084800000

First 10 hyperfactorials:
1
1
4
108
27648
86400000
4031078400000
3319766398771200000
55696437941726556979200000
21577941222941856209168026828800000

First 10 alternating factorials:
0 1 1 5 19 101 619 4421 35899 326981

First 5 exponential factorials:
1 1 2 9 262144

The number of digits in 5\$ is 183231

Reverse factorials:
0 <- rf(1)
2 <- rf(2)
3 <- rf(6)
4 <- rf(24)
5 <- rf(120)
6 <- rf(720)
7 <- rf(5040)
8 <- rf(40320)
9 <- rf(362880)
10 <- rf(3628800)
none <- rf(119)
```

## jq

Translation of: Wren

Works with gojq, the Go implementation of jq

The standard version of jq does not maintain integer precision sufficiently for some of the specific tasks (some of the hyperfactorials and the computation of 5\$) but can otherwise be used.

```# for integer precision:
def power(\$b): . as \$a | reduce range(0;\$b) as \$i (1; . * \$a);

def sf:
. as \$n
| if \$n < 2 then 1
else  {sfact: 1, fact: 1}
| reduce range (2;1+\$n) as \$i (.;
.fact *=  \$i
| .sfact *= .fact)
| .sfact
end;

def H:
. as \$n
| if \$n < 2 then 1
else
reduce range(2;1+\$n) as \$i ( {hfact: 1};
.hfact *= (\$i | power(\$i)))
| .hfact
end;

def af:
. as \$n
| if \$n < 1 then 0
else {afact: 0, fact: 1, sign: (if \$n%2 == 0 then -1 else 1 end)}
| reduce range(1; 1+\$n) as \$i (.;
.fact *= \$i
| .afact += .fact * .sign
| .sign *= -1)
| .afact
end;

def ef: # recursive
. as \$n
| if \$n < 1 then 1
else \$n | power( (\$n-1)|ef )
end;

def rf:
. as \$n
| {i: 0, fact: 1}
| until( .fact >= \$n;
.i += 1
| .fact = .fact * .i)
| if .fact > \$n then null else .i end;```

```"First 10 superfactorials:",
(range(0;10) | sf),

"\nFirst 10 hyperfactorials:",
(range(0; 10) | H),

"\nFirst 10 alternating factorials:",
(range(0;10) | af),

"\n\nFirst 5 exponential factorials:",
(range(0;5) | ef),

"\nThe number of digits in 5\$ is \(5 | ef | tostring | length)",

"\nReverse factorials:",
( 1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800, 119 | "\(rf) <- rf(\(.))")```
Output:

Invocation: gojq -nr -f special-factorials.jq

The output is essentially as for Wren and so is not repeated here.

## Julia

No recursion.

```superfactorial(n) = n < 1 ? 1 : mapreduce(factorial, *, 1:n)
sf(n) = superfactorial(n)

hyperfactorial(n) = n < 1 ? 1 : mapreduce(i -> i^i, *, 1:n)
H(n) = hyperfactorial(n)

alternating_factorial(n) = n < 1 ? 0 : mapreduce(i -> (-1)^(n - i) * factorial(i), +, 1:n)
af(n) = alternating_factorial(n)

exponential_factorial(n) = n < 1 ? 1 : foldl((x, y) -> y^x, 1:n)
n＄(n) = exponential_factorial(n)

function reverse_factorial(n)
n == 1 && return 0
fac = one(n)
for i in 2:10000
fac *= i
fac == n && return i
fac > n && break
end
return nothing
end
rf(n) = reverse_factorial(n)

println("N  Superfactorial    Hyperfactorial", " "^18, "Alternating Factorial   Exponential Factorial\n", "-"^98)
for n in 0:9
print(n, "  ")
for f in [sf, H, af, n＄]
if n < 5 || f != n＄
print(rpad(f(Int128(n)), f == H ? 37 : 24))
end
end
println()
end

println("\nThe number of digits in n＄(5) is ", length(string(n＄(BigInt(5)))))

println("\n\nN  Reverse Factorial\n", "-"^25)
for n in [1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800, 119]
end
```
Output:
```N  Superfactorial    Hyperfactorial                  Alternating Factorial   Exponential Factorial
--------------------------------------------------------------------------------------------------
0  1                       1                                    0                       1
1  1                       1                                    1                       1
2  2                       4                                    1                       2
3  12                      108                                  5                       9
4  288                     27648                                19                      262144
5  34560                   86400000                             101
6  24883200                4031078400000                        619
7  125411328000            3319766398771200000                  4421
8  5056584744960000        55696437941726556979200000           35899
9  1834933472251084800000  21577941222941856209168026828800000  326981

The number of digits in n＄(5) is 183231

N  Reverse Factorial
-------------------------
1         0
2         2
6         3
24        4
120       5
720       6
5040      7
40320     8
362880    9
3628800   10
119       nothing
```

## Kotlin

Translation of: C
```import java.math.BigInteger
import java.util.function.Function

