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%I #11 Mar 30 2012 18:55:49
%S 4,2,1,8,14,4,36,18,2,9,28,6,64,32,338,16,3,392,46,8,100,50,14,25,20,
%T 3038
%N Initialized continued fraction for sqrt(n-th nonsquare) has period (1,a(n)).
%C The definition of initialized continued fraction depends on the usual recurrences for ordinary simple continued fractions:
%C ...
%C (eq 1).... p(i)=a(i)*p(i-1)+p(i-2) for i>=2;
%C (eq 2)... q(i)=a(i)*q(i-1)+q(i-2) for i>=2.
%C ...
%C Suppose that p(0), q(0), p(1), q(1) are nonnegative integers such that {p(0), p(1)} is not {0} and {q(0), q(1)} is not {0}. Suppose [b(0),b(1),b(2),...] is an ordinary simple continued fraction of a number B. The equations (1) and (2) with b(i) substituted for a(i) are used to define "initialized convergents" p(i)/q(i). The limit of these exists and defines the "initialized continued fraction" of B, denoted by [(p(0),q(0),p(1),q(1)) ; b(0),b(1),b(2),...].
%C ...
%C Examples:
%C sqrt(2)=[(1,1,6,4) ; 1,4,1,4,1,4,1,4,1,4,...]
%C sqrt(3)=[(1,1,4,2) ; 1,2,1,2,1,2,1,2,1,2,...]
%C sqrt(5)=[(1,1,3,1) ; 1,1,1,1,1,1,1,1,1,1,...]
%C sqrt(76)=[(2,2,3302,377) ; 1,338,1,338,1,338,...]
%C ...
%C Formerly, an initialized continued fraction was here called a pseudo-continued fraction. [From Clark Kimberling, Jun 23 2011]
%D C. Kimberling, "Initialized continued fractions and Fibonacci numbers," in Proceedings of the Twelfth International Conference on Fibonacci Numbers and Their Applications, Congressus Numerantium 200 (2010) 269-284.
%e a(1)=4 because sqrt(2)=[(1,1,6,4) ; 1,4,1,4,1,4,...]
%e a(2)=2 because sqrt(3)=[(1,1,4,2) ; 1,2,1,2,1,2,...]
%K nonn,more
%O 1,1
%A _Clark Kimberling_
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