How about

my ($longest) = reduce { length($a) >= length($b) ? $a : $b  } /
   (
      [0-9]
      (?:
         (?: (?<=0)1 | (?<=1)2 | (?<=2)3 | (?<=3)4 | (?<=4)5
         |   (?<=5)6 | (?<=6)7 | (?<=7)8 | (?<=8)9 | (?<=9)0
         )+
      |
         (?: (?<=0)9 | (?<=1)0 | (?<=2)1 | (?<=3)2 | (?<=4)3
         |   (?<=5)4 | (?<=6)5 | (?<=7)6 | (?<=8)7 | (?<=9)8
         )+
      )
   )
/xg;
[download]

And I thought contiguous would be easier than non-contiguous.

Update: No, that's not right. Endpoints can belong to an ascending and a descending series.

my ($longest) = reduce { length($a) >= length($b) ? $a : $b  }
   /
      (
         [0-9]
         (?: (?<=0)1 | (?<=1)2 | (?<=2)3 | (?<=3)4 | (?<=4)5
         |   (?<=5)6 | (?<=6)7 | (?<=7)8 | (?<=8)9 | (?<=9)0
         )+
      )
   /xg,
   /
      (
         [0-9]
         (?: (?<=0)9 | (?<=1)0 | (?<=2)1 | (?<=3)2 | (?<=4)3
         |   (?<=5)4 | (?<=6)5 | (?<=7)6 | (?<=8)7 | (?<=9)8
         )+
      )
   /xg;
[download]

Update: This can be simplified if you just want the length of the longest rather than the sequence itself.

my $longest =
   max
   map length,
   /
      (
         (?: 1(?=2) | 2(?=3) | 3(?=4) | 4(?=5) | 5(?=6)
         |   6(?=7) | 7(?=8) | 8(?=9) | 9(?=0) | 0(?=1)
         )+
      |
         (?: 1(?=0) | 2(?=1) | 3(?=2) | 4(?=3) | 5(?=4)
         |   6(?=5) | 7(?=6) | 8(?=7) | 9(?=8) | 0(?=9)
         )+
      )
   /xg;

if (defined($longest)) {
    ++$longest;
} else {
    $longest = /[0-9]/ ? 1 : 0;
}
[download]

Ok, maybe not simpler.

I could save typing by using reverse, but it would be a bit slower.