/* n! = 1 * 2 * 3 * ... * n */
fun factorial(n: Int): BigInteger {
val bn = BigInteger.valueOf(n.toLong())
var result = BigInteger.ONE
var i = BigInteger.TWO
while (i <= bn) {
result *= i++
}
return result
}

/* if(n!) = n */
fun inverseFactorial(f: BigInteger): Int {
if (f == BigInteger.ONE) {
return 0
}

var p = BigInteger.ONE
var i = BigInteger.ONE

while (p < f) {
p *= i++
}

if (p == f) {
return i.toInt() - 1
}
return -1
}

/* sf(n) = 1! * 2! * 3! * ... . n! */
fun superFactorial(n: Int): BigInteger {
var result = BigInteger.ONE
for (i in 1..n) {
result *= factorial(i)
}
return result
}

/* H(n) = 1^1 * 2^2 * 3^3 * ... * n^n */
fun hyperFactorial(n: Int): BigInteger {
var result = BigInteger.ONE
for (i in 1..n) {
val bi = BigInteger.valueOf(i.toLong())
result *= bi.pow(i)
}
return result
}

/* af(n) = -1^(n-1)*1! + -1^(n-2)*2! + ... + -1^(0)*n! */
fun alternatingFactorial(n: Int): BigInteger {
var result = BigInteger.ZERO
for (i in 1..n) {
if ((n - i) % 2 == 0) {
result += factorial(i)
} else {
result -= factorial(i)
}
}
return result
}

/* n\$ = n ^ (n - 1) ^ ... ^ (2) ^ 1 */
fun exponentialFactorial(n: Int): BigInteger {
var result = BigInteger.ZERO
for (i in 1..n) {
result = BigInteger.valueOf(i.toLong()).pow(result.toInt())
}
return result
}

fun testFactorial(count: Int, f: Function<Int, BigInteger>, name: String) {
println("First \$count \$name:")
for (i in 0 until count) {
print("\${f.apply(i)} ")
}
println()
}

fun testInverse(f: Long) {
val n = inverseFactorial(BigInteger.valueOf(f))
if (n < 0) {
println("rf(\$f) = No Solution")
} else {
println("rf(\$f) = \$n")
}
}

fun main() {
testFactorial(10, ::superFactorial, "super factorials")
println()

testFactorial(10, ::hyperFactorial, "hyper factorials")
println()

testFactorial(10, ::alternatingFactorial, "alternating factorials")
println()

testFactorial(5, ::exponentialFactorial, "exponential factorials")
println()

testInverse(1)
testInverse(2)
testInverse(6)
testInverse(24)
testInverse(120)
testInverse(720)
testInverse(5040)
testInverse(40320)
testInverse(362880)
testInverse(3628800)
testInverse(119)
}
```
Output:
```First 10 super factorials:
1 1 2 12 288 34560 24883200 125411328000 5056584744960000 1834933472251084800000

First 10 hyper factorials:
1 1 4 108 27648 86400000 4031078400000 3319766398771200000 55696437941726556979200000 21577941222941856209168026828800000

First 10 alternating factorials:
0 1 1 5 19 101 619 4421 35899 326981

First 5 exponential factorials:
0 1 2 9 262144

rf(1) = 0
rf(2) = 2
rf(6) = 3
rf(24) = 4
rf(120) = 5
rf(720) = 6
rf(5040) = 7
rf(40320) = 8
rf(362880) = 9
rf(3628800) = 10
rf(119) = No Solution```

## Lua

Translation of: C
```-- n! = 1 * 2 * 3 * ... * n-1 * n
function factorial(n)
local result = 1
local i = 1
while i <= n do
result = result * i
i = i + 1
end
return result
end

-- if(n!) = n
function inverse_factorial(f)
local p = 1
local i = 1

if f == 1 then
return 0
end

while p < f do
p = p * i
i = i + 1
end

if p == f then
return i - 1
end
return -1
end

-- sf(n) = 1! * 2! * 3! * ... * (n-1)! * n!
function super_factorial(n)
local result = 1
local i = 1
while i <= n do
result = result * factorial(i)
i = i + 1
end
return result
end

-- H(n) = 1^1 * 2^2 * 3^3 * ... * (n-1)^(n-1) * n^n
function hyper_factorial(n)
local result = 1
for i=1, n do
result = result * i ^ i
end
return result
end

-- af(n) = -1^(n-1)*1! + -1^(n-1)*2! + ... + -1^(1)*(n-1)! + -1^(0)*n!
function alternating_factorial(n)
local result = 0
for i=1, n do
if (n - i) % 2 == 0 then
result = result + factorial(i)
else
result = result - factorial(i)
end
end
return result
end

-- n\$ = n ^ (n-1) ^ ... ^ 2 ^ 1
function exponential_factorial(n)
local result = 0
for i=1, n do
result = i ^ result
end
return result
end

function test_factorial(count, f, name)
print("First " .. count .. " " .. name)
for i=1,count do
io.write(math.floor(f(i - 1)) .. "  ")
end
print()
print()
end

function test_inverse(f)
local n = inverse_factorial(f)
if n < 0 then
print("rf(" .. f .. " = No Solution")
else
print("rf(" .. f .. " = " .. n)
end
end

test_factorial(9, super_factorial, "super factorials")
test_factorial(8, hyper_factorial, "hyper factorials")
test_factorial(10, alternating_factorial, "alternating factorials")
test_factorial(5, exponential_factorial, "exponential factorials")

test_inverse(1)
test_inverse(2)
test_inverse(6)
test_inverse(24)
test_inverse(120)
test_inverse(720)
test_inverse(5040)
test_inverse(40320)
test_inverse(362880)
test_inverse(3628800)
test_inverse(119)
```
Output:
```First 9 super factorials
1  1  2  12  288  34560  24883200  125411328000  5056584744960000

First 8 hyper factorials
1  1  4  108  27648  86400000  4031078400000  3319766398771200000

First 10 alternating factorials
0  1  1  5  19  101  619  4421  35899  326981

First 5 exponential factorials
0  1  2  9  262144

rf(1 = 0
rf(2 = 2
rf(6 = 3
rf(24 = 4
rf(120 = 5
rf(720 = 6
rf(5040 = 7
rf(40320 = 8
rf(362880 = 9
rf(3628800 = 10
rf(119 = No Solution```

## Mathematica/Wolfram Language

