Factorions: Difference between revisions
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Factorions in base 12:
1 2</pre>
Alternatively, a more efficient approach:
{{trans|Sidef}}
{{libheader|ntheory}}
<lang perl>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)";
}</lang>
{{out}}
<pre>
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)
</pre>
=={{header|Perl 6}}==
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