Factorions

From Rosetta Code
Task
Factorions
You are encouraged to solve this task according to the task description, using any language you may know.


Definition

A factorion is a natural number that equals the sum of the factorials of its digits.


Example

145   is a factorion in base 10 because:

          1! + 4! + 5!   =   1 + 24 + 120   =   145 


It can be shown (see talk page) that no factorion in base 10 can exceed   1,499,999.


Task

Write a program in your language to demonstrate, by calculating and printing out the factorions, that:

  •   There are   3   factorions in base   9
  •   There are   4   factorions in base 10
  •   There are   5   factorions in base 11
  •   There are   2   factorions in base 12     (up to the same upper bound as for base 10)


See also



11l

Translation of: Python
V fact = [1]
L(n) 1..11
   fact.append(fact[n-1] * n)

L(b) 9..12
   print(‘The factorions for base ’b‘ are:’)
   L(i) 1..1'499'999
      V fact_sum = 0
      V j = i
      L j > 0
         V d = j % b
         fact_sum += fact[d]
         j I/= b
      I fact_sum == i
         print(i, end' ‘ ’)
   print("\n")
Output:
The factorions for base 9 are:
1 2 41282 

The factorions for base 10 are:
1 2 145 40585 

The factorions for base 11 are:
1 2 26 48 40472 

The factorions for base 12 are:
1 2 

360 Assembly

*        Factorions                26/04/2020
FACTORIO CSECT
         USING  FACTORIO,R13       base register
         B      72(R15)            skip savearea
         DC     17F'0'             savearea
         SAVE   (14,12)            save previous context
         ST     R13,4(R15)         link backward
         ST     R15,8(R13)         link forward
         LR     R13,R15            set addressability
         XR     R4,R4              ~
         LA     R5,1               f=1
         LA     R3,FACT+4          @fact(1)
         LA     R6,1               i=1
       DO WHILE=(C,R6,LE,=A(NN2))  do i=1 to nn2
         MR     R4,R6                fact(i-1)*i
         ST     R5,0(R3)             fact(i)=fact(i-1)*i
         LA     R3,4(R3)             @fact(i+1)
         LA     R6,1(R6)             i++
       ENDDO    ,                  enddo i
         LA     R7,NN1             base=nn1
       DO WHILE=(C,R7,LE,=A(NN2))  do base=nn1 to nn2
	     MVC    PG,PGX               init buffer
         LA     R3,PG+6              @buffer
         XDECO  R7,XDEC              edit base
         MVC    PG+5(2),XDEC+10      output base
         LA     R3,PG+10             @buffer
         LA     R6,1                 i=1
       DO WHILE=(C,R6,LE,LIM)        do i=1 to lim 
         LA     R9,0                   s=0
         LR     R8,R6                  t=i
       DO WHILE=(C,R8,NE,=F'0')        while t<>0
         XR     R4,R4                    ~
         LR     R5,R8                    t 
         DR     R4,R7                    r5=t/base; r4=d=(t mod base)
         LR     R1,R4                    d
         SLA    R1,2                     ~
         L      R2,FACT(R1)              fact(d)
         AR     R9,R2                    s=s+fact(d)
         LR     R8,R5                    t=t/base
       ENDDO    ,                      endwhile
       IF    CR,R9,EQ,R6 THEN          if s=i then
         XDECO  R6,XDEC                  edit i
         MVC    0(6,R3),XDEC+6           output i
         LA     R3,7(R3)                 @buffer
       ENDIF    ,                      endif
         LA     R6,1(R6)               i++
       ENDDO    ,                    enddo i
         XPRNT  PG,L'PG              print buffer
         LA     R7,1(R7)             base++
       ENDDO    ,                  enddo base
         L      R13,4(0,R13)       restore previous savearea pointer
         RETURN (14,12),RC=0       restore registers from calling save
NN1      EQU    9                  nn1=9
NN2      EQU    12                 nn2=12
LIM      DC     f'1499999'         lim=1499999
FACT     DC     (NN2+1)F'1'        fact(0:12)
PG       DS     CL80               buffer
PGX      DC     CL80'Base .. : '   buffer init
XDEC     DS     CL12               temp fo xdeco
         REGEQU
         END    FACTORIO
Output:
Base  9 :      1      2  41282
Base 10 :      1      2    145  40585
Base 11 :      1      2     26     48  40472
Base 12 :      1      2

ALGOL 68

Translation of: C
BEGIN
    # cache factorials from 0 to 11 #
    [ 0 : 11 ]INT fact;
    fact[0] := 1;
    FOR n TO 11 DO
        fact[n] := fact[n-1] * n
    OD;
    FOR b FROM 9 TO 12 DO
        print( ( "The factorions for base ", whole( b, 0 ), " are:", newline ) );
        FOR i TO 1500000 - 1 DO
            INT sum := 0;
            INT j := i;
            WHILE j > 0 DO
                sum +:= fact[ j MOD b ];
                j OVERAB b
            OD;
            IF sum = i THEN print( ( whole( i, 0 ), " " ) ) FI
        OD;
        print( ( newline ) )
    OD
END
Output:
The factorions for base 9 are:
1 2 41282
The factorions for base 10 are:
1 2 145 40585
The factorions for base 11 are:
1 2 26 48 40472
The factorions for base 12 are:
1 2

Arturo

factorials: [1 1 2 6 24 120 720 5040 40320 362880 3628800 39916800]

factorion?: function [n, base][
    try? [
        n = sum map digits.base:base n 'x -> factorials\[x]
    ]
    else [
        print ["n:" n "base:" base]
        false
    ]
]

loop 9..12 'base ->
    print ["Base" base "factorions:" select 1..45000 'z -> factorion? z base]
]
Output:
Base 9 factorions: [1 2 41282] 
Base 10 factorions: [1 2 145 40585] 
Base 11 factorions: [1 2 26 48 40472] 
Base 12 factorions: [1 2]

AutoHotkey

Translation of: C
fact:=[]
fact[0] := 1
while (A_Index < 12)
	fact[A_Index] := fact[A_Index-1] * A_Index
b := 9
while (b <= 12) {
	res .= "base " b " factorions:  `t"
	while (A_Index < 1500000){
		sum := 0
		j := A_Index
		while (j > 0){
			d := Mod(j, b)
			sum += fact[d]
			j /= b
		}
		if (sum = A_Index) 
			res .= A_Index "  "
	}
	b++
	res .= "`n"
}
MsgBox % res
return
Output:
base 9 factorions:  	1  2  41282  
base 10 factorions:  	1  2  145  40585  
base 11 factorions:  	1  2  26  48  40472  
base 12 factorions:  	1  2  