```ClearAll[sf, expf]
sf[n_] := BarnesG[2 + n]
expf[n_] := Power @@ Range[n, 1, -1]

sf /@ Range[0, 9]
Hyperfactorial /@ Range[0, 9]
AlternatingFactorial /@ Range[0, 9]
expf /@ Range[0, 4]
```
Output:
```{1, 1, 2, 12, 288, 34560, 24883200, 125411328000, 5056584744960000, 1834933472251084800000}
{1, 1, 4, 108, 27648, 86400000, 4031078400000, 3319766398771200000, 55696437941726556979200000, 21577941222941856209168026828800000}
{0, 1, 1, 5, 19, 101, 619, 4421, 35899, 326981}
{1, 1, 2, 9, 262144}```

## Nim

Library: bignum
```import math, strformat, strutils, sugar
import bignum

proc pow(a: int; n: Int): Int =
## Compute a^n for "n" big integer.
var n = n
var a = newInt(a)
if a > 0:
result = newInt(1)
while not n.isZero:
if (n and 1) != 0:
result *= a
n = n shr 1
a *= a

func sf(n: Natural): Int =
result = newInt(1)
for i in 2..n:
result *= fac(i)

func hf(n: Natural): Int =
result = newInt(1)
for i in 2..n:
result *= pow(i, uint(i))

func af(n: Natural): Int =
result = newInt(0)
var m = (n and 1) shl 1 - 1
for i in 1..n:
result += m * fac(i)
m = -m

func ef(n: Natural): Int =
result = newInt(1)
for k in 2..n:
result = pow(k, result)

func rf(n: int | Int): int =
if n == 1: return 0
result = 1
var p = newInt(1)
while p < n:
inc result
p *= result
if p > n: result = -1

let sfs = collect(newSeq, for n in 0..9: sf(n))
echo &"First {sfs.len} superfactorials: ", sfs.join(" ")

let hfs = collect(newSeq, for n in 0..9: hf(n))
echo &"First {hfs.len} hyperfactorials: ", hfs.join(" ")

let afs = collect(newSeq, for n in 0..9: af(n))
echo &"First {afs.len} alternating factorials: ", afs.join(" ")

let efs = collect(newSeq, for n in 0..4: ef(n))
echo &"First {efs.len} exponential factorials: ", efs.join(" ")

echo "\nNumber of digits of ef(5): ", len(\$ef(5))

echo "\nReverse factorials:"
for n in [1, 2, 6, 24, 119, 120, 720, 5040, 40320, 362880, 3628800]:
let r = rf(n)
echo &"{n:7}: ", if r >= 0: &"{r:2}" else: "undefined"
```
Output:
```First 10 superfactorials: 1 1 2 12 288 34560 24883200 125411328000 5056584744960000 1834933472251084800000
First 10 hyperfactorials: 1 1 4 108 27648 86400000 4031078400000 3319766398771200000 55696437941726556979200000 21577941222941856209168026828800000
First 10 alternating factorials: 0 1 1 5 19 101 619 4421 35899 326981
First 5 exponential factorials: 1 1 2 9 262144

Number of digits of ef(5): 183231

Reverse factorials:
1:  0
2:  2
6:  3
24:  4
119: undefined
120:  5
720:  6
5040:  7
40320:  8
362880:  9
3628800: 10```

## Perl

Library: ntheory
```use strict;
use warnings;
use feature qw<signatures say>;
no warnings qw<experimental::signatures>;
use bigint try => 'GMP';
use ntheory qw<vecprod vecsum vecreduce vecfirstidx>;

