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

%I #23 Jul 14 2023 09:04:29

%S 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 22,22,23,24,25,26,26,27,28,29,30,30,31,32,33,34,34,35,36,37,37,38,39,

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

%C For k >= 1, a(1)=1; a(n) = n - floor((1/k)*a(a(n-1))). - _Yalcin Aktar_, Jul 13 2008

%C Conjecture: a(n) = floor(r(p)*(n+1)) with r(p)=(1/2)*(sqrt(p*(p+4))-p). - _Yalcin Aktar_, Jul 13 2008

%C From _Michel Dekking_, Jul 13 2023: (Start)

%C Here is a correction of these two comments from July 13, 2008:

%C Consider the following generalization of (a(n)).

%C Let p>2 be a natural number, and define the sequence b_p by

%C b_p = 1, b_p(n) = n - floor((1/p)*b_p (b_p (n-1)) for n>1.

%C Conjecture: b_p(n) = floor(r(p)*(n+1)) where r(p) =(1/2)*(sqrt(p*(p+4))-p).(End)

%H Michel Dekking, <a href="/A138466/a138466.png">Proof by Benoit Cloitre</a>

%F For n>=1, a(n) = floor(r*(n+1)) where r=(3/2)*(sqrt(7/3)-1).

%o (PARI) a(n)=floor((3/2)*(sqrt(7/3)-1)*(n+1))

%Y Cf. A005206, A135414.

%K nonn

%O 1,2

%A _Benoit Cloitre_, May 09 2008

%E More terms from _Yalcin Aktar_, Jul 13 2008