A146331 Numbers k such that continued fraction of (1 + sqrt(k))/2 has period 6.

18, 19, 22, 38, 39, 44, 54, 57, 58, 59, 66, 68, 70, 74, 86, 102, 105, 107, 111, 112, 114, 115, 130, 131, 146, 147, 148, 150, 159, 164, 175, 178, 183, 186, 187, 198, 203, 253, 258, 260, 264, 267, 273, 275, 278, 294, 303, 308, 309, 326, 327, 330, 333, 341, 346

Offset: 1

Author

Artur Jasinski, Oct 30 2008

Keywords

• nonn

Comments

For primes in this sequence see A146351.

Examples

			a(2) = 19 because continued fraction of (1+sqrt(19))/2 = 2, 1, 2, 8, 2, 1, 3, 1, 2, 8, 2, 1, 3, 1, 2, 8, 2, 1, 3, 1, 2, 8, 2, 1, 3, 1, 2, 8, 2, 1 ... has period (1, 2, 8, 2, 1, 3) length 6.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Harvey P. Dale)

Crossrefs

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

Programs

Maple

A146326 := proc(n) if not issqr(n) then numtheory[cfrac]( (1+sqrt(n))/2, 'periodic','quotients') ; nops(%[2]) ; else 0 ; fi; end: isA146331 := proc(n) RETURN(A146326(n) = 6) ; end: for n from 2 to 380 do if isA146331(n) then printf("%d,",n) ; fi; od: # R. J. Mathar, Sep 06 2009

Mathematica

cf6Q[n_]:=Module[{c=(1+Sqrt[n])/2},!IntegerQ[c]&&Length[ContinuedFraction[ c][[2]]]==6]; Select[Range[400],cf6Q] (* Harvey P. Dale, May 30 2012 *)

Extensions

39, 68, 150, 203, etc. added by R. J. Mathar, Sep 06 2009