AWK

# syntax: GAWK -f FACTORIONS.AWK
# converted from C
BEGIN {
    fact[0] = 1 # cache factorials from 0 to 11
    for (n=1; n<12; ++n) {
      fact[n] = fact[n-1] * n
    }
    for (b=9; b<=12; ++b) {
      printf("base %d factorions:",b)
      for (i=1; i<1500000; ++i) {
        sum = 0
        j = i
        while (j > 0) {
          d = j % b
          sum += fact[d]
          j = int(j/b)
        }
        if (sum == i) {
          printf(" %d",i)
        }
      }
      printf("\n")
    }
    exit(0)
}
Output:
base 9 factorions: 1 2 41282
base 10 factorions: 1 2 145 40585
base 11 factorions: 1 2 26 48 40472
base 12 factorions: 1 2

BASIC

Applesoft BASIC

100 DIM FACT(12)
110 FACT(0) = 1
120 FOR N = 1 TO 11
130     FACT(N) = FACT(N - 1) * N
140 NEXT 
200 FOR B = 9 TO 12
210     PRINT "THE FACTORIONS ";
215     PRINT "FOR BASE "B" ARE:"
220     FOR I = 1 TO 1499999
230         SUM = 0
240         FOR J = I TO 0 STEP 0
245             M =  INT (J / B)
250             D = J - M * B
260             SUM = SUM + FACT(D)
270             J = M
280         NEXT J
290         IF SU = I THEN  PRINT I" ";
300     NEXT I
310     PRINT : PRINT 
320 NEXT B

C

Translation of: Go
#include <stdio.h>

int main() {    
    int n, b, d;
    unsigned long long i, j, sum, fact[12];
    // cache factorials from 0 to 11
    fact[0] = 1;
    for (n = 1; n < 12; ++n) {
        fact[n] = fact[n-1] * n;
    }

    for (b = 9; b <= 12; ++b) {
        printf("The factorions for base %d are:\n", b);
        for (i = 1; i < 1500000; ++i) {
            sum = 0;
            j = i;
            while (j > 0) {
                d = j % b;
                sum += fact[d];
                j /= b;
            }
            if (sum == i) printf("%llu ", i);
        }
        printf("\n\n");
    }
    return 0;
}
Output:
The factorions for base 9 are:
1 2 41282 

The factorions for base 10 are:
1 2 145 40585 

The factorions for base 11 are:
1 2 26 48 40472 

The factorions for base 12 are:
1 2 

C++

Translation of: C
#include <iostream>

class factorion_t {
public:
    factorion_t() {
        f[0] = 1u;
        for (uint n = 1u; n < 12u; n++)
            f[n] = f[n - 1] * n;
    }

    bool operator()(uint i, uint b) const {
        uint sum = 0;
        for (uint j = i; j > 0u; j /= b)
            sum += f[j % b];
        return sum == i;
    }

private:
    ulong f[12];  //< cache factorials from 0 to 11
};

int main() {
    factorion_t factorion;
    for (uint b = 9u; b <= 12u; ++b) {
        std::cout << "factorions for base " << b << ':';
        for (uint i = 1u; i < 1500000u; ++i)
            if (factorion(i, b))
                std::cout << ' ' << i;
        std::cout << std::endl;
    }
    return 0;
}
Output:
factorions for base 9: 1 2 41282
factorions for base 10: 1 2 145 40585
factorions for base 11: 1 2 26 48 40472
factorions for base 12: 1 2

Common Lisp

(defparameter *bases* '(9 10 11 12))
(defparameter *limit* 1500000)

(defun ! (n) (apply #'* (loop for i from 2 to n collect i)))

(defparameter *digit-factorials* (mapcar #'! (loop for i from 0 to (1- (apply #'max *bases*)) collect i)))

(defun fact (n) (nth n *digit-factorials*))

(defun digit-value (digit)
  (let ((decimal (digit-char-p digit)))
    (cond ((not (null decimal)) decimal)
          ((char>= #\Z digit #\A) (+ (char-code digit) (- (char-code #\A)) 10))
          ((char>= #\z digit #\a) (+ (char-code digit) (- (char-code #\a)) 10))
          (t nil))))

(defun factorionp (n &optional (base 10))
  (= n (apply #'+
            (mapcar #'fact
                    (map 'list #'digit-value
                         (write-to-string n :base base))))))

(loop for base in *bases* do
  (let ((factorions
        (loop for i from 1 while (< i *limit*) if (factorionp i base) collect i)))
    (format t "In base ~a there are ~a factorions:~%" base (list-length factorions))
    (loop for n in factorions do
      (format t "~c~a" #\Tab (write-to-string n :base base))
      (if (/= base 10) (format t " (decimal ~a)" n))
      (format t "~%"))
    (format t "~%")))
Output:
In base 9 there are 3 factorions:
        1 (decimal 1)
        2 (decimal 2)
        62558 (decimal 41282)

In base 10 there are 4 factorions:
        1
        2
        145
        40585

In base 11 there are 5 factorions:
        1 (decimal 1)
        2 (decimal 2)
        24 (decimal 26)
        44 (decimal 48)
        28453 (decimal 40472)

In base 12 there are 2 factorions:
        1 (decimal 1)
        2 (decimal 2)


Delphi

Translation of: C
program Factorions;

{$APPTYPE CONSOLE}

uses
  System.SysUtils;

begin
  var fact: TArray<UInt64>;
  SetLength(fact, 12);

  fact[0] := 0;
  for var n := 1 to 11 do
    fact[n] := fact[n - 1] * n;

  for var b := 9 to 12 do
  begin
    writeln('The factorions for base ', b, ' are:');
    for var i := 1 to 1499999 do
    begin
      var sum := 0;
      var j := i;
      while j > 0 do
      begin
        var d := j mod b;
        sum := sum + fact[d];
        j := j div b;
      end;
      if sum = i then
        writeln(i, ' ');
    end;
    writeln(#10);
  end;
  readln;
end.

F#

//  Factorians. Nigel Galloway: October 22nd., 2021
let N=[|let mutable n=1 in yield n; for g in 1..11 do n<-n*g; yield n|]
let fG n g=let rec fN g=function i when i<n->g+N.[i] |i->fN(g+N.[i%n])(i/n) in fN 0 g 
{9..12}|>Seq.iter(fun n->printf $"In base %d{n} Factorians are:"; {1..1500000}|>Seq.iter(fun g->if g=fG n g then printf $" %d{g}"); printfn "")
Output:
In base 9 Factorians are: 1 2 41282
In base 10 Factorians are: 1 2 145 40585
In base 11 Factorians are: 1 2 26 48 40472
In base 12 Factorians are: 1 2

Factor

USING: formatting io kernel math math.parser math.ranges memoize
prettyprint sequences ;
IN: rosetta-code.factorions

! Memoize factorial function
MEMO: factorial ( n -- n! ) [ 1 ] [ [1,b] product ] if-zero ;

: factorion? ( n base -- ? )
    dupd >base string>digits [ factorial ] map-sum = ;

: show-factorions ( limit base -- )
    dup "The factorions for base %d are:\n" printf
    [ [1,b) ] dip [ dupd factorion? [ pprint bl ] [ drop ] if ]
    curry each nl ;

1,500,000 9 12 [a,b] [ show-factorions nl ] with each
Output:
The factorions for base 9 are:
1 2 41282 

The factorions for base 10 are:
1 2 145 40585 

The factorions for base 11 are:
1 2 26 48 40472 

The factorions for base 12 are:
1 2 

Fōrmulæ

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Programs in Fōrmulæ are created/edited online in its website.

