OFFSET
1
COMMENTS
Let u(n) be the initial (2^n)-letter subword of A010060, and let v(n) be the word remaining when {0100->1} is applied to u(n). Then for n>=2, |v(n)| = 3*2^(n-2) - 1 if n is odd, and |v(n)| = 3*2^(n-2)+1 if n is even.
LINKS
Clark Kimberling, Table of n, a(n) for n = 1..10000
EXAMPLE
As a word, A010060 = 0110100110010110100101100..., and replacing each 0100 by 1 gives 0111110010111101100111110010...
MATHEMATICA
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, May 09 2017
STATUS
approved