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a(1) = 1, a(n) = n - a(ceiling(n/2)).
3

%I #17 Sep 10 2020 06:36:25

%S 1,1,2,3,3,4,4,5,6,7,7,8,9,10,10,11,11,12,12,13,14,15,15,16,16,17,17,

%T 18,19,20,20,21,22,23,23,24,25,26,26,27,27,28,28,29,30,31,31,32,33,34,

%U 34,35,36,37,37,38,38,39,39,40,41,42,42,43,43,44,44,45,46,47,47,48,48

%N a(1) = 1, a(n) = n - a(ceiling(n/2)).

%H Amiram Eldar, <a href="/A076895/b076895.txt">Table of n, a(n) for n = 1..10000</a>

%F a(n) is asymptotic to (2/3)*n.

%F a(n) = A050292(n-1) + (1+(1-2*(C(n)-F(n)))*(-1)^F(n))/2 where C(n) = ceiling(log_2(n)); F(n) = floor(log_2(n)) and A050292(n) (with A050292(0)=0) is the maximum cardinality of a double-free subset of {1, 2, ..., n}. So using Wang's asymptotic formula for A050292, a(n) = (2/3)*n + O(log_4(n)). Series expansion: (1/(x-1)) * ( 1/(x-1) + Sum_{i>=1} (-1)^i*( x^(2^i)/(x^(2^i)-1) - x^(2^(i-1)) ). - Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Feb 17 2003

%F a(n+1) = b(n) with b(0)=0, b(2n) = -b(n) + 2n+1, b(2n+1) = -b(n) + 2n + 2 -[n==0]. Also a(n+1) = A050292(n) + A030301(n). - _Ralf Stephan_, Oct 28 2003

%t a[1] = 1; a[n_] := a[n] = n - a[Ceiling[n/2]]; Array[a, 100] (* _Amiram Eldar_, Sep 10 2020 *)

%o (PARI) a(n)=if(n<2,1,n-a(ceil(n/2)))

%Y Cf. A050292, A079411.

%K nonn

%O 1,3

%A _Benoit Cloitre_, Nov 26 2002

%E Edited by _Ralf Stephan_, Sep 01 2004