sub  f (\$n) { vecreduce     { \$a * \$b } 1,              1..\$n }
sub sf (\$n) { vecprod   map { f(\$_) }                   1..\$n }
sub  H (\$n) { vecprod   map { \$_ ** \$_ }                1..\$n }
sub af (\$n) { vecsum    map { (-1) ** (\$n-\$_) * f(\$_) } 1..\$n }
sub ef (\$n) { vecreduce     { \$b ** \$a }                1..\$n }
sub rf (\$n) {
my \$v =  vecfirstidx { f(\$_) >= \$n  } 0..1E6;
\$n == f(\$v) ? \$v : 'Nope'
}

say 'sf : ' . join ' ', map { sf \$_ } 0..9;
say 'H  : ' . join ' ', map {  H \$_ } 0..9;
say 'af : ' . join ' ', map { af \$_ } 0..9;
say 'ef : ' . join ' ', map { ef \$_ } 1..4;
say '5\$ has ' . length(5**4**3**2) . ' digits';
say 'rf : ' . join ' ', map { rf \$_ } <1 2 6 24 120 720 5040 40320 362880 3628800>;
say 'rf(119) = ' . rf(119);
```
Output:
```sf : 1 1 2 12 288 34560 24883200 125411328000 5056584744960000 1834933472251084800000
H  : 1 1 4 108 27648 86400000 4031078400000 3319766398771200000 55696437941726556979200000 21577941222941856209168026828800000
af : 0 1 1 5 19 101 619 4421 35899 326981
ef : 1 2 9 262144
5\$ has 183231 digits
rf : 0 2 3 4 5 6 7 8 9 10
rf(119) = Nope```

## Phix

Library: Phix/mpfr

Rather than leave this as four somewhat disjoint tasks with quite a bit of repetition, I commoned-up init/loop/print into test(), which means sf/H/af/ef aren't usable independently as-is, but reconstitution should be easy enough (along with somewhat saner and thread-safe routine-local vars).

```with javascript_semantics
include mpfr.e

mpz {r,fn} = mpz_inits(2) -- res, scratch var

procedure sf(integer i)
mpz_fac_ui(fn, i)
mpz_mul(r,r,fn)
end procedure

procedure H(integer i)
mpz_ui_pow_ui(fn, i, i)
mpz_mul(r,r,fn)
end procedure

integer sgn = 0
procedure af(integer i)
mpz_fac_ui(fn, i)
mpz_mul_si(fn,fn,sgn)
sgn *= -1
end procedure

procedure ef(integer i)
integer e = mpz_get_integer(r)
mpz_set_si(r,i)
mpz_pow_ui(r, r, e)
end procedure

procedure test(string fmt, integer fn, init=1, m=9)
sequence res = {}
for n=0 to m do
mpz_set_si(r,init)
sgn = iff(and_bits(n,1)?1:-1) -- (af only)
for i=1 to n do
fn(i)       -- or papply(tagset(n),fn)
end for
res = append(res,mpz_get_str(r))
end for
printf(1,fmt,{join(res)})
end procedure
test("First 10 superfactorials: %s\n",sf)
test("First 10 hyperfactorials: %s\n",H)
test("First 10 alternating factorials: %s\n",af,0)
test("First 5 exponential factorials: %s\n",ef,1,4)
ef(5) -- (nb now only works because ef(4) was just called)
printf(1,"Number of digits in 5\$: %,d\n",mpz_sizeinbase(r,10))

function rf(integer n)
if n=1 then return "0" end if
integer fac = 1, i = 1
while fac<n do
fac *= i
if fac=n then return sprint(i) end if
i += 1
end while
return "undefined"
end function
printf(1,"Reverse factorials: %s\n",{join(apply({1,2,6,24,120,720,5040,40320,362880,3628800,119},rf))})
```
Output:
```First 10 superfactorials: 1 1 2 12 288 34560 24883200 125411328000 5056584744960000 1834933472251084800000
First 10 hyperfactorials: 1 1 4 108 27648 86400000 4031078400000 3319766398771200000 55696437941726556979200000 21577941222941856209168026828800000
First 10 alternating factorials: 0 1 1 5 19 101 619 4421 35899 326981
First 5 exponential factorials: 1 1 2 9 262144
Number of digits in 5\$: 183,231
Reverse factorials: 0 2 3 4 5 6 7 8 9 10 undefined
```

