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a(1)=1; a(2n)=a(n)+1, a(2n+1)=a(n) mod 2.
1

%I #11 Aug 26 2020 07:06:13

%S 1,2,1,3,0,2,1,4,1,1,0,3,0,2,1,5,0,2,1,2,1,1,0,4,1,1,0,3,0,2,1,6,1,1,

%T 0,3,0,2,1,3,0,2,1,2,1,1,0,5,0,2,1,2,1,1,0,4,1,1,0,3,0,2,1,7,0,2,1,2,

%U 1,1,0,4,1,1,0,3,0,2,1,4,1,1,0,3,0,2,1,3,0,2,1,2,1,1,0,6,1,1,0,3,0,2,1,3,0

%N a(1)=1; a(2n)=a(n)+1, a(2n+1)=a(n) mod 2.

%H Robert Israel, <a href="/A091829/b091829.txt">Table of n, a(n) for n = 1..10000</a>

%F n>0 a(2*A059009(n)+1)=0; n>0 a(n) mod 2 = 1-A059448(n); a(2^n)=n+1

%p f:= proc(n) option remember;

%p if n::even then procname(n/2)+1 else procname((n-1)/2) mod 2 fi

%p end proc:

%p f(1):= 1:

%p map(f, [$1..100]); # _Robert Israel_, Jan 22 2018

%t a[1] = 1;

%t a[n_] := a[n] = If[OddQ[n], Mod[a[n-1]+1, 2], a[n/2]+1];

%t Array[a, 100] (* _Jean-François Alcover_, Aug 26 2020 *)

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

%K nonn

%O 1,2

%A _Benoit Cloitre_, Mar 09 2004