A020347 Numbers k such that the continued fraction for sqrt(k) has period 6.

19, 21, 22, 45, 52, 54, 57, 59, 70, 77, 88, 107, 111, 112, 114, 117, 131, 164, 165, 175, 178, 183, 187, 208, 216, 221, 232, 267, 270, 273, 275, 278, 280, 285, 294, 296, 303, 308, 350, 357, 371, 372, 374, 381, 387, 407, 418, 437, 456, 470, 498, 499, 507, 510, 514, 518

Offset: 1

Author

David W. Wilson

Keywords

• nonn
• changed

Comments

Includes A157265, corresponding to continued fractions [6*k+4,1,1,2,1,1,12*k+8,1,1,2,1,1,12*k+8,...]. - Robert Israel, Nov 21 2019

Examples

			The continued fraction for sqrt(19) is 4 + 1/(2 + 1/(1 + 1/(3 + 1/(1 + 1/(2 + 1/(8 + 1/(2 + 1/(1 + 1/(3 + 1/(1 + 1/(2 + 1/(8 + ..., which has period 6, so 19 is in the sequence.
The continued fraction for sqrt(20) is 4 + 1/(2 + 1/(8 + 1/(2 + 1/(8 + 1/(2 + 1/(8 + ..., which has a period of 2, so 20 is not in the sequence.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..2500 from Robert Israel)

Crossrefs

Cf. A157265.

Programs

Maple

filter:= proc(n)
  not issqr(n) and nops(numtheory:-cfrac(sqrt(n),periodic,quotients)[2])=6
end proc:
select(filter, [$1..1000]); # Robert Israel, Nov 21 2019

Mathematica

Select[Range[500], Length[Last[ContinuedFraction[Sqrt[#]]]] == 6 &] (* Alonso del Arte, Mar 04 2018 *)