It makes a small  insignificant  difference to the outcome.
A quick check  because quicker than trying to derive a formula  shows that of the 729,000 3value combinations, 11,839 contain consecutive numbers:
use Algorithm::Combinatorics qw[ variations_with_repetition ];;
@c = variations_with_repetition( [ 1 .. 90 ], 3 );;
print scalar @c;;
729000
$d = 0; $_>[0]+1 == $_>[1] or $_>[1]+1 == $_>[2] and ++$d for @c;
+print $d;;
7922
And there are 8010 combinations of 2 sets of three that must be excluded because the last digit of the first set of 3 is one less that the first digit of the second set: $d = 0; $_>[0] == $_>[2]+1 and ++$d for @c; print $d;;
8010
Which means that instead of 531,441,000,000 6value combinations, there are only (729,000  7922 )**2  8010 = 519,953,474,074, which means it would still take 40bits to represent any legal 6value string; so the math doesn't change: ( 1  40/48 ) *100 = 16.67% is best possible for *any* dataset.
Any algorithm that achieves better on any given dataset; will not achieve the same results for all datasets; nor even a high percentage of them.
Ie. To achieve better, you'd need to reduce the size of the domain so that you could compress 36bits into 48 which would give 25% compression. But that would require throwing away 87% of the possible datasets
With the rise and rise of 'Social' network sites: 'Computers are making people easier to use everyday'
Examine what is said, not who speaks  Silence betokens consent  Love the truth but pardon error.
In the absence of evidence, opinion is indistinguishable from prejudice.
Suck that fhit