In this page you can see and run the program(s) related to this task and their results. You can also change either the programs or the parameters they are called with, for experimentation, but remember that these programs were created with the main purpose of showing a clear solution of the task, and they generally lack any kind of validation.

Solution

Definitions:

The following calculates factorion lists from bases 9 to 12, with a limit of 1,499,999

FreeBASIC

Dim As Integer fact(12), suma, d, j
fact(0) = 1
For n As Integer = 1 To 11
    fact(n) = fact(n-1) * n
Next n
For b As Integer = 9 To 12
    Print "Los factoriones para base " & b & " son: "
    For i As Integer = 1 To 1499999
        suma = 0
        j = i
        While j > 0
            d = j Mod b
            suma += fact(d)
            j \= b
        Wend
        If suma = i Then Print i & " ";
    Next i
    Print : Print
Next b
Sleep
Output:
Los factoriones para base 9 son:
1 2 41282

Los factoriones para base 10 son:
1 2 145 40585

Los factoriones para base 11 son:
1 2 26 48 40472

Los factoriones para base 12 son:
1 2

Frink

factorion[n, base] := sum[map["factorial", integerDigits[n, base]]]

for base = 9 to 12
{
   for n = 1 to 1_499_999
      if n == factorion[n, base]
         println["$base\t$n"]
}
Output:
9	1
9	2
9	41282
10	1
10	2
10	145
10	40585
11	1
11	2
11	26
11	48
11	40472
12	1
12	2

Go

package main

import (
    "fmt"
    "strconv"
)

func main() {
    // cache factorials from 0 to 11
    var fact [12]uint64
    fact[0] = 1
    for n := uint64(1); n < 12; n++ {
        fact[n] = fact[n-1] * n
    }

    for b := 9; b <= 12; b++ {
        fmt.Printf("The factorions for base %d are:\n", b)
        for i := uint64(1); i < 1500000; i++ {
            digits := strconv.FormatUint(i, b)
            sum := uint64(0)
            for _, digit := range digits {
                if digit < 'a' {
                    sum += fact[digit-'0']
                } else {
                    sum += fact[digit+10-'a']
                }
            }
            if sum == i {
                fmt.Printf("%d ", i)
            }
        }
        fmt.Println("\n")
    }
}
Output:
The factorions for base 9 are:
1 2 41282 

The factorions for base 10 are:
1 2 145 40585 

The factorions for base 11 are:
1 2 26 48 40472 

The factorions for base 12 are:
1 2 

Haskell

import Text.Printf (printf)
import Data.List (unfoldr)
import Control.Monad (guard)

factorion :: Int -> Int -> Bool
factorion b n = f b n == n
 where
  f b = sum . map (product . enumFromTo 1) . unfoldr (\x -> guard (x > 0) >> pure (x `mod` b, x `div` b))

main :: IO ()
main = mapM_ (uncurry (printf "Factorions for base %2d: %s\n") . (\(a, b) -> (b, result a b))) 
  [(3,9), (4,10), (5,11), (2,12)]
 where 
  factorions b = filter (factorion b) [1..]
  result n = show . take n . factorions
Output:
Factorions for base  9: [1,2,41282]
Factorions for base 10: [1,2,145,40585]
Factorions for base 11: [1,2,26,48,40472]
Factorions for base 12: [1,2]

J

   index=: $ #: I.@:,
   factorion=: 10&$: :(] = [: +/ [: ! #.^:_1)&>

   FACTORIONS=: 9 0 +"1 index Q=: 9 10 11 12 factorion/ i. 1500000

   NB. columns: base, factorion in base 10, factorion in base
   (,. ".@:((Num_j_,26}.Alpha_j_) {~ #.inv/)"1) FACTORIONS
 9     1     1
 9     2     2
 9 41282 62558
10     1     1
10     2     2
10   145   145
10 40585 40585
11     1     1
11     2     2
11    26    24
11    48    44
11 40472 28453
12     1     1
12     2     2

   NB. tallies of factorions in the bases
   (9+i.4),.+/"1 Q
 9 3
10 4
11 5
12 2

Java

public class Factorion {
    public static void main(String [] args){
        System.out.println("Base 9:");
        for(int i = 1; i <= 1499999; i++){
            String iStri = String.valueOf(i);
            int multiplied = operate(iStri,9);
            if(multiplied == i){
                System.out.print(i + "\t");
            }
        }
        System.out.println("\nBase 10:");
        for(int i = 1; i <= 1499999; i++){
            String iStri = String.valueOf(i);
            int multiplied = operate(iStri,10);
            if(multiplied == i){
                System.out.print(i + "\t");
            }
        }
        System.out.println("\nBase 11:");
        for(int i = 1; i <= 1499999; i++){
            String iStri = String.valueOf(i);
            int multiplied = operate(iStri,11);
            if(multiplied == i){
                System.out.print(i + "\t");
            }
        }
        System.out.println("\nBase 12:");
        for(int i = 1; i <= 1499999; i++){
            String iStri = String.valueOf(i);
            int multiplied = operate(iStri,12);
            if(multiplied == i){
                System.out.print(i + "\t");
            }
        }
    }
    public static int factorialRec(int n){
        int result = 1;
        return n == 0 ? result : result * n * factorialRec(n-1);
    }

    public static int operate(String s, int base){
        int sum = 0;
        String strx = fromDeci(base, Integer.parseInt(s));
        for(int i = 0; i < strx.length(); i++){
            if(strx.charAt(i) == 'A'){
                sum += factorialRec(10);
            }else if(strx.charAt(i) == 'B') {
                sum += factorialRec(11);
            }else if(strx.charAt(i) == 'C') {
                sum += factorialRec(12);
            }else {
                sum += factorialRec(Integer.parseInt(String.valueOf(strx.charAt(i)), base));
            }
        }
        return sum;
    }
    // Ln 57-71 from Geeks for Geeks @ https://www.geeksforgeeks.org/convert-base-decimal-vice-versa/
    static char reVal(int num) {
        if (num >= 0 && num <= 9)
            return (char)(num + 48);
        else
            return (char)(num - 10 + 65);
    }
    static String fromDeci(int base, int num){
        StringBuilder s = new StringBuilder();
        while (num > 0) {
            s.append(reVal(num % base));
            num /= base;
        }
        return new String(new StringBuilder(s).reverse());
    }
}
Output:
Base 9:
1	2	41282	
Base 10:
1	2	145	40585	
Base 11:
1	2	26	48	40472	
Base 12:
1	2	

jq

Works with: jq

Also works with gojq, the Go implementation of jq, and with fq.

