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A146334
Numbers k such that continued fraction of (1 + sqrt(k))/2 has period 10.
2
43, 67, 116, 129, 134, 161, 162, 184, 218, 242, 243, 246, 270, 274, 297, 301, 314, 338, 339, 345, 354, 356, 407, 411, 451, 452, 459, 465, 475, 498, 515, 517, 532, 534, 561, 563, 590, 591, 595, 597, 603, 611, 638, 648, 657, 665, 669, 671, 690, 705, 715
OFFSET
1,1
COMMENTS
For primes in this sequence see A146355.
LINKS
EXAMPLE
a(1) = 43 because continued fraction of (1+Sqrt[43])/2 = 3, 1, 3, 1, 1, 12, 1, 1, 3, 1, 5, 1, 3, 1, 1, 12, 1, 1, 3, 1, 5, 1, 3, 1, 1, 12, 1, 1, 3, 1, ... has period (1, 3, 1, 1, 12, 1, 1, 3, 1, 5) length 10.
MAPLE
A146326 := proc(n) if not issqr(n) then numtheory[cfrac]( (1+sqrt(n))/2, 'periodic', 'quotients') ; nops(%[2]) ; else 0 ; fi; end: isA146334 := proc(n) RETURN(A146326(n) = 10) ; end: for n from 2 to 715 do if isA146334(n) then printf("%d, ", n) ; fi; od: # R. J. Mathar, Sep 06 2009
MATHEMATICA
cf10Q[n_]:=Module[{s=(1+Sqrt[n])/2, x}, x=If[IntegerQ[s], 1, Length[ ContinuedFraction[ s][[2]]]]; x==10]; Select[Range[750], cf10Q] (* Harvey P. Dale, Sep 22 2015 *)
KEYWORD
nonn
AUTHOR
Artur Jasinski, Oct 30 2008
EXTENSIONS
284 removed by R. J. Mathar, Sep 06 2009
STATUS
approved