A146340 revision #12
|
| |
|
|
A146340
|
|
Numbers k such that continued fraction of (1 + sqrt(k))/2 has period 17.
|
|
3
|
|
|
|
521, 617, 709, 1433, 1597, 2549, 2909, 2965, 3161, 3581, 3821, 4013, 4285, 4649
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,1
|
|
|
COMMENTS
|
For primes in this sequence see A146362.
|
|
|
LINKS
|
|
|
|
EXAMPLE
|
a(1) = 521 because continued fraction of (1+sqrt(521))/2 = 11, 1, 10, 2, 5, 4, 2, 1, 1, 1, 1, 2, 4, 5, 2, 10, 1, 21, 1, 10, 2, 5, 4, 2, 1, 1, 1, 1, 2, 4, 5, 2, 10, 1, 21, 1, 10, 2, 5, ... has period (1, 10, 2, 5, 4, 2, 1, 1, 1, 1, 2, 4, 5, 2, 10, 1, 21) length 17.
|
|
|
MAPLE
|
A146326 := proc(n) if not issqr(n) then numtheory[cfrac]( (1+sqrt(n))/2, 'periodic', 'quotients') ; nops(%[2]) ; else 0 ; fi; end: isA146340 := proc(n) RETURN(A146326(n) = 17) ; end: for n from 2 do if isA146340(n) then printf("%d, \n", n) ; fi; od: # R. J. Mathar, 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++ ]; AppendTo[aa, m]], {n, 1, 500}]; bb = {}; Do[If[aa[[n]] == 17, AppendTo[bb, n]], {n, 1, Length[aa]}]; bb
cf17Q[n_]:=Module[{s=(1+Sqrt[n])/2}, If[IntegerQ[s], 1, Length[ ContinuedFraction[ s][[2]]]]==17]; Select[Range[5000], cf17Q] (* Harvey P. Dale, Dec 20 2017 *)
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
more,nonn
|
|
|
AUTHOR
|
|
|
|
EXTENSIONS
|
998 and 1006 removed, sequence extended by R. J. Mathar, Sep 06 2009
|
|
|
STATUS
|
approved
|
| |
|
|
