[seqfan] Further observations on the parity A233312

[Jeremy Gardiner]

Run lengths in the parity of A233312:
1,1,1,1,1,3,1,1,1,5,1,1,1,1,1,11,1,1,1,1,1,3,1,1,1,21,1,1,1,1,1,3,1,1,1,5,1,
1,1,1,1,43,1,1,1,1,1,3,1,1,1,5,1,1,1,1,1,11,1,1,1,1,1,3,1,1,1,85,1,1,1,1,1,3
,...

Run lengths with repeated 1's removed: call this a(n)
3,5,11,3,21,3,5,43,3,5,11,3,85,3,5,11,3,21,3,5,171,3,5,11,3,21,3,5,43,3,5,11
,3,341,3,5,11,3,21,3,5,43,3,5,11,3,85,3,5,11,3,21,3,5,683,3,5,11,3,21,3,5,43
,...

Cf. Jacobsthal numbers A001045.

Now b(n)=(a(n)-1)/2:
1,2,5,1,10,1,2,21,1,2,5,1,42,1,2,5,1,10,1,2,85,1,2,5,1,10,1,2,21,1,2,5,1,170
,1,2,5,1,10,1,2,21,1,2,5,1,42,1,2,5,1,10,1,2,341,1,2,5,1,10,1,2,21,1,2,5,1,4
2,1,2,5,1,10,1,2,85,1,2,5,1,10,1,2,21,1,2,5,1,682,1,2,5,1,10,1,2,21,1,2,5,1,
...

We have apparently A085358 Runs of zeros in binomial(3k,k)/(2k+1) (Mod 2):
relates ternary trees (A001764) to the infinite Fibonacci word (A003849).

Jeremy