## PureBasic

Translation of: FreeBASIC
```Procedure.q factorial(n.i)
If n < 2
ProcedureReturn 1
Else
ProcedureReturn n * Factorial(n - 1)
EndIf
EndProcedure

Procedure.q sf(n.i)
p.i = 1
For k.i = 1 To n
p = p * factorial(k)
Next k
ProcedureReturn p
EndProcedure

Procedure.q H(n.i)
p.i = 1
For k.i = 1 To n
p = p * Pow(k, k)
Next k
ProcedureReturn p
EndProcedure

Procedure.q af(n.i)
s.i = 0
For i.i = 1 To n
s = s + Pow((-1), (n-i)) * factorial(i)
Next i
ProcedureReturn s
EndProcedure

Procedure.q ef(n.i)
If n < 2
ProcedureReturn 1
Else
ProcedureReturn Pow(n, ef(n-1))
EndIf
EndProcedure

Procedure.i rf(n.i)
r.i = 0
While #True
rr.i = factorial(r)
If rr > n : ProcedureReturn -1 : EndIf
If rr = n : ProcedureReturn r : EndIf
r + 1
Wend
EndProcedure

OpenConsole()
PrintN("First 8 ...")
PrintN(" superfactorials      hyperfactorials  alternating factorials")
For n.i = 0 To 7 ;con 8 o más necesitaríamos BigInt
PrintN(RSet(Str(sf(n)),16) + "  " + RSet(Str(H(n)),19) + "     " + RSet(Str(af(n)),19))
Next n

PrintN(#CRLF\$ + #CRLF\$ + "First 5 exponential factorials:")
For n.i = 0 To 4
Print(Str(ef(n)) + " ")
Next n

PrintN(#CRLF\$ + #CRLF\$ + "Reverse factorials:")
For n.i = 1 To 10
PrintN(RSet(Str(rf(factorial(n))),2) + " <- rf(" + Str(factorial(n)) + ")")
Next n
PrintN(RSet(Str(rf(factorial(119))),2) + " <- rf(119)")

PrintN(#CRLF\$ + "Press ENTER to exit"): Input()
CloseConsole()```

## Python

Writing a custom factorial instead of math.prod is trivial but using a standard library tool is always a nice change.

It also means less code :)

```#Aamrun, 5th October 2021

from math import prod

def superFactorial(n):
return prod([prod(range(1,i+1)) for i in range(1,n+1)])

def hyperFactorial(n):
return prod([i**i for i in range(1,n+1)])

def alternatingFactorial(n):
return sum([(-1)**(n-i)*prod(range(1,i+1)) for i in range(1,n+1)])

def exponentialFactorial(n):
if n in [0,1]:
return 1
else:
return n**exponentialFactorial(n-1)

def inverseFactorial(n):
i = 1
while True:
if n == prod(range(1,i)):
return i-1
elif n < prod(range(1,i)):
return "undefined"
i+=1

print("Superfactorials for [0,9] :")
print({"sf(" + str(i) + ") " : superFactorial(i) for i in range(0,10)})

print("\nHyperfactorials for [0,9] :")
print({"H(" + str(i) + ") "  : hyperFactorial(i) for i in range(0,10)})

print("\nAlternating factorials for [0,9] :")
print({"af(" + str(i) + ") " : alternatingFactorial(i) for i in range(0,10)})

print("\nExponential factorials for [0,4] :")
print({str(i) + "\$ " : exponentialFactorial(i) for i in range(0,5)})

print("\nDigits in 5\$ : " , len(str(exponentialFactorial(5))))

factorialSet = [1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800]

print("\nInverse factorials for " , factorialSet)
print({"rf(" + str(i) + ") ":inverseFactorial(i) for i in factorialSet})

print("\nrf(119) : " + inverseFactorial(119))
```
Output:
```Superfactorials for [0,9] :
{'sf(0) ': 1, 'sf(1) ': 1, 'sf(2) ': 2, 'sf(3) ': 12, 'sf(4) ': 288, 'sf(5) ': 34560, 'sf(6) ': 24883200, 'sf(7) ': 125411328000, 'sf(8) ': 5056584744960000, 'sf(9) ': 1834933472251084800000}

Hyperfactorials for [0,9] :
{'H(0) ': 1, 'H(1) ': 1, 'H(2) ': 4, 'H(3) ': 108, 'H(4) ': 27648, 'H(5) ': 86400000, 'H(6) ': 4031078400000, 'H(7) ': 3319766398771200000, 'H(8) ': 55696437941726556979200000, 'H(9) ': 21577941222941856209168026828800000}

Alternating factorials for [0,9] :
{'af(0) ': 0, 'af(1) ': 1, 'af(2) ': 1, 'af(3) ': 5, 'af(4) ': 19, 'af(5) ': 101, 'af(6) ': 619, 'af(7) ': 4421, 'af(8) ': 35899, 'af(9) ': 326981}

Exponential factorials for [0,4] :
{'0\$ ': 1, '1\$ ': 1, '2\$ ': 2, '3\$ ': 9, '4\$ ': 262144}

Digits in 5\$ :  183231

Inverse factorials for  [1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800]
{'rf(1) ': 0, 'rf(2) ': 2, 'rf(6) ': 3, 'rf(24) ': 4, 'rf(120) ': 5, 'rf(720) ': 6, 'rf(5040) ': 7, 'rf(40320) ': 8, 'rf(362880) ': 9, 'rf(3628800) ': 10}

rf(119) : undefined
```

## Quackery

```  [ 1 & ]                  is odd        ( n --> b )

[ 1 swap
[ 10 / dup 0 > while
dip 1+ again ]
drop ]                 is digitcount ( n --> n )

