>>12677521Hey same guy here. I thought about it again while having a snack. Im not going to try to give a formal mathematical solution because I tried it earlier and was not a fcking good idea.
Anyways the thing is that when n is odd, 2^n+n-1 is even. So when the sequence reaches its maximun length without repeated blocks last turn was from second player and thus, first player is going to loose. That means second player will always win if he can "force" this situation. As the two earlier posts say, its relatively intuitive to see that if the second player puts always the "opposite" digit he will force this situation and win because he is kind of forcing the sequency to continually change, hence avoiding the repetition of blocks.
This is one of those problems that seems simple at first, but the more you think about it the more tricky it gets. Nice.
