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A138467 a(1)=1, then for n>=2 a(n) = n - floor((1/3)*a(a(n-1))).

%I A138467 #23 Jul 14 2023 09:04:29
%S A138467 1,2,3,3,4,5,6,7,7,8,9,10,11,11,12,13,14,15,15,16,17,18,18,19,20,21,
%T A138467 22,22,23,24,25,26,26,27,28,29,30,30,31,32,33,34,34,35,36,37,37,38,39,
%U A138467 40,41,41,42,43,44,45,45,46,47,48,49,49,50,51,52,53,53,54,55,56,56,57,58
%N A138467 a(1)=1, then for n>=2 a(n) = n - floor((1/3)*a(a(n-1))).
%C A138467 For k >= 1, a(1)=1; a(n) = n - floor((1/k)*a(a(n-1))). - _Yalcin Aktar_, Jul 13 2008
%C A138467 Conjecture: a(n) = floor(r(p)*(n+1)) with r(p)=(1/2)*(sqrt(p*(p+4))-p). - _Yalcin Aktar_, Jul 13 2008
%C A138467 From _Michel Dekking_, Jul 13 2023: (Start)
%C A138467 Here is a correction of these two comments from July 13, 2008:
%C A138467     Consider the following generalization of (a(n)).
%C A138467   Let p>2 be a natural number, and define the sequence b_p by
%C A138467   b_p = 1, b_p(n) = n - floor((1/p)*b_p (b_p (n-1))   for   n>1.
%C A138467   Conjecture: b_p(n) = floor(r(p)*(n+1)) where r(p) =(1/2)*(sqrt(p*(p+4))-p).(End)
%H A138467 Michel Dekking, <a href="/A138466/a138466.png">Proof by Benoit Cloitre</a>
%F A138467 For n>=1, a(n) = floor(r*(n+1)) where r=(3/2)*(sqrt(7/3)-1).
%o A138467 (PARI) a(n)=floor((3/2)*(sqrt(7/3)-1)*(n+1))
%Y A138467 Cf. A005206, A135414.
%K A138467 nonn
%O A138467 1,2
%A A138467 _Benoit Cloitre_, May 09 2008
%E A138467 More terms from _Yalcin Aktar_, Jul 13 2008