cp's OEIS Frontend

A268696 a(1)=1; thereafter a(n+1) = floor(m/2), where m = number of occurrences of a(n) in [a(1),...,a(n)].

%I A268696 #37 Feb 12 2016 17:33:34
%S A268696 1,0,0,1,1,1,2,0,1,2,1,3,0,2,1,3,1,4,0,2,2,2,3,1,4,1,5,0,3,2,3,2,4,1,
%T A268696 5,1,6,0,3,3,3,4,2,4,2,5,1,6,1,7,0,4,3,4,3,5,2,5,2,6,1,7,1,8,0,4,4,4,
%U A268696 5,3,5,3,6,2,6,2,7,1,8,1,9,0,5,4,5,4,6,3,6,3,7,2,7,2,8,1,9,1,10,0
%N A268696 a(1)=1; thereafter a(n+1) = floor(m/2), where m = number of occurrences of a(n) in [a(1),...,a(n)].
%C A268696 It appears that a(n) <= ceiling(sqrt(n)).
%C A268696 Comments from _N. J. A. Sloane_, Feb 12 2016: (Start)
%C A268696 In fact it appears that a(n) <= floor(sqrt(n)) except when n belongs to the sequence S := [99, 120, 142, 167, 193, 222, 252, 285, 319, ...], which has second differences 1,3,1,3,1,3,... and is the sequence {99; A035608(k)+21*k+120, k>=0}. For these values of n it appears that a(n) = ceiling(sqrt(n)). The first example is a(99) = 10 = ceiling(sqrt(99)).
%C A268696 The zeros occur at positions [2, 3, 8, 13, 19, 28, 38, 51, 65, 82, 100, 121, 143, 168, 194, 223, 253, 286, 320, ...], which apart from the initial terms appears to be S+1.
%C A268696 Without the division by 2 in the definition (that is, if a(n+1)=m), we get A158416. (End)
%H A268696 N. J. A. Sloane, <a href="/A268696/b268696.txt">Table of n, a(n) for n = 1..3000</a>, Feb 12 2016 [First 998 terms from David Seelmann]
%e A268696 a(2) is equal to the number of times a(1) = 1 appears in the sequence before, divided by two, rounding down. Since 1 appears once before, a(2) = floor(1/2) = 0.
%e A268696 a(3) is  equal to the number of times 0 appears in the sequence before, which is again once, divided by two, rounding down. So a(3) = floor(1/2) = 0.
%e A268696 a(4) is the number of times 0 appears before (twice) divided by two, which gives us 1.
%t A268696 a = {1}; Do[AppendTo[a, Floor[Count[a, n_ /; n == a[[k - 1]]]/2]], {k, 2, 120}]; a (* _Michael De Vlieger_, Feb 11 2016 *)
%Y A268696 Cf. A035608, A158416.
%K A268696 nonn,look,hear
%O A268696 1,7
%A A268696 _David Seelmann_, Feb 11 2016
%E A268696 More terms from _Michael De Vlieger_, Feb 11 2016