The main difficulty in computing the factorions of an arbitrary base is obtaining a tight limit on the maximum value a factorion can have in that base. The present entry accordingly does at least provide a function, `sufficient`, for computing an upper bound with respect to a particular base, and uses it to compute the factorions of all bases from 2 through 9.

However, the algorithm used by `sufficient` is too simplistic to be of much practical use for bases 10 or higher. For base 10, the task description provides a value with a link to a justification. For bases 11 and 12, we use limits that are known to be sufficient, as per (*) [1].

# A stream of factorials
# [N|factorials][n] is n!
def factorials:
   select(. > 0)
   | 1,
     foreach range(1; .) as $n(1; . * $n);

# The base-$b factorions less than or equal to $max 
def factorions($b; $max):
  ($max // 1500000) as $max
  | [$b|factorials] as $fact
  | range(1; $max) as $i
  | {sum: 0, j: $i}
  | until( .j == 0 or .sum > $i; 
       ( .j % $b) as $d
       | .sum += $fact[$d]
       | .j = ((.j/$b)|floor) )
  | select(.sum == $i)
  | $i ;

# input: base
# output: an upper bound for the factorions in that base
def sufficient:
  . as $base
  | [12|factorials] as $fact
  | $fact[$base-1] as $f
  | { digits: 1, value: $base}
  | until ( (.value > ($f * .digits) );
     .digits += 1
     | .value *= $base )  ;

# Show the factorions for all based from 2 through 12:
(range(2;10)
 | . as $base
 | sufficient.value as $max
 | {$base, factorions: ([factorions($base; $max)] | join(" "))}),
  {base: 10, factorions: ([factorions(10; 1500000)] | join(" "))},  # limit per the task description
  {base: 11, factorions: ([factorions(11; 50000)] | join(" "))},    # a limit known to be sufficient per (*)
  {base: 12, factorions: ([factorions(12; 50000)] | join(" "))}     # a limit known to be sufficient per (*)
Output:
{"base":2,"factorions":"1 2"}
{"base":3,"factorions":"1 2"}
{"base":4,"factorions":"1 2 7"}
{"base":5,"factorions":"1 2 49"}
{"base":6,"factorions":"1 2 25 26"}
{"base":7,"factorions":"1 2"}
{"base":8,"factorions":"1 2"}
{"base":9,"factorions":"1 2 41282"}
{"base":10,"factorions":"1 2 145 40585"}
{"base":11,"factorions":"1 2 26 48 40472"}
{"base":12,"factorions":"1 2"}

Julia

isfactorian(n, base) = mapreduce(factorial, +, map(c -> parse(Int, c, base=16), split(string(n, base=base), ""))) == n

printallfactorian(base) = println("Factorians for base $base: ", [n for n in 1:100000 if isfactorian(n, base)])

foreach(printallfactorian, 9:12)
Output:
Factorians for base 9: [1, 2, 41282]
Factorians for base 10: [1, 2, 145, 40585]
Factorians for base 11: [1, 2, 26, 48, 40472]
Factorians for base 12: [1, 2]

Lambdatalk

{def facts
 {S.first
  {S.map {{lambda {:a :i} 
                  {A.addlast! {* {A.get {- :i 1} :a} :i} :a}
          } {A.new 1}}
         {S.serie 1 11}}}}
-> facts

{def sumfacts
 {def sumfacts.r
  {lambda {:base :sum :i}
   {if {> :i 0}
    then {sumfacts.r :base
                     {+ :sum {A.get {% :i :base} {facts}}}
                     {floor {/ :i :base}}}
    else :sum }}}
 {lambda {:base :n}
  {sumfacts.r :base 0 :n}}} 
-> sumfacts

{def show
 {lambda {:base}
  {S.replace \s by space in
   {S.map {{lambda {:base :i}
                   {if {= {sumfacts :base :i} :i} then :i else}
           } :base}
          {S.serie 1 50000}}}}}
-> show

{S.map {lambda {:base}
               {div}factorions for base :base: {show :base}} 
       9 10 11 12}
->
factorions for base 9: 1 2 41282 
factorions for base 10: 1 2 145 40585 
factorions for base 11: 1 2 26 48 40472 
factorions for base 12: 1 2

Lang

Translation of: Python
# Enabling raw variable names boosts the performance massivly [DO NOT RUN WITHOUT enabling raw variable names]
lang.rawVariableNames = 1

# Cache factorials from 0 to 11
&fact = fn.listOf(1)
$n = 1
while($n < 12) {
	&fact += &fact[-|$n] * $n
	
	$n += 1
}

$b = 9
while($b <= 12) {
	fn.printf(The factorions for base %d are:%n, $b)
	
	$i = 1
	while($i < 1500000) {
		$sum = 0
		
		$j = $i
		while($j > 0) {
			$d $= $j % $b
			$sum += &fact[$d]
			$j //= $b
		}
		
		if($sum == $i) {
			fn.print($i\s)
		}
		
		$i += 1
	}
	
	fn.println(\n)
	
	$b += 1
}
Output:
The factorions for base 9 are:
1 2 41282 

The factorions for base 10 are:
1 2 145 40585 

The factorions for base 11 are:
1 2 26 48 40472 

The factorions for base 12 are:
1 2 

Mathematica / Wolfram Language

ClearAll[FactorionQ]
FactorionQ[n_,b_:10]:=Total[IntegerDigits[n,b]!]==n
Select[Range[1500000],FactorionQ[#,9]&]
Select[Range[1500000],FactorionQ[#,10]&]
Select[Range[1500000],FactorionQ[#,11]&]
Select[Range[1500000],FactorionQ[#,12]&]
Output:
{1, 2, 41282}
{1, 2, 145, 40585}
{1, 2, 26, 48, 40472}
{1, 2}

Nim

Note that the library has precomputed the values of factorial, so there is no need for caching.

