%I #14 Jul 09 2017 08:01:20
%S 1,2,2,5,1,2,6,2,4,5,4,4,1,2,3,8,6,2,6,5,2,6,4,11,1,2,8,2,7,12,6,2,2,
%T 5,6,5,8,10,4,11,1,2,2,8,15,6,9,10,6,2,16,5,4,10,2,16,4,9,4,4,1,2,9,2,
%U 8,2,17,8,10,6,6,2,16,5,4,8,4,21
%N Length of the period of the continued fraction for the square root of D, the discriminant of indefinite binary quadratic forms. D is given in A079896.
%C This is a subsequence of A003285.
%C If a(n) is even then the smallest positive integer solution of the Pell equation x^2 - D(n)*y^2 = +1 with D(n) = A079896(n) is given by (x0, y0) = (P,Q) with P/Q = [a,b[1], ..., b[a(n)-1]. If a(n) is odd then the smallest positive integer solution of the Pell equation x^2 - D(n)*y^2 = +1 is given by (x0, y0) = (P^2 + D(n)*Q^2, 2*P*Q). See e.g., the Silverman reference Theorem 40.4 on p. 351.
%C For positive integer d, d not a square, the Pell equations X^2 - d*Y^2 = +4 and X^2 - d*Y^2 = -4 have no proper solutions. For D(n) = A079896(n) there are solutions for X^2 - D(n)*Y^2 = +4 or -4 (inclusive or). See the Wolfdieter Lang link under A225953 for Pell +4 or -4 solutions.
%D J. H. Silverman, A Friendly Introduction to Number Theory, 3rd ed., Pearson Education, Inc, 2006, p. 351.
%e a(0) = 1 because sqrt(5) = [2,repeat(4)].
%e a(1) = 2 because sqrt(8) = [2,repeat(1,4)].
%e a(23) = 11 because sqrt(61) = [7,repeat(1,4,3,1,2 2,1,3,4,1,14)].
%e Pell +1 equation: n = 23 with D = 61 has odd a(23)
%e P/Q = [7,1,4,3,1,2,2,1,3,4,1] = 29718/3805 (in lowest terms). Therefore (x0, y0) = (1766319049, 226153980), see A174762 (Of course, (1, 0) is the smallest nonnegative solution.)
%p See the _Robert Israel_ program under A003285, adapted to n -> a(n).
%t Length[Last@ #] & /@ ContinuedFraction@ Sqrt@ Select[Range@ 200, And[MemberQ[{0, 1}, Mod[#, 4]], ! IntegerQ@ Sqrt@ #] &] (* _Michael De Vlieger_, Feb 11 2016, after A079896 *)
%Y Cf. A003285, A033313, A033317, A079896, A225953.
%K nonn,easy
%O 0,2
%A _Wolfdieter Lang_, Feb 03 2016