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A146338 Numbers k such that continued fraction of (1+sqrt(k))/2 has period 15. 2
193, 281, 481, 1417, 1861, 1933, 2089, 2141, 2197, 2437, 2741, 2837, 3037, 3065, 3121, 3413, 3589, 3625, 3785, 3925, 3977 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

COMMENTS

For primes in this sequence see A146360.

LINKS

Table of n, a(n) for n=1..21.

EXAMPLE

a(1) = 193 because continued fraction of (1+sqrt(193))/2 = 7, 2, 4, 6, 1, 2, 1, 1, 1, 1, 2, 1, 6, 4, 2, 13, 2, 4, 6, 1, 2, 1, 1, 1, 1, 2, 1, 6, 4, 2, 13... has period (2, 4, 6, 1, 2, 1, 1, 1, 1, 2, 1, 6, 4, 2, 13) length 15.

MAPLE

A146326 := proc(n) if not issqr(n) then numtheory[cfrac]( (1+sqrt(n))/2, 'periodic', 'quotients') ; nops(%[2]) ; else 0 ; fi; end:

isA146338 := proc(n) RETURN(A146326(n) = 15) ; end:

for n from 2 to 4000 do if isA146338(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++ ]; 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]] == 14, AppendTo[bb, n]], {n, 1, Length[aa]}]; bb

CROSSREFS

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

Sequence in context: A142925 A060333 A020352 * A201858 A146360 A050964

Adjacent sequences:  A146335 A146336 A146337 * A146339 A146340 A146341

KEYWORD

more,nonn

AUTHOR

Artur Jasinski, Oct 30 2008

EXTENSIONS

Extended beyond 3 terms by R. J. Mathar, Sep 06 2009

STATUS

approved

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Last modified March 26 22:42 EDT 2019. Contains 321565 sequences. (Running on oeis4.)