from math import fac
from strutils import join

iterator digits(n, base: Natural): Natural =
  ## Yield the digits of "n" in base "base".
  var n = n
  while true:
    yield n mod base
    n = n div base
    if n == 0: break

func isFactorion(n, base: Natural): bool =
  ## Return true if "n" is a factorion for base "base".
  var s = 0
  for d in n.digits(base):
    inc s, fac(d)
  result = s == n

func factorions(base, limit: Natural): seq[Natural] =
  ## Return the list of factorions for base "base" up to "limit".
  for n in 1..limit:
    if n.isFactorion(base):
      result.add(n)


for base in 9..12:
  echo "Factorions for base ", base, ':'
  echo factorions(base, 1_500_000 - 1).join(" ")
Output:
Factorions for base 9:
1 2 41282
Factorions for base 10:
1 2 145 40585
Factorions for base 11:
1 2 26 48 40472
Factorions for base 12:
1 2

OCaml

Translation of: C
let () =
  (* cache factorials from 0 to 11 *)
  let fact = Array.make 12 0 in
  fact.(0) <- 1;
  for n = 1 to pred 12 do
    fact.(n) <- fact.(n-1) * n;
  done;

  for b = 9 to 12 do
    Printf.printf "The factorions for base %d are:\n" b;
    for i = 1 to pred 1_500_000 do
      let sum = ref 0 in
      let j = ref i in
      while !j > 0 do
        let d = !j mod b in
        sum := !sum + fact.(d);
        j := !j / b;
      done;
      if !sum = i then (print_int i; print_string " ")
    done;
    print_string "\n\n";
  done

Pascal

modified munchhausen numbers#Pascal. output in base and 0! == 1!, so in Base 10 40585 has the same digits as 14558.

program munchhausennumber;
{$IFDEF FPC}{$MODE objFPC}{$Optimization,On,all}{$ELSE}{$APPTYPE CONSOLE}{$ENDIF}
uses
  sysutils;
type
  tdigit  = byte;
const
  MAXBASE = 17;

var
  DgtPotDgt : array[0..MAXBASE-1] of NativeUint;
  dgtCnt : array[0..MAXBASE-1] of NativeInt;
  cnt: NativeUint;

function convertToString(n:NativeUint;base:byte):AnsiString;
const
  cBASEDIGITS = '0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvxyz';
var
  r,dgt: NativeUint;
begin
  IF base > length(cBASEDIGITS) then
    EXIT('Base to big');
  result := '';
  repeat
    r := n div base;
    dgt := n-r*base;
    result := cBASEDIGITS[dgt+1]+result;
    n := r;
  until n =0;
end;

function CheckSameDigits(n1,n2,base:NativeUInt):boolean;
var

  i : NativeUInt;
Begin
  fillchar(dgtCnt,SizeOf(dgtCnt),#0);
  repeat
    //increment digit of n1
    i := n1;n1 := n1 div base;i := i-n1*base;inc(dgtCnt[i]);
    //decrement digit of n2
    i := n2;n2 := n2 div base;i := i-n2*base;dec(dgtCnt[i]);
  until (n1=0) AND (n2= 0);
  result := true;
  For i := 2 to Base-1 do
    result := result AND (dgtCnt[i]=0);
  result := result AND (dgtCnt[0]+dgtCnt[1]=0);

end;

procedure Munch(number,DgtPowSum,minDigit:NativeUInt;digits,base:NativeInt);
var
  i: NativeUint;
  s1,s2: AnsiString;
begin
  inc(cnt);
  number := number*base;
  IF digits > 1 then
  Begin
    For i := minDigit to base-1 do
      Munch(number+i,DgtPowSum+DgtPotDgt[i],i,digits-1,base);
  end
  else
    For i := minDigit to base-1 do
      //number is always the arrangement of the digits leading to smallest number
      IF (number+i)<= (DgtPowSum+DgtPotDgt[i]) then
        IF CheckSameDigits(number+i,DgtPowSum+DgtPotDgt[i],base) then
          iF number+i>0 then
          begin
            s1 := convertToString(DgtPowSum+DgtPotDgt[i],base);
            s2 := convertToString(number+i,base);
            If length(s1)= length(s2) then
              writeln(Format('%*d %*s  %*s',[Base-1,DgtPowSum+DgtPotDgt[i],Base-1,s1,Base-1,s2]));
          end;
end;

//factorions
procedure InitDgtPotDgt(base:byte);
var
  i: NativeUint;
Begin
  DgtPotDgt[0]:= 1;
  For i := 1 to Base-1 do
    DgtPotDgt[i] := DgtPotDgt[i-1]*i;
  DgtPotDgt[0]:= 0;
end;
{
//Munchhausen numbers
procedure InitDgtPotDgt;
var
  i,k,dgtpow: NativeUint;
Begin
  // digit ^ digit ,special case 0^0 here 0
  DgtPotDgt[0]:= 0;
  For i := 1 to Base-1 do
  Begin
    dgtpow := i;
    For k := 2 to i do
      dgtpow := dgtpow*i;
    DgtPotDgt[i] := dgtpow;
  end;
end;
}
var
  base : byte;
begin
  cnt := 0;
  For base := 2 to MAXBASE do
  begin
    writeln('Base = ',base);
    InitDgtPotDgt(base);
    Munch(0,0,0,base,base);
  end;
  writeln('Check Count ',cnt);
end.
Output:
TIO.RUN Real time: 45.701 s User time: 44.968 s Sys. time: 0.055 s CPU share: 98.51 %
Base = 2
1 1  1
Base = 3
 1  1   1
 2  2   2
Base = 4
  1   1    1
  2   2    2
  7  13   13
Base = 5
   1    1     1
   2    2     2
  49  144   144
Base = 6
    1     1      1
    2     2      2
   25    41     14
   26    42     24
Base = 7
     1      1       1
     2      2       2
Base = 8
      1       1        1
      2       2        2
Base = 9
       1        1         1
       2        2         2
   41282    62558     25568
Base = 10
        1         1          1
        2         2          2
      145       145        145
    40585     40585      14558
Base = 11
         1          1           1
         2          2           2
        26         24          24
        48         44          44
     40472      28453       23458
Base = 12
          1           1            1
          2           2            2
Base = 13
           1            1             1
           2            2             2
   519326767     83790C5B      135789BC
Base = 14
            1             1              1
            2             2              2
  12973363226     8B0DD409C      11489BCDD
Base = 15
             1              1               1
             2              2               2
          1441            661             166
          1442            662             266
Base = 16
              1               1                1
              2               2                2
  2615428934649     260F3B66BF9      1236669BBFF
Base = 17
               1                1                 1
               2                2                 2
           40465             8405              1458
  43153254185213     146F2G8500G4      111244568FGG
  43153254226251     146F2G8586G4      124456688FGG
Check Count 1571990934

