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

1,1

Comments

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

Links

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

Example

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.

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]]] &]

Crossrefs

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

Sequence in context: A007961 A212067 A060811 * A344939 A278742 A250174

Adjacent sequences: A146324 A146325 A146326 * A146328 A146329 A146330

Keyword

nonn

Author

Artur Jasinski, Oct 30 2008

Extensions

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

Status

approved