%I A115384
%S A115384 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,
%T A115384 15,16,17,17,17,18,18,19,20,20,20,21,22,22,23,23,23,24,24,25,26,26,27,
%U A115384 27,27,28,29,29,29,30,30,31,32,32,33,33,33,34,34,35,36,36,36,37,38,38
%N A115384 Partial sums of squares of Thue-Morse numbers A010060(n).
%C A115384 n/a(n)-->1/2; a(n) = number of odious numbers <= n, cf. A000069. - Reinhard 
               Zumkeller (reinhard.zumkeller(AT)gmail.com), Aug 26 2007
%C A115384 a(n) = n + 1 - A159481(n). [From Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), 
               Apr 16 2009]
%F A115384 a(n)=sum{k=0..n, A010060(k)^2}; a(n+1)=A115382(2n, n).
%F A115384 a(n) = sum_{i=1..n} S2(n) mod 2, where S@ = 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 (ctibor.zizka(AT)seznam.cz), Feb 25 2008
%F A115384 a(n)=floor((n+1)/2)+(1+(-1)^n)*(1-(-1)^A000120(n))/4. [From Vladimir 
               Shevelev (shevelev(AT)bgu.ac.il), May 27 2009]
%Y A115384 Sequence in context: A055037 A125186 A140473 this_sequence A131411 A125059 
               A029112
%Y A115384 Adjacent sequences: A115381 A115382 A115383 this_sequence A115385 A115386 
               A115387
%K A115384 easy,nonn
%O A115384 0,3
%A A115384 Paul Barry (pbarry(AT)wit.ie), Jan 21 2006