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# 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
```

## 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.functionsmath.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:" print10 [ sf ] .show nl "First 10 hyperfactorials:" print10 [ H ] .show nl "First 10 alternating factorials:" print10 [ af ] .show nl "First 5 exponential factorials:" print5 [ \$ ] .show nl "Number of digits in 5\$:" print5 \$ 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 pend 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 pend 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 send 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    wendend function for n as uinteger = 0 to 7    print sf(n), H(n), af(n)next nprintfor n as uinteger = 0 to 4    print ef(n);" ";next nprint : printfor n as uinteger =0 to 9    print rf(factorial(n));" ";next nprint 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)))        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

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 nothingendrf(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

Translation of: C
`import java.math.BigIntegerimport 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 * nfunction factorial(n)    local result = 1    local i = 1    while i <= n do        result = result * i        i = i + 1    end    return resultend -- if(n!) = nfunction 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 -1end -- 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 resultend -- H(n) = 1^1 * 2^2 * 3^3 * ... * (n-1)^(n-1) * n^nfunction hyper_factorial(n)    local result = 1    for i=1, n do        result = result * i ^ i    end    return resultend -- 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 resultend -- n\$ = n ^ (n-1) ^ ... ^ 2 ^ 1function exponential_factorial(n)    local result = 0    for i=1, n do        result = i ^ result    end    return resultend 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)    endend 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, sugarimport 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

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
```

## 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
```

## 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 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 tellcall hdr 'hyper';             do j=0  to 9;    \$= \$ hf(j);    end;         call tellcall hdr 'alternating ';      do j=0  to 9;    \$= \$ af(j);    end;         call tellcall 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+1sf:     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:";              returntell:   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

`\$ 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)} 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

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 BigIntimport "/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 // recursiveef = 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)
```