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

1,1

Comments

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

Links

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

Example

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.

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 *)

Crossrefs

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

Sequence in context: A328118 A047275 A047590 * A272344 A133893 A101647

Adjacent sequences: A146326 A146327 A146328 * A146330 A146331 A146332

Keyword

nonn

Author

Artur Jasinski, Oct 30 2008

Extensions

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

Status

approved