If you look at the first problem in the sequence as hinted by Gamer (pid=249) and the hints provided by Dustin, the reasoning behind this sequence is similar to the prior problem combined with a "look back" not just of the previous term but all the prior terms.
For each pair of digits, the first is the number of occurrences of the second digit in all of the lines above. The even digits are listed in increasing order. If no instances of a digit have occurred then rather than a zero followed by that digit, these digits are skipped.
Only every other term is being shown.
The entire sequence is as follows:
1
11
31
4113
612314
8112332416
B13253342618
D15283441536281B
G172A3643546482B1D
I192C394456617581A3B2D1G
M1B2E3B465862768292A4B1C3D2G1I

Posted by Larry
on 20130803 21:10:58 