A059855 Period of continued fraction for sqrt(n^2+4), n >= 1.

1, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2

Offset: 1

Author

Labos Elemer, Feb 27 2001

Keywords

• nonn
• easy

Comments

From Jianing Song, May 01 2021: (Start)

The old name was "Quotient cycle length of sqrt(n^2+4)."

Essentially the same as A010695 and A021400. (End)

Examples

			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.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A003285, A010695, A021400.

Period of continued fraction for sqrt(n^2+k): A059853 (k=3), this sequence (k=4), A059854 (k=5).

Programs

Maple

with(numtheory): [seq(nops(cfrac(sqrt(k^2+4), 'periodic', 'quotients')[2]), k=1..100)];

Mathematica

a[n_] := Length @ ContinuedFraction[Sqrt[n^2 + 4]][[2]]; Array[a, 100] (* Amiram Eldar, May 13 2020 *)

Formula

a(n) = 2 for even n, a(n) = 5 for odd n > 1.

a(n) = A003285(n^2+4). - Jianing Song, May 01 2021