

A249973


Positive integers A when the positive roots of r^2 = Ar + B are listed in increasing order.


1



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, 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, 5, 2, 4, 1, 3, 2, 5, 1, 4, 3
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OFFSET

1,4


COMMENTS

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.
In case of a tie in r values, then sort in increasing order of sqrt(A*A + B*B).
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.
For the A and B values leading to a positive integer limit r see a comment in A063929.  Wolfdieter Lang, Jan 12 2015


LINKS



EXAMPLE

a(6) = 2 because the 6th smallest value of r (approximately 2.732050808) is that for A=2, B=2.


PROG



CROSSREFS



KEYWORD

easy,nonn


AUTHOR



EXTENSIONS



STATUS

approved



