|
| |
|
|
A146327
|
|
Numbers n such that continued fraction of (1+Sqrt[n])/2 has period 2.
|
|
1
| |
|
|
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
(list; graph; refs; listen; history; internal format)
|
|
|
|
OFFSET
| 1,1
|
|
|
COMMENTS
| For primes in this sequence see A056899, primes of the form n^2+2.
|
|
|
EXAMPLE
| a(1) = 2 because continued fraction of (1+Sqrt[2])/2 = 1, 4, 1, 4, 1, 4, 1, ...
has period (1,4) length 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: [From R. J. Mathar (mathar(AT)strw.leidenuniv.nl), 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: A061909 A007961 A060811 * A081706 A032804 A047473
Adjacent sequences: A146324 A146325 A146326 * A146328 A146329 A146330
|
|
|
KEYWORD
| nonn
|
|
|
AUTHOR
| Artur Jasinski (grafix(AT)csl.pl), Oct 30 2008
|
|
|
EXTENSIONS
| Added 226, 227, 290, 291 - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Sep 06 2009
|
| |
|
|
