Here's a first pass at it.

It does partial math (same number of digits on both sides) to cut off further consideration because of duplicates or 0.

It does base 18 in under 3 minutes. 

#!/usr/bin/perl -l

# https://perlmonks.org/?node_id=1218964

use strict;
use warnings;
use POSIX;

my $base = shift // 10;
my @numbers = (0..9, 'a'..'z')[0..$base-1];
my $numberlength = $base - 1 >> 1;
my $pattern = '-' x $numberlength . '=' . '-' x $numberlength . '*';

sub inbase
  {
  my ($n, $b) = @_;
  my $ans = '';
  while($n > 0)
    {
    $ans = $numbers[$n % $b] . $ans;
    $n = int $n / $base;
    }
  $ans || 0;
  }

# dddd=dddd*d   # for base 10

my $solutions = my $count = 0;
my @queue = map "$pattern$_", @numbers[2..$#numbers];
while( @queue )
  {
  $count++;
  $_ = shift @queue;
  /(\w).*\1/ and next;
  if( !/.*=.*\K-/ )
    {
    print;
    $solutions++;
    next;
    }
  my ($before, $after) = ($`, $');
  my $mul = POSIX::strtol( substr( $after, -1 ), $base );

  for my $d ( @numbers[1..$#numbers] )
    {
    $_ =~ $d and next;
    my $new = "$before$d$after";
    $new =~ /=-*(\w+)/ or next;;
    my $len = length $1;
    my $prod = POSIX::strtol( $1, $base ) * $mul;
    my $baseprod = inbase( $prod, $base );
    length $baseprod > $numberlength and next;

    substr $new, $numberlength - $len, $len, substr $baseprod, -$len;
    $new =~ /0|(\w).*\1/ and next;
    push @queue, $new;
    }
  }
print "\nsteps $count  solutions $solutions";

[download]

In reply to Re: Efficient enumeration of pandigital fractions by tybalt89
in thread Efficient enumeration of pandigital fractions by kikuchiyo

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