Factorions: Difference between revisions
Thundergnat (talk | contribs) (→{{header|Perl 6}}: Add a Perl 6 example) |
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say "\n\nFactorions in base $b:"; |
say "\n\nFactorions in base $b:"; |
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for ^$b { if $_ == @f[$_] { print " |
for ^$b { if $_ == @f[$_] { print "$_ " } }; |
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hyper for 1 .. $limit div $b -> $i { |
hyper for 1 .. $limit div $b -> $i { |
Revision as of 19:19, 12 August 2019
- Definition
A factorion is a natural number that equals the sum of the factorials of its digits. For example 145 is a factorion in base 10 because:
1! + 4! + 5! = 1 + 24 + 120 = 145.
- Task
It can be shown (see Wikipedia article below) that no factorion in base 10 can exceed 1,499,999.
Write a program in your language to demonstrate, by calculating and printing out the factorions, that:
1. There are 4 factorions in base 10.
2. There are 3 factorions in base 9, 5 factorions in base 11 but only 2 factorions in base 12 up to the same upper bound as for base 10.
- See also
C
<lang c>#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;
}</lang>
- 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
<lang 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</lang>
- 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
Go
<lang 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") }
}</lang>
- 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
Perl 6
<lang perl6>constant @f = 1, |[\*] 1..*;
constant $limit = 1500000;
for 9 .. 12 -> $b {
say "\n\nFactorions in base $b:";
for ^$b { if $_ == @f[$_] { print "$_ " } };
hyper for 1 .. $limit div $b -> $i { my $sum; my $prod = $i * $b;
for $i.polymod($b xx *) { $sum += @f[$_]; $sum = 0 and last if $sum > $prod }
next if $sum == 0;
print "{$sum + @f[$_]} " and last if $sum + @f[$_] == $prod + $_ for ^$b; }
}</lang>
- 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
zkl
<lang zkl></lang> <lang zkl></lang>
- Output: