A146333 Numbers k such that continued fraction of (1 + sqrt(k))/2 has period 8.
31, 40, 46, 71, 76, 88, 91, 92, 96, 104, 108, 152, 153, 155, 176, 188, 192, 200, 206, 207, 234, 238, 261, 266, 276, 279, 280, 282, 320, 328, 335, 336, 348, 366, 378, 383, 386, 392, 408, 414, 450, 476, 477, 480, 488, 501, 503, 504, 505, 540, 542, 555, 558, 581
Offset: 1
Keywords
Examples
a(1) = 31 because continued fraction of (1+sqrt(31))/2 = 3, 3, 1, 1, 10, 1, 1, 3, 5, 3, 1, 1, 10, 1, 1, 3, 5, 3, 1, 1, 10, 1, ... has period (3, 1, 1, 10, 1, 1, 3, 5) length 8.
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Harvey P. Dale)
Programs
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Maple
A146326 := proc(n) if not issqr(n) then numtheory[cfrac]( (1+sqrt(n))/2, 'periodic','quotients') ; nops(%[2]) ; else 0 ; fi; end: isA146333 := proc(n) RETURN(A146326(n) = 8) ; end: for n from 2 to 700 do if isA146333(n) then printf("%d,",n) ; fi; od: # R. J. Mathar, Sep 06 2009
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Mathematica
cf8Q[n_]:=Module[{sqrt=Sqrt[n]},!IntegerQ[sqrt]&&Length[ ContinuedFraction[ (1+sqrt)/2][[2]]]==8]; Select[Range[600],cf8Q] (* Harvey P. Dale, Sep 06 2012 *)
Extensions
155 and 279 etc. added, 311 etc. removed by R. J. Mathar, Sep 06 2009
Comments