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%I #10 Feb 22 2013 14:38:48
%S 5,7,13,19,21,27,29,35,37,43,49,51,57,59,65,67,71,73,79,81,87,89,95,
%T 97,101,103,109,111,117,119,125,131,133,139,141,147,149,155,161,163,
%U 169,171,177,179,183,185,191,193,199,201,207,213,215,221,223,229,231,237
%N Odd numbers m such that the sequence defined by b(1) = m; for k>1, b(k) = floor((1+sqrt(3))*b(k-1)) contains only odd numbers.
%C Conjecture: let r(z)= (1/2) *(z+sqrt(z^2+4*z)), for any integral z>=1. Then the sequence a(n)-4n (where a(n) is the sequence of odd numbers m such that the sequence defined by b(1) = m; for k>1, b(k) = floor(r(z)*b(k-1)) contains only odd numbers) becomes ultimately periodic. _Benoit Cloitre_, Aug 10 2003
%F Observation: a(n+1)-a(n) = 2, 4 or 6 for every n, a(n)=4n+O(1) and more precisely it seems that abs(a(n)-4n)<=9. Is the sequence a(n)-4n ultimately periodic ? _Benoit Cloitre_, Aug 10 2003
%e For m = 5 we get 5, 13, 35, 95, 259, 707, 1931, 5275, 14411, 39371, ... (cf. A057960).
%Y Cf. A086843.
%K nonn
%O 1,1
%A _Philippe Deléham_, Aug 09 2003
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