%I #20 Jan 12 2015 03:05:41
%S 1,1,1,2,1,2,1,2,1,1,2,3,1,2,1,3,2,1,3,2,1,3,2,1,1,2,3,4,1,2,3,1,4,2,
%T 1,3,2,4,1,3,2,1,4,3,2,1,4,3,2,1,1,2,3,4,5,1,2,3,4,1,2,5,3,1,4,2,1,3,
%U 5,2,4,1,3,2,5,1,4,3
%N Positive integers A when the positive roots of r^2 = Ar + B are listed in increasing order.
%C Generalize the Fibonacci sequence recurrence equation as: F_(n+1) = A*F_n + B*F_(n1), where A and B are positive integers. As n goes to infinity, the ratio F_n / F_(n1) approaches the positive real number r = (A + sqrt(A*A + 4B))/2. This sequence gives the A values in increasing order of r.
%C In case of a tie in r values, then sort in increasing order of sqrt(A*A + B*B).
%C This A sequence appears to be the ordinal transform of the B sequence (A249974) and vice versa. The associative arrays of A and B are transposes. The first row of A's associative array seems to be A006000.
%C For the A and B values leading to a positive integer limit r see a comment in A063929.  _Wolfdieter Lang_, Jan 12 2015
%e a(6) = 2 because the 6th smallest value of r (approximately 2.732050808) is that for A=2, B=2.
%o (PARI) \\ see A249974
%Y Cf. A249974, A006000.
%K easy,nonn
%O 1,4
%A _Kerry Mitchell_, Nov 09 2014
%E Edited.  _Wolfdieter Lang_, Jan 11 2015
