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

[Frank Adams-Watters]

That looks like a definite maybe to me. Interesting graph.

Franklin T. Adams-Watters

-----Original Message-----
From: Hans Havermann <gladhobo at teksavvy.com>

> 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