A115384 Partial sums of Thue-Morse numbers A010060.
0, 1, 2, 2, 3, 3, 3, 4, 5, 5, 5, 6, 6, 7, 8, 8, 9, 9, 9, 10, 10, 11, 12, 12, 12, 13, 14, 14, 15, 15, 15, 16, 17, 17, 17, 18, 18, 19, 20, 20, 20, 21, 22, 22, 23, 23, 23, 24, 24, 25, 26, 26, 27, 27, 27, 28, 29, 29, 29, 30, 30, 31, 32, 32, 33, 33, 33, 34, 34, 35, 36, 36, 36, 37, 38, 38
Offset: 0
Links
- G. C. Greubel, Table of n, a(n) for n = 0..10000
- Hassan Tarfaoui, Concours Général 1990 - Exercice 1 (in French).
- Index to sequences related to Olympiads and other Mathematical competitions.
Programs
-
Mathematica
Accumulate[Nest[Flatten[#/.{0->{0,1},1->{1,0}}]&,{0},7]] (* Peter J. C. Moses, Apr 15 2013 *) Accumulate[ThueMorse[Range[0,100]]] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Jun 02 2017 *) -
PARI
a(n)=n\2 + (n%2 || hammingweight(n+1)%2==0) \\ Charles R Greathouse IV, Mar 22 2013
-
Python
def A115384(n): return (n>>1)+(n&1|((n+1).bit_count()&1^1)) # Chai Wah Wu, Mar 01 2023
Formula
a(n) = Sum_{k=0..n} A010060(k)^2.
a(n+1) = A115382(2n, n).
a(n)/n -> 1/2; a(n) = number of odious numbers <= n, see A000069. - Reinhard Zumkeller, Aug 26 2007, corrected by M. F. Hasler, May 22 2017.
a(n) = Sum_{i=1..n} (S2(n) mod 2), where S2 = binary weight; lim a(n)/n = 1/2. More generally, consider a(n) = Sum_{i=1..n} (F(Sk(n)) mod m), where Sk(n) is sum of digits of n, n in base k; F(t) is an arithmetic function; m integer. How does lim a(n)/n depend on F(t)? - Ctibor O. Zizka, Feb 25 2008
a(n) = n + 1 - A159481(n). - Reinhard Zumkeller, Apr 16 2009
a(n) = floor((n+1)/2)+(1+(-1)^n)*(1-(-1)^A000120(n))/4. - Vladimir Shevelev, May 27 2009
G.f.: (1/(1 - x)^2 - Product_{k>=1} (1 - x^(2^k)))/2. - Ilya Gutkovskiy, Apr 03 2019
a(n) = A026430(n+1) - n - 1. - Michel Dekking, Sep 17 2019
a(2n+1) = n+1 (see Hassan Tarfaoui link, Concours Général 1990). - Bernard Schott, Jan 21 2022
Extensions
Edited by M. F. Hasler, May 22 2017