|
| |
|
|
A146330
|
|
Numbers k such that continued fraction of (1+Sqrt[k])/2 has period 5
|
|
2
| |
|
|
41, 149, 157, 181, 269, 397, 425, 493, 565, 697, 761, 941, 1013, 1037, 1325, 1565, 1781, 1825
(list; graph; refs; listen; history; internal format)
|
|
|
|
OFFSET
| 1,1
|
|
|
COMMENTS
| For primes in this sequence see A146350.
|
|
|
EXAMPLE
| a(1) = 41 because continued fraction of (1+Sqrt[41])/2 = 3, 1, 2, 2, 1, 5, 1, 2, 2, 1, 5, 1, 2, 2, 1, 5, 1, 2, ...
has period (1,2,2,1,5) length 5
|
|
|
MAPLE
| A146326 := proc(n) if not issqr(n) then numtheory[cfrac]( (1+sqrt(n))/2, 'periodic', 'quotients') ; nops(%[2]) ; else 0 ; fi; end: isA146330 := proc(n) RETURN(A146326(n) = 5) ; end: for n from 2 to 2000 do if isA146330(n) then printf("%d, ", n) ; fi; od: [From R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Sep 06 2009]
|
|
|
MATHEMATICA
| s = 10; aa = {}; Do[k = ContinuedFraction[(1 + Sqrt[n])/2, 1000]; If[Length[k] < 190, AppendTo[aa, 0], m = 1; While[k[[s ]] != k[[s + m]] || k[[s + m]] != k[[s + 2 m]] || k[[s + 2 m]] != k[[s + 3 m]] || k[[s + 3 m]] != k[[s + 4 m]], m++ ]; s = s + 1; While[k[[s ]] != k[[s + m]] || k[[s + m]] != k[[s + 2 m]] || k[[s + 2 m]] != k[[s + 3 m]] || k[[s + 3 m]] != k[[s + 4 m]], m++ ]; s = s + 1; While[k[[s ]] != k[[s + m]] || k[[s + m]] != k[[s + 2 m]] || k[[s + 2 m]] != k[[s + 3 m]] || k[[s + 3 m]] != k[[s + 4 m]], m++ ]; AppendTo[aa, m]], {n, 1, 500}]; bb = {}; Do[If[aa[[n]] == 5, AppendTo[bb, n]], {n, 1, Length[aa]}]; bb (*Artur Jasinski*)
|
|
|
CROSSREFS
| A000290, A078370, A146326-A146345, A146348-A146360.
Sequence in context: A142449 A044373 A044754 * A146350 A050954 A192821
Adjacent sequences: A146327 A146328 A146329 * A146331 A146332 A146333
|
|
|
KEYWORD
| nonn
|
|
|
AUTHOR
| Artur Jasinski (grafix(AT)csl.pl), Oct 30 2008
|
|
|
EXTENSIONS
| Removed 259 and 1026 - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Sep 06 2009
|
| |
|
|
