cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A074072 Numerators of iterations of Thue-Morse sequence.

Original entry on oeis.org

0, 1, 3, 105, 13515, 1771476585, 3804217000364127435, 140350834813144189858090274002849666665, 23879457134773177491341539034414728352082211931046871698790017205810876429515
Offset: 0

Views

Author

Eric W. Weisstein, Aug 18 2002

Keywords

Examples

			Fractions begin with 0, 1/4, 3/8, 105/256, 13515/32768, 1771476585/4294967296, ...
		

Crossrefs

Cf. A074073 (denominators).

Programs

  • Mathematica
    f[0] = 0; f[n_] := (2^(-2^n))*(2^(2^(n - 1)) - 1)*(2^(2^(n - 1))*f[n - 1] + 1); Numerator@ Array[f, 9, 0] (* Amiram Eldar, Oct 05 2023 *)

Extensions

a(8) from Amiram Eldar, Oct 05 2023

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

Original entry on oeis.org

1, 2, 7, 106, 27031, 1771476586, 7608434000728254871, 140350834813144189858090274002849666666, 47758914269546354982683078068829456704164423862093743397580034411621752859031
Offset: 0

Views

Author

Vladimir Shevelev, Jul 08 2009, Jul 14 2009

Keywords

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.

Crossrefs

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

A122497 Let f(S) denote the interchange of 1's and 2's in S. Let S_0 = 1, S_{N+1} = f(S_N).S_N, where the dot indicates concatenation. Sequence gives S_0.S_1.S_2.S_3....

Original entry on oeis.org

1, 2, 1, 1, 2, 2, 1, 2, 1, 1, 2, 1, 2, 2, 1, 1, 2, 2, 1, 2, 1, 1, 2, 2, 1, 1, 2, 1, 2, 2, 1, 2, 1, 1, 2, 1, 2, 2, 1, 1, 2, 2, 1, 2, 1, 1, 2, 1, 2, 2, 1, 2, 1, 1, 2, 2, 1, 1, 2, 1, 2, 2, 1, 1, 2, 2, 1, 2, 1, 1, 2, 2, 1, 1, 2, 1, 2, 2, 1, 2, 1, 1, 2, 1, 2, 2, 1, 1, 2, 2, 1, 2, 1, 1, 2, 2, 1, 1, 2, 1, 2, 2, 1, 1, 2
Offset: 1

Views

Author

Roger L. Bagula, Sep 15 2006

Keywords

Comments

An alternating triangular Morse-Thue sequence based on A010060 using {1,2} instead of {0,1} substitutions.

Examples

			The first few S_i are:
1
2, 1
1, 2, 2, 1
2, 1, 1, 2, 1, 2, 2, 1
1, 2, 2, 1, 2, 1, 1, 2, 2, 1, 1, 2, 1, 2, 2, 1
		

Crossrefs

Programs

  • Mathematica
    ThueMorse[n_, b_] := Nest[Flatten[ # /. {1 -> {1, 2}, 2 -> {2, 1}}] &, {b}, n] a = Table[ThueMorse[n, 1 + Mod[n, 2]], {n, 0, 7}] Flatten[a]

Formula

a(n) = A059448(n) + 1. - Filip Zaludek, Dec 10 2016

Extensions

Edited by N. J. A. Sloane, May 22 2007
Showing 1-3 of 3 results.