[ 1 1 rot times
[ i^ 1+ *
tuck * swap ]
drop ]                 is s!         ( n --> n )

[ 1 swap times
[ i^ 1+ dup ** * ] ] is h!         ( n --> n )

[ 0 1 rot times
[ i^ 1+ * tuck
i odd iff - else +
swap ]
drop ]                 is a!         ( n --> n )

[ dup 0 = if done
dup 1 - recurse ** ]   is **!        ( n --> n )

[ this ]                 is undefined  (   --> t )

[ dup 1 = iff
[ drop 0 ] done
1 swap
[ over /mod 0 != iff
[ drop undefined
swap ]
done
dip 1+
dup 1 = until
dip [ 1 - ] ]
drop ]               is i!         ( n --> n )

say "Superfactorials:" sp
10 times [ i^ s! echo sp ]
cr cr
say "Hyperfactorials:" sp
10 times [ i^ h! echo sp ]
cr cr
say "Alternating factorials: "
10 times [ i^ a! echo sp ]
cr cr
say "Exponential factorials: "
5 times [ i^ **! echo sp ]
cr cr
say "Number of digits in \$5: "
5 **! digitcount echo
cr cr
say "Inverse factorials: "
' [ 1 2 6 24 119 120 720 5040
40320 362880 3628800 ]
witheach [ i! echo sp ]```
Output:
```Superfactorials: 1 1 2 12 288 34560 24883200 125411328000 5056584744960000 1834933472251084800000

Hyperfactorials: 1 1 4 108 27648 86400000 4031078400000 3319766398771200000 55696437941726556979200000 21577941222941856209168026828800000

Alternating factorials: 0 1 1 5 19 101 619 4421 35899 326981

Exponential factorials: 0 1 2 9 262144

Number of digits in \$5: 183231

Inverse factorials: 0 2 3 4 undefined 5 6 7 8 9 10
```

## Raku

```sub postfix:<!> (\$n) {   [*] 1 .. \$n }
sub postfix:<\$> (\$n) { [R**] 1 .. \$n }

sub sf (\$n) { [*] map {                      \$_! }, 1 .. \$n }
sub H  (\$n) { [*] map {                \$_ ** \$_  }, 1 .. \$n }
sub af (\$n) { [+] map { (-1) ** (\$n - \$_) *  \$_! }, 1 .. \$n }
sub rf (\$n) {
state @f = 1, |[\*] 1..*;
\$n == .value ?? .key !! Nil given @f.first: :p, * >= \$n;
}

say 'sf : ', map &sf , 0..9;
say 'H  : ', map &H  , 0..9;
say 'af : ', map &af , 0..9;
say '\$  : ', map *\$  , 1..4;

say '5\$ has ', 5\$.chars, ' digits';

say 'rf : ', map &rf, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800;
say 'rf(119) = ', rf(119).raku;
```
Output:
```sf : (1 1 2 12 288 34560 24883200 125411328000 5056584744960000 1834933472251084800000)
H  : (1 1 4 108 27648 86400000 4031078400000 3319766398771200000 55696437941726556979200000 21577941222941856209168026828800000)
af : (0 1 1 5 19 101 619 4421 35899 326981)
\$  : (1 2 9 262144)
5\$ has 183231 digits
rf : (0 2 3 4 5 6 7 8 9 10)
rf(119) = Nil```

## REXX

```/* REXX program to calculate Special factorials */

numeric digits 35

line = "superfactorials 0-9:       "
do n = 0 to 9
line = line superfactorial(n)
end
say line

line = "hyperfactorials 0-9:       "
do n = 0 to 9
line = line hyperfactorial(n)
end
say line

line = "alternating factorials 0-9:"
do n = 0 to 9
line = line alternatingfactorial(n)
end
say line

line = "exponential factorials 0-4:"
do n = 0 to 4
line = line exponentialfactorial(n)
end
say line

say "exponential factorial 5:   ",
length(format(exponentialfactorial(5), , , 0)) "digits"

line = "inverse factorials:        "
numbers = "1 2 6 24 120 720 5040 40320 362880 3628800 119"
do i = 1 to words(numbers)
line = line inversefactorial(word(numbers,i))
end
say line

return

superfactorial: procedure
parse arg n
sf = 1
f = 1
do k = 1 to n
f = f * k
sf = sf * f
end
return sf

hyperfactorial: procedure
parse arg n
hf = 1
do k = 1 to n
hf = hf * k ** k
end
return hf

alternatingfactorial: procedure
parse arg n
af = 0
f = 1
do i = 1 to n
f = f * i
af = af + (-1) ** (n - i) * f
end
return af

exponentialfactorial: procedure
parse arg n
ef = 1
do i = 1 to n
ef = i ** ef
end
return ef

inversefactorial: procedure
parse arg f
n = 1
do i = 2 while n < f
n = n * i
end
if n = f then
if i > 2 then
return i - 1
else
return 0
else
return "undefined"
```
Output:
```superfactorials 0-9:        1 1 2 12 288 34560 24883200 125411328000 5056584744960000 1834933472251084800000
hyperfactorials 0-9:        1 1 4 108 27648 86400000 4031078400000 3319766398771200000 55696437941726556979200000 21577941222941856209168026828800000
alternating factorials 0-9: 0 1 1 5 19 101 619 4421 35899 326981
exponential factorials 0-4: 1 1 2 9 262144
exponential factorial 5:    183231 digits
inverse factorials:         0 2 3 4 5 6 7 8 9 10 undefined
```

## RPL

Super factorial:

