Factorions
From Rosetta Code
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 the Wikipedia article below) 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)
Contents
360 Assembly[edit]
* 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[edit]
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
AWK[edit]
# 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
C[edit]
#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
Factor[edit]
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[edit]
In this page you can see the solution of this task.
Fōrmulæ programs are not textual, visualization/edition of programs is done showing/manipulating structures but not text (more info). Moreover, there can be multiple visual representations of the same program. Even though it is possible to have textual representation —i.e. XML, JSON— they are intended for transportation effects more than visualization and edition.
The option to show Fōrmulæ programs and their results is showing images. Unfortunately images cannot be uploaded in Rosetta Code.
Frink[edit]
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[edit]
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[edit]
import Text.Printf (printf)
import Data.List (unfoldr, foldr)
import Control.Monad (guard)
factorion :: Int -> Int -> Bool
factorion b n = f b n == n
where
f b = foldr (\x y -> y + product [1..x]) 0 . 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[edit]
index=: $ #: [email protected]:,
factorion=: 10&$: :(] = [: +/ [: ! #.^:_1)&>
FACTORIONS=: 9 0 +"1 index Q=: 9 10 11 12 factorion/ i. 1500000
NB. base, factorion expressed in bases 10, and base
(,. "[email protected]:((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[edit]
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
Julia[edit]
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]
OCaml[edit]
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
Perl[edit]
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..$b-1;
}
print "\n\n";
}
- Output:
Factorions in base 9: 1 2 41282 Factorions in base 10: 1 2 145 40585 Factorions in base 11: 1 2 26 48 40472 Factorions in base 12: 1 2
Alternatively, a more efficient approach:
use 5.020;
use ntheory qw(:all);
use experimental qw(signatures);
use Algorithm::Combinatorics qw(combinations_with_repetition);
sub max_power ($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[edit]
-- cache factorials from 0 to 11
sequence fact = repeat(1,12)
for n=2 to length(fact) do
fact[n] = fact[n-1]*(n-1)
end for
for b=9 to 12 do
printf(1,"The factorions for base %d are:\n", b)
for i=1 to 1499999 do
atom total = 0, j = i, d
while j>0 and total<=i do
d = remainder(j,b)
total += fact[d+1]
j = floor(j/b)
end while
if total==i then printf(1,"%d ", i) end if
end for
printf(1,"\n\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
Python[edit]
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(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()
- 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
Racket[edit]
#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[edit]
(formerly Perl 6)
constant @factorial = 1, |[\*] 1..*;
constant $limit = 1500000;
constant $bases = 9 .. 12;
my @result;
$bases.race(:1batch).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[edit]
/*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
Ring[edit]
load "stdlib.ring"
for n = 1 to 100000
fac = 0
numStr = string(n)
for m = 1 to len(numStr)
num = number(numStr[m])
fac = fac + factorial(num)
next
if n = fac
see "Factorion: " + n + nl
ok
next
Factorion: 1 Factorion: 2 Factorion: 145 Factorion: 40585
Sidef[edit]
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[edit]
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
Wren[edit]
// 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
VBScript[edit]
' 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
zkl[edit]
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
