A059854 Period of continued fraction for sqrt(n^2+5), n >= 3.

4, 6, 2, 3, 6, 8, 10, 2, 4, 9, 4, 14, 2, 16, 6, 12, 12, 2, 16, 22, 10, 24, 2, 24, 12, 24, 16, 2, 6, 26, 30, 26, 2, 7, 20, 12, 18, 2, 18, 11, 20, 64, 2, 20, 30, 19, 22, 2, 40, 20, 10, 50, 2, 10, 38, 74, 14, 2, 22, 64, 50, 72, 2, 48, 10, 30, 48, 2, 22, 51, 10, 36, 2, 34, 12, 47, 46, 2

Offset: 3

Author

Labos Elemer, Feb 27 2001

Keywords

• nonn

Comments

The old name was "Quotient cycle length of sqrt(n^2+5) for n>=3." - Jianing Song, May 01 2021

Examples

			sqrt(13^2+5) = [13; 5, 4, 5, 26], so a(13) = 4.
sqrt(14^2+5) = [14; 5, 1, 1, 1, 2, 1, 8, 1, 2, 1, 1, 1, 5, 28], so a(14) = 14.
sqrt(15^2+5) = [15; 6, 30], so a(15) = 2.
sqrt(16^2+5) = [16; 6, 2, 3, 7, 1, 3, 1, 2, 1, 3, 1, 7, 3, 2, 6, 32], so a(16) = 16.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 3..10000

Crossrefs

Cf. A003285.

Period of continued fraction for sqrt(n^2+k): A059853 (k=3), A059855 (k=4), this sequence (k=5).

Programs

Maple

with(numtheory): [seq(nops(cfrac(sqrt(k^2+5), 'periodic', 'quotients')[2]), k=3..256)];

Mathematica

a[n_] := Length[ContinuedFraction[Sqrt[n^2 + 5]][[2]]]; Array[a, 100, 3] (* Amiram Eldar, Jul 10 2024 *)

Formula

If n is a multiple of 5 then a(n) = 2.

a(n) = A003285(n^2+5). - Jianing Song, May 01 2021

Extensions

New name by Jianing Song, May 01 2021