A146327 Numbers k such that the continued fraction of (1 + sqrt(k))/2 has period 2.

2, 3, 10, 11, 12, 15, 21, 26, 27, 30, 35, 45, 50, 51, 56, 63, 77, 82, 83, 84, 87, 90, 93, 99, 117, 122, 123, 132, 143, 165, 170, 171, 182, 195, 221, 226, 227, 228, 230, 231, 235, 237, 240, 245, 255, 285, 290, 291, 306, 323, 357, 362, 363, 380, 399, 437, 442, 443

Offset: 1

Author

Artur Jasinski, Oct 30 2008

Keywords

• nonn

Comments

For primes in this sequence see A056899, primes of the form k^2 + 2.

Examples

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

Links

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

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: isA146327 := proc(n) RETURN(A146326(n) = 2) ; end: for n from 2 to 450 do if isA146327(n) then printf("%d,",n) ; fi; od: # R. J. Mathar, Sep 06 2009

Mathematica

Select[Range[1000], 2 == Length[ContinuedFraction[(1 + Sqrt[#])/2][[2]]] &]

Extensions

226, 227, 290, 291 added by R. J. Mathar, Sep 06 2009