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Amiram Eldar, <a href="/A059855/b059855_1.txt">Table of n, a(n) for n = 1..10000</a>
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Amiram Eldar, <a href="/A059855/b059855_1.txt">Table of n, a(n) for n = 1..10000</a>
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Amiram Eldar, <a href="/A059855/b059855_1.txt">Table of n, a(n) for n = 1..10000</a>
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a(n) = 2 for even n, a(n) = 5 for odd n > 1.
For even numbers 2, for odds 5 is the length of cycles: n=96,97 the integer parts and cycles are: [96],[48,192]] and [97],[48, 1, 1, 48, 194] resp. Inside cycles floor(n/2),1,1 and 2n arise.
For even n, sqrt(n^2+4) = [n; n/2, 2*n], hence a(n) = 2.
For odd n > 1, sqrt(n^2+4) = [n; (n-1)/2, 1, 1, (n-1)/2, 2*n], hence a(n) = 5.
Quotient cycle lengths in Period of continued fraction expansion of for sqrt(n^2+4), n >= 1.
From Jianing Song, May 01 2021: (Start)
The old name was "Quotient cycle length of sqrt(n^2+4)."
a(n) = A003285(n^2+4). - Jianing Song, May 01 2021
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