0,2

One can prove that if in the sequence of numbers N for which A010060(N+2^n)= A010060(N) to replace the odious (evil) terms by 1's (0's), then we obtain 2^(n+1)-periodic (0,1)-sequence; if to write it in the form .xx...,i.e., as a binary infinite fraction, then the corresponding fraction has the form a(n)/A000215(n). These fractions very fast coverge to the Thue-Morse constant .4124540336401...; e.g a(5)/(2^32+1)approximates this constant up to 10^(-9). These approximations differ from A074072-A074073. Conjecture. For n>=1, the fraction a(n)/A000215(n) is a convergent corresponding to the continued fraction for the Thue-Morse constant.

V. Shevelev, Equations of the form t(x+a)=t(x) and t(x+a)=1-t(x) for Thue-Morse sequence,

a(1)=2, and, for n>=1, a(n+1)=1+(2^(2^n)-1)a(n).

Cf. A010060 A000215 A085394 A08395 A081706 A161627 A161639 A074072 A074073

Sequence in context: A102598 A102747 A122524 this_sequence A072664 A045310 A000157

Adjacent sequences: A162631 A162632 A162633 this_sequence A162635 A162636 A162637

nonn,uned

Vladimir Shevelev (shevelev(AT)bgu.ac.il), Jul 08 2009, Jul 14 2009