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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, 2081, 2153, 2165, 2173, 2465, 2477 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

COMMENTS

For primes in this sequence see A146350.

LINKS

Table of n, a(n) for n=1..24.

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

isA146330 := proc(n) RETURN(A146326(n) = 5) ; end:

for n from 2 to 2000 do if isA146330(n) then printf("%d, ", 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++ ]; 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

CROSSREFS

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

Sequence in context: A254898 A217087 A303910 * A146350 A050954 A192821

Adjacent sequences:  A146327 A146328 A146329 * A146331 A146332 A146333

KEYWORD

nonn

AUTHOR

Artur Jasinski, Oct 30 2008

EXTENSIONS

259 and 1026 removed by R. J. Mathar, Sep 06 2009

STATUS

approved

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Last modified July 22 10:03 EDT 2019. Contains 325219 sequences. (Running on oeis4.)