Perl

Raku version

Translation of: Raku
Library: ntheory
use strict;
use warnings;
use ntheory qw/factorial todigits/;

my $limit = 1500000;

for my $b (9 .. 12) {
    print "Factorions in base $b:\n";
    $_ == factorial($_) and print "$_ " for 0..$b-1;

    for my $i (1 .. int $limit/$b) {
        my $sum;
        my $prod = $i * $b;

        for (reverse todigits($i, $b)) {
            $sum += factorial($_);
            $sum = 0 && last if $sum > $prod;
        }

        next if $sum == 0;
        ($sum + factorial($_) == $prod + $_) and print $prod+$_ . ' ' for 0..$b-1;
    }
    print "\n\n";
}
Output:
Factorions in base 9:
1 2 41282

Factorions in base 10:
1 2 145 40585

Factorions in base 11:
1 2 26 48 40472

Factorions in base 12:
1 2

Sidef version

Alternatively, a more efficient approach:

Translation of: Sidef
Library: ntheory
use 5.020;
use ntheory qw(:all);
use experimental qw(signatures);
use Algorithm::Combinatorics qw(combinations_with_repetition);

sub max_power ($base = 10) {
    my $m = 1;
    my $f = factorial($base - 1);
    while ($m * $f >= $base**($m-1)) {
        $m += 1;
    }
    return $m-1;
}

sub factorions ($base = 10) {

    my @result;
    my @digits    = (0 .. $base-1);
    my @factorial = map { factorial($_) } @digits;

    foreach my $k (1 .. max_power($base)) {
        my $iter = combinations_with_repetition(\@digits, $k);
        while (my $comb = $iter->next) {
            my $n = vecsum(map { $factorial[$_] } @$comb);
            if (join(' ', sort { $a <=> $b } todigits($n, $base)) eq join(' ', @$comb)) {
                push @result, $n;
            }
        }
    }

    return @result;
}

foreach my $base (2 .. 14) {
    my @r = factorions($base);
    say "Factorions in base $base are (@r)";
}
Output:
Factorions in base 2 are (1 2)
Factorions in base 3 are (1 2)
Factorions in base 4 are (1 2 7)
Factorions in base 5 are (1 2 49)
Factorions in base 6 are (1 2 25 26)
Factorions in base 7 are (1 2)
Factorions in base 8 are (1 2)
Factorions in base 9 are (1 2 41282)
Factorions in base 10 are (1 2 145 40585)
Factorions in base 11 are (1 2 26 48 40472)
Factorions in base 12 are (1 2)
Factorions in base 13 are (1 2 519326767)
Factorions in base 14 are (1 2 12973363226)

Phix

Translation of: C

As per talk page (ok, and the task description), this is incorrectly using the base 10 limit for bases 9, 11, and 12.

with javascript_semantics
for base=9 to 12 do
    printf(1,"The factorions for base %d are: ", base)
    for i=1 to 1499999 do
        atom total = 0, j = i, d
        while j>0 and total<=i do
            d = remainder(j,base)
            total += factorial(d)
            j = floor(j/base)
        end while
        if total==i then printf(1,"%d ", i) end if
    end for
    printf(1,"\n")
end for
Output:
The factorions for base 9 are: 1 2 41282
The factorions for base 10 are: 1 2 145 40585
The factorions for base 11 are: 1 2 26 48 40472
The factorions for base 12 are: 1 2
Translation of: Sidef

Using the correct limits and much faster, or at least it was until I upped the bases to 14.

with javascript_semantics
function max_power(integer base = 10)
    integer m = 1
    atom f = factorial(base-1)
    while m*f >= power(base,m-1) do
        m += 1
    end while
    return m-1
end function
 
constant digits = "0123456789abcd"

function fcomb(sequence res, integer base, n, at=1, atom fsum=0, string chosen="")
    if length(chosen)=n then
        string fs = sort(sprintf("%a",{{base,fsum}}))
        if fs=chosen then
            res = append(res,sprintf("%d",fsum))
        end if
    else
        for i=at to base do
            res = fcomb(res,base,n,i,fsum+factorial(i-1),chosen&digits[i])
        end for
    end if
    return res
end function

function factorions(integer base = 10)
    sequence result = {}
    for k=1 to max_power(base) do
        result &= fcomb({},base,k)
    end for
    return result
end function

for base=2 to 14 do
    printf(1,"Base %2d factorions: %s\n",{base,join(factorions(base))})
end for
Output:
Base  2 factorions: 1 2
Base  3 factorions: 1 2
Base  4 factorions: 1 2 7
Base  5 factorions: 1 2 49
Base  6 factorions: 1 2 25 26
Base  7 factorions: 1 2
Base  8 factorions: 1 2
Base  9 factorions: 1 2 41282
Base 10 factorions: 1 2 145 40585
Base 11 factorions: 1 2 26 48 40472
Base 12 factorions: 1 2
Base 13 factorions: 1 2 519326767
Base 14 factorions: 1 2 12973363226

It will in fact go all the way to 17, though I don't recommend it:

Base 15 factorions: 1 2 1441 1442
Base 16 factorions: 1 2 2615428934649
Base 17 factorions: 1 2 40465 43153254185213 43153254226251

PureBasic

Translation of: C
Declare main()

If OpenConsole() : main() : Else : End 1 : EndIf
Input() : End

Procedure main()
  Define.i n,b,d,i,j,sum
  Dim fact.i(12)
  
  fact(0)=1
  For n=1 To 11 : fact(n)=fact(n-1)*n : Next
  
  For b=9 To 12
    PrintN("The factorions for base "+Str(b)+" are: ")
    For i=1 To 1500000-1
      sum=0 : j=i
      While j>0
        d=j%b : sum+fact(d) : j/b
      Wend
      If sum=i : Print(Str(i)+" ") : EndIf
    Next
    Print(~"\n\n")
  Next
EndProcedure
Output:
The factorions for base 9 are: 
1 2 41282 

The factorions for base 10 are: 
1 2 145 40585 

The factorions for base 11 are: 
1 2 26 48 40472 

The factorions for base 12 are: 
1 2 

Python

Translation of: C
fact = [1] # cache factorials from 0 to 11
for n in range(1, 12):
    fact.append(fact[n-1] * n)

for b in range(9, 12+1):
    print(f"The factorions for base {b} are:")
    for i in range(1, 1500000):
        fact_sum = 0
        j = i
        while j > 0:
            d = j % b
            fact_sum += fact[d]
            j = j//b
        if fact_sum == i:
            print(i, end=" ")
    print("\n")
Output:
The factorions for base 9 are:
1 2 41282

The factorions for base 10 are:
1 2 145 40585

The factorions for base 11 are:
1 2 26 48 40472

The factorions for base 12 are:
1 2

Quackery

  [ table ]              is results   (   n --> s )
  4 times 
    [ ' [ stack [ ] ]
      copy
      ' results put ]

