[seqfan] Re: Look and say -- my parity

[Hans Havermann]

> Jean-Marc Falcoz just computed this 62-term chain, telling me that this might be beaten.

Without the "extend W with the smallest available integer" restriction, there's a chance of infinite chains. Start with '22'. Except I'm not going to look at the numerical representations, just the (odd,even) digit counts. So, start with (0,2):

# 1   1:  (0,2).
# 2   2:  (0,4), (3,3).
# 3   3:  (0,6), (3,5), (4,6).
# 4   4:  (0,8), (3,7), (4,8), (7,7).
# 5   6:  (3,9), (3,10), (7,9), (7,10), (10,9), (10,10).
# 6   6:  (6,11), (7,11), (10,12), (11,12), (14,11), (15,11).
# 7   7:  (8,14), (10,13), (10,14), (15,13), (15,14), (18,13), (18,14).
# 8   7:  (13,15), (14,15), (15,15), (18,16), (19,16), (20,18), (21,16).
# 9   8:  (16,18), (18,17), (18,18), (21,20), (22,19), (23,20), (23,21), (24,19).
#10   8:  (18,22), (20,21), (21,21), (23,24), (24,23), (24,26), (25,24), (26,23).
...

#62  16:  (213,205), (228,189), (229,188), (230,187), (231,186), (231,187), (232,185), (232,186), (233,184), (234,183), (234,184), (235,182), (235,183), (236,181)*, (236,182), (239,179).

(236,181)* corresponds to Falcoz's final term 18122361 so I'm hopeful that I've set this up correctly and that my program works as expected. I'm not interested in chaining the counts across iterations, only in how many counts there are, because if that number drops to zero we are done and the chain is finite. Here's a graph of the counts:

http://chesswanks.com/num/EvenOdd.png