%I
%S 3,7,9,11,15,19,23,25,27,31,33,35,39,41,43,47,51,55,57,59,63,67,71,73,
%T 75,79,83,87,89,91,95,97,99,103,105,107,111,115,119,121,123,127,129,
%U 131,135,137,139,143,147,151,153,155,159,161,163,167,169,171,175,179
%N Define a sequence c depending on n by: c(1)=1 and c(2)=n; c(k+2) = (c(k+1) + c(k))/2 if c(k+1) and c(k) have the same parity; otherwise c(k+2)=abs(c(k+1)2*c(k)); sequence gives values of n such that lim k > infinity c(k) = infinity.
%C If c(2) is even then c(k) = 1 for k >= 2*c(2), hence there is no even value in the sequence. If n is in the sequence, there is an integer k(n) and an integer m(n) such that for any k >= k(n) c(2k)c(2k1) = 2*m(n) and c(2k+1)c(2k)=m(n). Sometimes m(n) = (n1)/2 but not always. If B(n) = a(n+1)a(n) then B(n) = 2 or 4, but B(n) does not seem to follow any pattern.
%C Conjecture: a(n) = A036554(n)+1.  _Vladeta Jovovic_, Apr 01 2003
%C a(n) = A036554(n)+1 = A079523(n)+2.  _Ralf Stephan_, Jun 09 2003
%C Conjecture: this sequence gives the positions of 0's in the limiting 0word of the morphism 0>11, 1>10, A285384.  _Clark Kimberling_, Apr 26 2017
%C Conjecture: This also gives the positions of the 1's in A328979.  _N. J. A. Sloane_, Nov 05 2019
%F Conjecture : lim n > infinity a(n)/n = 3.
%e if c(2)=41 > c(3)= 21 >c(4)= 31 >c(5)= 26 >c(6)= 36 >c(7)= 31 >c(8)= 41 >c(9)= 36 then c(2k)c(2k1)= 10 c(2k+1)c(2k) =  5 for k >=2 implies that c(k) > infinity hence 41 is in the sequence.
%Y Cf. A036554, A079523, A285384, A328979.
%K nonn
%O 1,1
%A _Benoit Cloitre_, Aug 12 2002
