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Special factorials

From Rosetta Code
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
Name Formula Example calculation Links
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


Task
  • 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.


See also



C[edit]

#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++[edit]

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[edit]

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[edit]

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[edit]

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[edit]

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)))
afact.Add(afact, t.Mul(fact, sign))
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[edit]

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;

The tasks:

 
"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[edit]

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]
println(rpad(n, 10), rf(n))
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[edit]

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[edit]

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[edit]

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[edit]

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)
# Start with Int values for "n".
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[edit]

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[edit]

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
    mpz_add(r,r,fn)
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

Python[edit]

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

Raku[edit]

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[edit]

/*REXX program to compute some special factorials:   superfactorials,   hyperfactorials,*/
/*───────────────────────────────────── alternating factorials, exponential factorials.*/
numeric digits 1000 /*allows humongous results to be shown.*/
call hdr 'super'; do j=0 to 9; $= $ sf(j); end; call tell
call hdr 'hyper'; do j=0 to 9; $= $ hf(j); end; call tell
call hdr 'alternating '; do j=0 to 9; $= $ af(j); end; call tell
call hdr 'exponential '; do j=0 to 5; $= $ ef(j); end; call tell
@= 'the number of decimal digits in the exponential factorial of '
say @ 5 " is:"; $= ' 'commas( efn( ef(5) ) ); call tell
@= 'the inverse factorial of'
do j=1 for 10; say @ right(!(j), 8) " is: " rf(!(j))
end /*j*/
say @ right(119, 8) " is: " rf(119)
exit 0 /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
!: procedure; parse arg x;  != 1; do #=2 to x;  != ! * #; end; return !
af: procedure; parse arg x; if x==0 then return 0; prev= 0; call af!; return !
af!: do #=1 for x;  != !(#) - prev; prev= !; end; return !
commas: parse arg ?; do jc=length(?)-3 to 1 by -3; ?=insert(',', ?, jc); end; return ?
ef: procedure; parse arg x; if x==0 | x==1 then return 1; return x**ef(x-1)
efn: procedure; parse arg x; numeric digits 9; x= x; parse var x 'E' d; return d+1
sf: procedure; parse arg x;  != 1; do #=2 to x;  != ! * !(#); end; return !
hf: procedure; parse arg x;  != 1; do #=2 to x;  != ! * #**#; end; return !
rf: procedure; parse arg x; do #=0 until f>=x; f=!(#); end; return rfr()
rfr: if x==f then return #;  ?= 'undefined'; return ?
hdr: parse arg ?,,$; say 'the first ten '?"factorials:"; return
tell: say substr($, 2); say; $=; return
output   when using the internal default input:
the first  10  superfactorials:
1 1 2 12 288 34560 24883200 125411328000 5056584744960000 1834933472251084800000

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

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

the first  5  exponential factorials:
1 1 2 9 262144

the number of decimal digits in the exponential factorial of  5  is:
183,231

the inverse factorial of  1  is:  0
the inverse factorial of  2  is:  2
the inverse factorial of  6  is:  3
the inverse factorial of  24  is:  4
the inverse factorial of  120  is:  5
the inverse factorial of  720  is:  6
the inverse factorial of  5040  is:  7
the inverse factorial of  40320  is:  8
the inverse factorial of  362880  is:  9
the inverse factorial of  3628800  is:  10
the inverse factorial of  119  is:  undefined

Seed7[edit]

$ 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[edit]

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)
}
 
func p_adic_inverse (p, k) {
 
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 n = p_adic_inverse(2, t)
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[edit]

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)