A162634 Numerators of fractions with denominators A000215(n) approximating the Thue-Morse constant.

1, 2, 7, 106, 27031, 1771476586, 7608434000728254871, 140350834813144189858090274002849666666, 47758914269546354982683078068829456704164423862093743397580034411621752859031

Offset: 0

Author

Vladimir Shevelev, Jul 08 2009, Jul 14 2009

Keywords

• nonn

Comments

One can prove that if in the sequence of numbers N for which A010060(N+2^n)= A010060(N) you replace the odious (evil) terms by 1's (0's), then we obtain 2^(n+1)-periodic (0,1)-sequence; if you 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 converge 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.

Links

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

Crossrefs

Cf. A010060, A000215, A085394, A085395, A081706, A161627, A161639, A074072, A074073.

Programs

PARI

a(n)=if(n<=1, [1,2][n+1], 1+(2^(2^(n-1))-1)*a(n-1)); /* Joerg Arndt, Mar 11 2013 */

Formula

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

Extensions

Added more terms, Joerg Arndt, Mar 11 2013