Special factorials

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	Wikipedia OEIS:A000178
Hyperfactorial	`n` H(`n`) = ∏ `k`^k `k`=1	H(4) = 1¹ × 2² × 3³ × 4⁴ = 27,648	Wikipedia OEIS:A002109
Alternating factorial	`n` af(`n`) = ∑ (-1)^n-i`i`! `i`=1	af(3) = -1²×1! + -1¹×2! + -1⁰×3! = 5	Wikipedia OEIS:A005165
Exponential factorial	n$ = `n`^{(n-1)^{(n-2)^...}}	4$ = 4^{3^2¹} = 262,144	Wikipedia OEIS:A049384

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

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

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)

   for i in 1:100000
       fac = factorial(typeof(n)(i))
       fac == n && return i
       fac > n && break
   end
   return nothing

end rf(n) = reverse_factorial(n)

println("N Superfactorial Hyperfactorial Alternating Factorial Exponential Factorial\n", "-"^88) for n in 0:9

   print(n, "  ")
   for f in [sf, H, af, n＄]
       if n < 5 || f != n＄
           print(rpad((f(n)), 25))
       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         -907465429310504960      35899
9  8705808953839190016      2649120435010011136      326981

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


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

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

<lang rexx>/*REXX program to compute some special factorials: superfactorials, hyperfactorials,*/ /*───────────────────────────────────── alternating factorials, exponential factorials.*/ numeric digits 100 /*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

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)