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proposed
Integers n k (not perfect squares) such that the continued fraction expansion of the square root of n k has period at most 2.
In a recent paper, Justin Thomas, Julian Rosen, and I show that this is equivalent to the following criterion: let d be the integer part of the square root. Then sqrt(nk) has period at most 2 if and only if 2d/(n k - d^2) is an integer.
The first term is 2 because sqrt(2) is irrational and for nk=2, d=1, 2d/(n k - d^2) = 1 is an integer.
In a recent paper , Justin Thomas, myself and Julian Rosen , and I show that this is equivalent to the following criterion: let d be the integer part of the square root. Then sqrt{(n} ) has period at most 2 if and only if 2d/(n - d^2) is an integer.
Justin Thomas, J., Krishnan Shankar, K., Julian Rosen, J., "Continued Fractions, Square Roots and the orbit of 1/0 on the boundary of the hyperbolic plane", preprint.
The first term is 2 because sqrt{(2} ) is irrational and for n=2, d=1, 2d/(n - d^2) = 1 is an integer.
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K. Shankar, <a href="http://www.math.ou.edu/~shankar/paperspubs.html">Square roots and continued fractions</a>.
K. Shankar, <a href="http://www.math.ou.edu/~shankar/research/cfrac.pdf">SQUARE ROOTS, CONTINUED FRACTIONS AND THE ORBIT OF 1/0 ON dH2</a>
nonn,new
nonn
Integers n (not perfect squares) such that the continued fraction expansion of the square root of n has period at most 2.
2, 3, 5, 6, 8, 10, 11, 15, 17, 18, 20, 24, 26, 27, 30, 35, 37, 38, 39, 40, 42, 48, 50, 51, 56, 63, 65, 66, 68, 72, 80, 82, 83, 84, 87, 90
1,1
In a recent paper Justin Thomas, myself and Julian Rosen show that this is equivalent to the following criterion: let d be the integer part of the square root. Then sqrt{n} has period at most 2 if and only if 2d/(n - d^2) is an integer.
Thomas, J., Shankar, K., Rosen, J., "Continued Fractions, Square Roots and the orbit of 1/0 on the boundary of the hyperbolic plane", preprint.
K. Shankar, <a href="http://www.math.ou.edu/~shankar/papers.html">Square roots and continued fractions</a>.
The first term is 2 because sqrt{2} is irrational and for n=2, d=1, 2d/(n - d^2) = 1 is an integer.
nonn,new
Krishnan Shankar (shankar(AT)math.ou.edu), Apr 17 2006
approved