Rosetta Code

Special factorials

Revision as of 03:51, 12 January 2022 by Alextretyak (talk | contribs) (Added 11l)

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

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.
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



11l

Translation of: Nim

<lang 11l>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’}’)</lang>
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

<lang c>#include <math.h>

  1. include <stdint.h>
  2. 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;

}</lang>

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

<lang cpp>#include <cmath>

  1. include <cstdint>
  2. include <iostream>
  3. 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;

}</lang>

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

<lang factor>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</lang>

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

<lang>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)</lang>

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. <lang freebasic>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)</lang>

Go

Translation of: Wren

<lang go>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)
   }

}</lang>

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. <lang jq>

  1. 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;</lang>

The tasks: <lang jq> "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(\(.))")

</lang>

Output:

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

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

Julia

No recursion. <lang julia>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

</lang>

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

<lang scala>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)

}</lang>

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

<lang lua>-- 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)</lang>

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

<lang Mathematica>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]</lang>

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

<lang Nim>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"</lang>
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

<lang perl>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);</lang>

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
    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

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 :) <lang Python>

  1. 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)) </lang>

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

<lang perl6>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;</lang>

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

<lang 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 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</lang>

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

<lang 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;</lang>
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

<lang ruby>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(', ' ) + ', ...')</lang>

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. <lang ecmascript>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)</lang>

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)
Retrieved from "https://rosettacode.org/wiki/Special_factorials?oldid=309432"