  [ results dup take 
    rot join swap put ]  is addresult ( n n -->   )

  [ table 9 10 11 12 ]   is radix     (   n --> n )

  [ table 1 ]            is !         (   n --> n )     
  1 11 times
    [ i^ 1+ * dup
      ' ! put ]
  drop 
    
  [ dip dup
    0 temp put
    [ tuck /mod !
      temp tally
      swap over 0 = 
      until ]
    2drop 
    temp take = ]       is factorion ( n n --> b )

  1500000 times
    [ i^ 4 times 
      [ dup 
        i^ radix
        factorion if
          [ dup i^ 
            addresult ] ]
      drop ]
  4 times 
    [ say "Factorions for base "
     i^ radix echo say ": "
     i^ results take echo cr ]
Output:
Factorions for base 9: [ 1 2 41282 ]
Factorions for base 10: [ 1 2 145 40585 ]
Factorions for base 11: [ 1 2 26 48 40472 ]
Factorions for base 12: [ 1 2 ]


Racket

Translation of: C
#lang racket

(define fact
  (curry list-ref (for/fold ([result (list 1)] #:result (reverse result))
                            ([x (in-range 1 20)])
                    (cons (* x (first result)) result))))

(for ([b (in-range 9 13)])
  (printf "The factorions for base ~a are:\n" b)
  (for ([i (in-range 1 1500000)])
    (let loop ([sum 0] [n i])
      (cond
        [(positive? n) (loop (+ sum (fact (modulo n b))) (quotient n b))]
        [(= sum i) (printf "~a " i)])))
  (newline))
Output:
The factorions for base 9 are:
1 2 41282 
The factorions for base 10 are:
1 2 145 40585 
The factorions for base 11 are:
1 2 26 48 40472 
The factorions for base 12 are:
1 2 

Raku

(formerly Perl 6)

Works with: Rakudo version 2019.07.1
constant @factorial = 1, |[\*] 1..*;

constant $limit = 1500000;

constant $bases = 9 .. 12;

my @result;

$bases.map: -> $base {

    @result[$base] = "\nFactorions in base $base:\n1 2";

    sink (1 .. $limit div $base).map: -> $i {
        my $product = $i * $base;
        my $partial;

        for $i.polymod($base xx *) {
            $partial += @factorial[$_];
            last if $partial > $product
        }

        next if $partial > $product;

        my $sum;

        for ^$base {
            last if ($sum = $partial + @factorial[$_]) > $product + $_;
            @result[$base] ~= " $sum" and last if $sum == $product + $_
        }
    }
}

.say for @result[$bases];
Output:
Factorions in base 9:
1 2 41282

Factorions in base 10:
1 2 145 40585

Factorions in base 11:
1 2 26 48 40472

Factorions in base 12:
1 2

REXX

Translation of: C
/*REXX program calculates and displays   factorions   in  bases  nine ───► twelve.      */
parse arg LOb HIb lim .                          /*obtain optional arguments from the CL*/
if LOb=='' | LOb==","  then LOb=       9         /*Not specified?  Then use the default.*/
if HIb=='' | HIb==","  then HIb=      12         /* "      "         "   "   "      "   */
if lim=='' | lim==","  then lim= 1500000  -  1   /* "      "         "   "   "      "   */

  do fact=0  to HIb;   !.fact= !(fact)           /*use memoization for factorials.      */
  end   /*fact*/

  do base=LOb  to  HIb                           /*process all the required bases.      */
  @= 1 2                                         /*initialize the list  (@)  to  1 & 2. */
          do j=3  for lim-2;  $= 0               /*initialize the sum   ($)  to  zero.  */
                                          t= j   /*define the target  (for the sum !'s).*/
                                 do until t==0;    d= t // base      /*obtain a "digit".*/
                                                   $= $ + !.d        /*add  !(d) to sum.*/
                                                   t= t % base       /*get a new target.*/
                                 end   /*until*/
          if $==j  then @= @ j                   /*Good factorial sum? Then add to list.*/
          end   /*i*/
  say
  say 'The factorions for base '      right( base, length(HIb) )        " are: "         @
  end   /*base*/
exit                                             /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
!: procedure; parse arg x;  !=1;    do j=2  to x;  !=!*j;  end;   return !  /*factorials*/
output   when using the default inputs:
The factorions for base   9  are:  1 2 41282

The factorions for base  10  are:  1 2 145 40585

The factorions for base  11  are:  1 2 26 48 40472

The factorions for base  12  are:  1 2

RPL

Translation of: C
Works with: Halcyon Calc version 4.2.7
Code Comments
≪ 
  { } 1 11 FOR n n FACT + NEXT → base fact 
  ≪ { } 1 1500000 FOR n 
        0 n WHILE DUP REPEAT 
           fact OVER base MOD 1 MAX GET 
           ROT + SWAP 
           base / IP
       END DROP 
       IF n == THEN n + END 
     NEXT 
≫ ≫ ‘FTRION’ STO 
( base -- { factorions } ) 
 Cache 1! to 11!
  
 Loop until all digits scanned 
    Get (last digit)! even if last digit = 0
    Add to sum of digits
    prepare next loop
 
 Store factorion


The following lines of command deliver what is required:

 9 FTRION
10 FTRION
11 FTRION
12 FTRION
Output:
4: { 1 2 41282 }
3: { 1 2 145 40585 }
2: { 1 2 26 48 40472 }
1: { 1 2 }

Ruby

def factorion?(n, base)
  n.digits(base).sum{|digit| (1..digit).inject(1, :*)} == n 
end

(9..12).each do |base|
  puts "Base #{base} factorions: #{(1..1_500_000).select{|n| factorion?(n, base)}.join(" ")} "
end
Output:
Base 9 factorions: 1 2 41282 
Base 10 factorions: 1 2 145 40585 
Base 11 factorions: 1 2 26 48 40472 
Base 12 factorions: 1 2 

Scala

Translation of: C++
object Factorion extends App {
    private def is_factorion(i: Int, b: Int): Boolean = {
        var sum = 0L
        var j = i
        while (j > 0) {
            sum +=  f(j % b)
            j /= b
        }
        sum == i
    }

    private val f = Array.ofDim[Long](12)
    f(0) = 1L
    (1 until 12).foreach(n => f(n) = f(n - 1) * n)
    (9 to 12).foreach(b => {
        print(s"factorions for base $b:")
        (1 to 1500000).filter(is_factorion(_, b)).foreach(i => print(s" $i"))
        println
    })
}

