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A062769
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Smallest number m such that the continued fraction expansion of sqrt(m) has period length 2n + 1.
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0
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2, 41, 13, 58, 106, 61, 193, 109, 157, 337, 181, 586, 457, 949, 821, 601, 613, 1061, 421, 541, 1117, 1153, 1249, 1069, 1021, 1201, 1669, 2381, 1453, 2137, 2053, 1801, 2293, 1381, 1549, 3733, 3541, 3217, 5857, 1621, 3169, 4657, 2689, 3049, 2389, 4057
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,1
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COMMENTS
| 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
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LINKS
| D. Alpern, Continued Fraction calculator
K. Matthews, Calculating the simple continued fraction of a quadratic irrational
U. Sondermann, Continued Fractions
G. Xiao, Contfrac,continued fraction expansion server with k-th convergent calculator.
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EXAMPLE
| 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], ...
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CROSSREFS
| Cf. A013646, A003814, A031396, A003654.
Sequence in context: A104134 A162868 A059476 * A033841 A107194 A154767
Adjacent sequences: A062766 A062767 A062768 * A062770 A062771 A062772
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KEYWORD
| nonn
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AUTHOR
| Lekraj Beedassy (blekraj(AT)yahoo.com), Jul 17 2001
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EXTENSIONS
| More terms from Naohiro Nomoto (n_nomoto(AT)yabumi.com), Jan 01 2002
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