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A146333
Numbers k such that continued fraction of (1 + sqrt(k))/2 has period 8.
3
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
1,1
COMMENTS
For primes in this sequence see A146353.
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Harvey P. Dale)
EXAMPLE
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.
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
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 *)
KEYWORD
nonn
AUTHOR
Artur Jasinski, Oct 30 2008
EXTENSIONS
155 and 279 etc. added, 311 etc. removed by R. J. Mathar, Sep 06 2009
STATUS
approved