```≪ 1 1 ROT FOR j j FACT * NEXT
≫ 'SFACT' STO
```

Hyper factorial:

```≪ 1 1 ROT FOR j j DUP ^ * NEXT
≫ 'HFACT' STO
```

Alternating factorial:

```≪ → n
≪ 0 IF n THEN
1 n FOR j -1 n j - ^ j FACT * + NEXT
≫ ≫ 'AFACT' STO
```

Exponential factorial:

```≪ IF DUP THEN
DUP 1 - 1 FOR j j ^ -1 STEP
ELSE NOT END
≫ 'EFACT' STO
```

Inverse factorial - this function actually inverses Gamma(x+1), so its result is always defined:

```≪ 'FACT(x)' OVER - 'x' ROT LN ROOT 'x' PURGE
≫ 'IFACT' STO
```
```≪ { } 0 9 FOR j j SFACT + NEXT ≫ EVAL
≪ { } 0 9 FOR j j HFACT + NEXT ≫ EVAL
≪ { } 0 9 FOR j j AFACT + NEXT ≫ EVAL
≪ { } 0 4 FOR j j EFACT + NEXT ≫ EVAL
3628800 IFACT
119 IFACT
```
Output:
```6: { 1 1 2 12 288 34560 24883200 125411328000 5.05658474496E+15 1.83493347225E+21 }
5: { 1 1 4 108 27648 86400000 4.0310784E+12 3.31976639877E+18 5.56964379417E+25 2.15779412229E+34 }
4: { 0 1 1 5 19 101 619 4421 35899 326981 }
3: { 0 1 2 9 262144 }
2: 10
1: 4.99509387149
```

## Seed7

```\$ include "seed7_05.s7i";
include "bigint.s7i";

const func bigInteger: superfactorial (in bigInteger: limit) is func
result
var bigInteger: product is 1_;
local
var bigInteger: k is 0_;
begin
for k range 1_ to limit do
product *:= !k;   # prefix ! calculates the factorial in Seed7
end for;
end func;

const func bigInteger: hyperfactorial (in bigInteger: limit) is func
result
var bigInteger: product is 1_;
local
var bigInteger: k is 0_;
begin
for k range 1_ to limit do
product *:= k ** ord(k);
end for;
end func;

const func bigInteger: alternating (in bigInteger: limit) is func
result
var bigInteger: sum is 0_;
local
var bigInteger: i is 0_;
begin
for i range 1_ to limit do
sum +:= (-1_) ** ord(limit - i) * !i;
end for;
end func;

const func bigInteger: exponential (in bigInteger: limit) is func
result
var bigInteger: pow is 0_;
local
var bigInteger: n is 0_;
begin
for n range 1_ to limit do
pow := n ** ord(pow);
end for;
end func;

const func integer: invFactorial (in integer: n) is func
result
var integer: r is 0;
local
var integer: a is 1;
var integer: b is 1;
begin
if n <> 1 then
while n > a do
incr(b);
a *:= b;
end while;
if a = n then
r := b;
else
r := -1;
end if;
end if;
end func;

const proc: show (in bigInteger: limit, inout bigInteger: aVariable,
ref func bigInteger: anExpression) is func

begin
for aVariable range 0_ to limit do
write(anExpression <& " ");
end for;
writeln; writeln;
end func;

const proc: main is func
local
var bigInteger: x is 0_;
var integer: n is 0;
begin
writeln("First 10 superfactorials:");
show(9_, x, superfactorial(x));
writeln("First 10 hyperfactorials:");
show(9_, x, hyperfactorial(x));
writeln("First 10 alternating factorials:");
show(9_, x, alternating(x));
writeln("First 5 exponential factorials:");
show(4_, x, exponential(x));
for n range [] (1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800, 119) do
writeln("invFactorial(" <& n <& ") = " <& invFactorial(n));
end for;
end func;```
Output:
```First 10 superfactorials:
1 1 2 12 288 34560 24883200 125411328000 5056584744960000 1834933472251084800000

First 10 hyperfactorials:
1 1 4 108 27648 86400000 4031078400000 3319766398771200000 55696437941726556979200000 21577941222941856209168026828800000

First 10 alternating factorials:
0 1 1 5 19 101 619 4421 35899 326981

First 5 exponential factorials:
0 1 2 9 262144

invFactorial(1) = 0
invFactorial(2) = 2
invFactorial(6) = 3
invFactorial(24) = 4
invFactorial(120) = 5
invFactorial(720) = 6
invFactorial(5040) = 7
invFactorial(40320) = 8
invFactorial(362880) = 9
invFactorial(3628800) = 10
invFactorial(119) = -1
```