Sidef

func max_power(b = 10) {
    var m = 1
    var f = (b-1)!
    while (m*f >= b**(m-1)) {
        m += 1
    }
    return m-1
}

func factorions(b = 10) {

    var result = []
    var digits = @^b
    var fact = digits.map { _! }

    for k in (1 .. max_power(b)) {
        digits.combinations_with_repetition(k, {|*comb|
            var n = comb.sum_by { fact[_] }
            if (n.digits(b).sort == comb) {
                result << n
            }
        })
    }

    return result
}

for b in (2..12) {
    var r = factorions(b)
    say "Base #{'%2d' % b} factorions: #{r}"
}
Output:
Base  2 factorions: [1, 2]
Base  3 factorions: [1, 2]
Base  4 factorions: [1, 2, 7]
Base  5 factorions: [1, 2, 49]
Base  6 factorions: [1, 2, 25, 26]
Base  7 factorions: [1, 2]
Base  8 factorions: [1, 2]
Base  9 factorions: [1, 2, 41282]
Base 10 factorions: [1, 2, 145, 40585]
Base 11 factorions: [1, 2, 26, 48, 40472]
Base 12 factorions: [1, 2]

Swift

Translation of: C
var fact = Array(repeating: 0, count: 12)

fact[0] = 1

for n in 1..<12 {
  fact[n] = fact[n - 1] * n
}

for b in 9...12 {
  print("The factorions for base \(b) are:")

  for i in 1..<1500000 {
    var sum = 0
    var j = i

    while j > 0 {
      sum += fact[j % b]
      j /= b
    }

    if sum == i {
      print("\(i)", terminator: " ")
      fflush(stdout)
    }
  }

  print("\n")
}
Output:
The factorions for base 9 are:
1 2 41282 

The factorions for base 10 are:
1 2 145 40585 

The factorions for base 11 are:
1 2 26 48 40472 

The factorions for base 12 are:
1 2

uBasic/4tH

Translation of: FreeBASIC

It will take some time, but it will get there.

Dim @f(12)

@f(0) = 1: For n = 1 To 11 : @f(n) = @f(n-1) * n : Next

For b = 9 To 12
  Print "The factorions for base ";b;" are: "
  For i = 1 To 1499999
    s = 0
    j = i
    Do While j > 0
      d = j % b
      s = s + @f(d)
      j = j / b
    Loop
    If s = i Then Print i;" ";
  Next
  Print : Print
Next
Output:
The factorions for base 9 are: 
1 2 41282 

The factorions for base 10 are: 
1 2 145 40585 

The factorions for base 11 are: 
1 2 26 48 40472 

The factorions for base 12 are: 
1 2 


0 OK, 0:379

V (Vlang)

Translation of: Go
import strconv

fn main() {
    // cache factorials from 0 to 11
    mut fact := [12]u64{}
    fact[0] = 1
    for n := u64(1); n < 12; n++ {
        fact[n] = fact[n-1] * n
    }
 
    for b := 9; b <= 12; b++ {
        println("The factorions for base $b are:")
        for i := u64(1); i < 1500000; i++ {
            digits := strconv.format_uint(i, b)
            mut sum := u64(0)
            for digit in digits {
                if digit < `a` {
                    sum += fact[digit-`0`]
                } else {
                    sum += fact[digit+10-`a`]
                }
            }
            if sum == i {
                print("$i ")
            }
        }
        println("\n")
    }
}
Output:
The factorions for base 9 are:
1 2 41282 

The factorions for base 10 are:
1 2 145 40585 

The factorions for base 11 are:
1 2 26 48 40472 

The factorions for base 12 are:
1 2  

VBScript

' Factorions - VBScript - PG - 26/04/2020
    Dim fact()
	nn1=9 : nn2=12
	lim=1499999
    ReDim fact(nn2)
	fact(0)=1
	For i=1 To nn2
		fact(i)=fact(i-1)*i
	Next
	For base=nn1 To nn2
		list=""
		For i=1 To lim
			s=0
			t=i
			Do While t<>0
				d=t Mod base
				s=s+fact(d)
				t=t\base
			Loop
			If s=i Then list=list &" "& i
		Next
		Wscript.Echo "the factorions for base "& right(" "& base,2) &" are: "& list
	Next
Output:
the factorions for base  9 are: 1 2 41282
the factorions for base 10 are: 1 2 145 40585
the factorions for base 11 are: 1 2 26 48 40472
the factorions for base 12 are: 1 2

Wren

Translation of: C
// cache factorials from 0 to 11
var fact = List.filled(12, 0)
fact[0] = 1
for (n in 1..11) fact[n] = fact[n-1] * n

for (b in 9..12) {
    System.print("The factorions for base %(b) are:")
    for (i in 1...1500000) {
        var sum = 0
        var j = i
        while (j > 0) {
            var d = j % b
            sum = sum + fact[d]
            j = (j/b).floor
        }
        if (sum == i) System.write("%(i) ")
    }
    System.print("\n")
}
Output:
The factorions for base 9 are:
1 2 41282 

The factorions for base 10 are:
1 2 145 40585 

The factorions for base 11 are:
1 2 26 48 40472 

The factorions for base 12 are:
1 2 

XPL0

Translation of: C
int N, Base, Digit, I, J, Sum, Factorial(12);
[Factorial(0):= 1;      \cache factorials from 0 to 11
for N:= 1 to 12-1 do
    Factorial(N):= Factorial(N-1)*N;
for Base:= 9 to 12 do
    [Text(0, "The factorions for base "); IntOut(0, Base); Text(0, " are:^m^j");
    for I:= 1 to 1_499_999 do
        [Sum:= 0;
        J:= I;
        while J > 0 do
            [Digit:= rem(J/Base);
            Sum:= Sum + Factorial(Digit);
            J:= J/Base;
            ];
        if Sum = I then [IntOut(0, I);  ChOut(0, ^ )];
        ];
    CrLf(0);  CrLf(0);
    ];
]
Output:
The factorions for base 9 are:
1 2 41282 

The factorions for base 10 are:
1 2 145 40585 

The factorions for base 11 are:
1 2 26 48 40472 

The factorions for base 12 are:
1 2 

zkl

Translation of: C
var facts=[0..12].pump(List,fcn(n){ (1).reduce(n,fcn(N,n){ N*n },1) }); #(1,1,2,6....)
fcn factorions(base){
   fs:=List();
   foreach n in ([1..1_499_999]){
      sum,j := 0,n;
      while(j){
	 sum+=facts[j%base];
	 j/=base;
      }
      if(sum==n) fs.append(n);
   }
   fs
}
foreach n in ([9..12]){
   println("The factorions for base %2d are: ".fmt(n),factorions(n).concat("  "));
}
Output:
The factorions for base  9 are: 1  2  41282
The factorions for base 10 are: 1  2  145  40585
The factorions for base 11 are: 1  2  26  48  40472
The factorions for base 12 are: 1  2