Search: id:A062769
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%I A062769
%S A062769 2,41,13,58,106,61,193,109,157,337,181,586,457,949,821,601,613,1061,
%T A062769 421,541,1117,1153,1249,1069,1021,1201,1669,2381,1453,2137,2053,1801,
%U A062769 2293,1381,1549,3733,3541,3217,5857,1621,3169,4657,2689,3049,2389,4057
%N A062769 Smallest number m such that the continued fraction expansion of sqrt(m)
has period length 2n + 1.
%C A062769 If continued fraction for sqrt(N) has period length (2k + 1) and k-th
convergent P(k)/Q(k) [taking P(-1)=1; Q(-1)=0 where necessary], then
the i-th positive solution V(i)=[x(i),y(i)] to the Pell equation
x^2-Ny^2=1 satisfies the recurrence V(i+2) = 2AV(i+1) - V(i) starting
with V(0)=(1,0); V(1)=(A,B) where A = 2S^2 + 1; B = 2ST and S = P(k)Q(k)
+ P(k-1)Q(k-1); T = Q(k)^2 + Q(k-1)^2
%H A062769 D. Alpern, Continued
Fraction calculator
%H A062769 K. Matthews, Calculating
the simple continued fraction of a quadratic irrational
%H A062769 U. Sondermann,
Continued Fractions
%H A062769 G. Xiao,
Contfrac,continued fraction expansion server with k-th convergent
calculator.
%e A062769 For n = 2, 2n+1 = 5. a(2) = 13 and we indeed have sqrt(13) = [3; 1, 1,
1, 1, 6] with period length 5, the first one in the sequence sqrt(29)
= [5; 2, 1, 1, 2, 10], sqrt(53) = [7; 3, 1, 1, 3, 14], sqrt(74) =
[8; 1, 1, 1, 1, 16], sqrt(85) = [9; 4, 1, 1, 4, 18], sqrt(89) = [9;
2, 3, 3, 2, 18], ...
%Y A062769 Cf. A013646, A003814, A031396, A003654.
%Y A062769 Sequence in context: A104134 A162868 A059476 this_sequence A033841 A107194
A154767
%Y A062769 Adjacent sequences: A062766 A062767 A062768 this_sequence A062770 A062771
A062772
%K A062769 nonn
%O A062769 0,1
%A A062769 Lekraj Beedassy (blekraj(AT)yahoo.com), Jul 17 2001
%E A062769 More terms from Naohiro Nomoto (n_nomoto(AT)yabumi.com), Jan 01 2002
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