## Sidef

```func sf(n) { 1..n -> prod {|k| k! } }
func H(n)  { 1..n -> prod {|k| k**k } }
func af(n) { 1..n -> sum  {|k| (-1)**(n-k) * k! } }
func ef(n) { 1..n -> reduce({|a,b| b**a }, 1) }

func factorial_valuation(n,p) {
(n - n.sumdigits(p)) / (p-1)
}

var n = (k * (p - 1))

while (factorial_valuation(n, p) < k) {
n -= (n % p)
n += p
}

return n
}

func rf(f) {

return nil if (f < 0)
return 0   if (f <= 1)

var t = valuation(f, 2) || return nil
var d = factor(n + 1)[-1]

if (f.valuation(d) == factorial_valuation(n+1, d)) {
++n
}

for p in (primes(2, n)) {
var v = factorial_valuation(n, p)
f.valuation(p) == v || return nil
f /= p**v
}

(f == 1) ? n : nil
}

say ('sf : ', 10.of(sf).join(' '))
say ('H  : ', 10.of(H).join(' '))
say ('af : ', 10.of(af).join(' '))
say ('ef : ', 5.of(ef).join(' '))

say "ef(5) has #{ef(5).len} digits"

say ('rf : ', [1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800].map(rf))
say ('rf(119) = ', rf(119) \\ 'nil')
say ('rf is defined for: ', 8.by { defined(rf(_)) }.join(', ' ) + ', ...')
```
Output:
```sf : 1 1 2 12 288 34560 24883200 125411328000 5056584744960000 1834933472251084800000
H  : 1 1 4 108 27648 86400000 4031078400000 3319766398771200000 55696437941726556979200000 21577941222941856209168026828800000
af : 0 1 1 5 19 101 619 4421 35899 326981
ef : 1 1 2 9 262144
ef(5) has 183231 digits
rf : [0, 2, 3, 4, 5, 6, 7, 8, 9, 10]
rf(119) = nil
rf is defined for: 0, 1, 2, 6, 24, 120, 720, 5040, ...
```

## Wren

Library: Wren-big
Library: Wren-fmt

We've little choice but to use BigInt here as Wren can only deal natively with integers up to 2^53.

```import "./big" for BigInt
import "./fmt" for Fmt

var sf = Fn.new { |n|
if (n < 2) return BigInt.one
var sfact = BigInt.one
var fact  = BigInt.one
for (i in 2..n) {
fact = fact * i
sfact = sfact * fact
}
return sfact
}

var H = Fn.new { |n|
if (n < 2) return BigInt.one
var hfact = BigInt.one
for (i in 2..n) hfact = hfact * BigInt.new(i).pow(i)
return hfact
}

var af = Fn.new { |n|
if (n < 1) return BigInt.zero
var afact = BigInt.zero
var fact  = BigInt.one
var sign  = (n%2 == 0) ? -1 : 1
for (i in 1..n) {
fact = fact * i
afact = afact + fact * sign
sign = -sign
}
return afact
}

var ef // recursive
ef = Fn.new { |n|
if (n < 1) return BigInt.one
return BigInt.new(n).pow(ef.call(n-1))
}

var rf = Fn.new { |n|
var i = 0
var fact = BigInt.one
while (true) {
if (fact == n) return i
if (fact > n)  return "none"
i = i + 1
fact = fact * i
}
}

System.print("First 10 superfactorials:")
for (i in 0..9) System.print(sf.call(i))

System.print("\nFirst 10 hyperfactorials:")
for (i in 0..9) System.print(H.call(i))

System.print("\nFirst 10 alternating factorials:")
for (i in 0..9) System.write("%(af.call(i)) ")

System.print("\n\nFirst 5 exponential factorials:")
for (i in 0..4) System.write("%(ef.call(i)) ")
System.print()

Fmt.print("\nThe number of digits in 5\$\$ is \$,d\n", ef.call(5).toString.count)

System.print("Reverse factorials:")
var facts = [1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800, 119]
for (fact in facts) Fmt.print("\$4s <- rf(\$d)", rf.call(fact), fact)
```
Output:
```First 10 superfactorials:
1
1
2
12
288
34560
24883200
125411328000
5056584744960000
1834933472251084800000

First 10 hyperfactorials:
1
1
4
108
27648
86400000
4031078400000
3319766398771200000
55696437941726556979200000
21577941222941856209168026828800000

First 10 alternating factorials:
0 1 1 5 19 101 619 4421 35899 326981

First 5 exponential factorials:
1 1 2 9 262144

The number of digits in 5\$ is 183,231

Reverse factorials:
0 <- rf(1)
2 <- rf(2)
3 <- rf(6)
4 <- rf(24)
5 <- rf(120)
6 <- rf(720)
7 <- rf(5040)
8 <- rf(40320)
9 <- rf(362880)
10 <- rf(3628800)
none <- rf(119)
```