

A267857


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.


1



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, 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, 8, 2, 17, 8, 10, 6, 6, 2, 16, 5, 4, 8, 4, 21
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OFFSET

0,2


COMMENTS

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.
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.


REFERENCES

J. H. Silverman, A Friendly Introduction to Number Theory, 3rd ed., Pearson Education, Inc, 2006, p. 351.


LINKS



EXAMPLE

a(0) = 1 because sqrt(5) = [2,repeat(4)].
a(1) = 2 because sqrt(8) = [2,repeat(1,4)].
a(23) = 11 because sqrt(61) = [7,repeat(1,4,3,1,2 2,1,3,4,1,14)].
Pell +1 equation: n = 23 with D = 61 has odd a(23)
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.)


MAPLE



MATHEMATICA

Length[Last@ #] & /@ ContinuedFraction@ Sqrt@ Select[Range@ 200, And[MemberQ[{0, 1}, Mod[#, 4]], ! IntegerQ@ Sqrt@ #] &] (* Michael De Vlieger, Feb 11 2016, after A079896 *)


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



STATUS

approved



