A062769 Smallest number m such that the continued fraction expansion of sqrt(m) has period 2n + 1.

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, 4549

Offset: 0

Author

Lekraj Beedassy, Jul 17 2001

Keywords

• nonn

Comments

If the continued fraction for sqrt(N) has period (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 - N*y^2 = 1 satisfies the recurrence V(i+2) = 2*A*V(i+1) - V(i) starting with V(0)=(1,0); V(1) = (A,B) where A = 2*S^2 + 1; B = 2*S*T and S = P(k)*Q(k) + P(k-1)*Q(k-1); T = Q(k)^2 + Q(k-1)^2.

Examples

			For n = 2, 2n+1 = 5. a(2) = 13 and we indeed have sqrt(13) = [3; 1, 1, 1, 1, 6] with period 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], ...
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 0..1000 (Terms 0..999 from T. D. Noe)
• Dario Alpern, Continued Fraction calculator.
• Keith Matthews, Calculating the simple continued fraction of a quadratic irrational.
• Ulrich Sondermann, Continued Fractions.
• Gang Xiao, Contfrac,continued fraction expansion server with k-th convergent calculator.

Crossrefs

Cf. A013646, A003814, A031396, A003654.

Programs

Mathematica

nn = 50; t = Table[0, {nn}]; n = 1; found = 0; While[found < nn, n++; If[! IntegerQ[Sqrt[n]], c = ContinuedFraction[Sqrt[n]]; len = Length[c[[2]]]; If[OddQ[len] && (len + 1)/2 <= nn && t[[(len + 1)/2]] == 0, t[[(len + 1)/2]] = n; found++]]]; t (* T. D. Noe, Apr 04 2014 *)

Extensions

More terms from Naohiro Nomoto, Jan 01 2002