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 A072939 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. 7

%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(2k-1) = 2*m(n) and c(2k+1)-c(2k)=-m(n). Sometimes m(n) = (n-1)/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 0-word 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(2k-1)= 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

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Last modified December 5 10:45 EST 2019. Contains 329751 sequences. (Running on oeis4.)