So here's my thinking. What you are really trying to figure out is how long your sequence is; the concept of start is arbitrary. Therefore, what you should be doing is a rolling bitwise xor; whenever your checksum is zero, you have potentially passed through two cycles. I don't know what your alphabet looks like, but presumably you could do better than ASCII encoding to lower your collision rate.

#!/usr/bin/perl
use strict;
use warnings;
use 5.10.0;

my $seq = 'deabcdefabcdefgAbcdeabcdefabcdefgAbcdeabcdefabcdefgAbcdeabc
+defabcdefgAbcdeabcdefabcdefgAbcdeabcdefabcdefg';
my %val = do{my $cnt = 0; map {$_ => ++$cnt} split //, 'Aabcdefg'};

my $xor = 0;
my $loop = 0;
for (split //, $seq) {
    $loop++;
    $xor ^= $val{$_};
    say $loop if !$xor;
}
[download]

outputs

9
36
45
72
81
[download]

Note that the only candidate that has both 2*x and 4*x in the list is the correct answer, 18. Setting the %val hash to one-hots instead of consecutive integers gets rid of the noise, but also chews up your bit vector pretty quickly for large alphabets.

my %val = do{my $cnt = 0; map {$_ => 2**(++$cnt)} split //, 'Aabcdefg'
+};
[download]