A146329 Numbers k such that continued fraction of (1 + sqrt(k))/2 has period 4.

6, 7, 8, 14, 20, 23, 24, 28, 32, 33, 34, 42, 47, 48, 52, 55, 60, 62, 69, 72, 75, 78, 79, 80, 95, 98, 110, 119, 120, 126, 133, 135, 136, 138, 140, 141, 142, 156, 167, 168, 174, 180, 189, 194, 205, 210, 213, 215, 219, 220, 222, 223, 224, 248, 252, 254, 272, 287, 288

Offset: 1

Author

Artur Jasinski, Oct 30 2008

Keywords

• nonn

Comments

For primes in this sequence see A028871 - {2}.

Examples

			a(2) = 7 because continued fraction of (1 + sqrt(7))/2 = 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4, ... has period (1,1,4,1) length 4.
		

Links

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

Crossrefs

Cf. A000290, A078370, A146326-A146345, A028871, A146348-A146360.

Programs

Maple

isA146329 := proc(n) RETURN(A146326(n) = 4) ; end:
for n from 2 to 400 do if isA146329(n) then printf("%d,",n) ; fi; od: # R. J. Mathar, Sep 06 2009

Mathematica

cf4Q[n_]:=Module[{s=(1+Sqrt[n])/2},If[IntegerQ[s],1,Length[ ContinuedFraction[ s][[2]]]]==4]; Select[Range[300],cf4Q] (* Harvey P. Dale, Dec 14 2017 *)

Extensions

39, 68, 150, 155, etc. removed by R. J. Mathar, Sep 06 2009