

A146331


Numbers k such that continued fraction of (1 + sqrt(k))/2 has period 6.


3



18, 19, 22, 38, 39, 44, 54, 57, 58, 59, 66, 68, 70, 74, 86, 102, 105, 107, 111, 112, 114, 115, 130, 131, 146, 147, 148, 150, 159, 164, 175, 178, 183, 186, 187, 198, 203, 253, 258, 260, 264, 267, 273, 275, 278, 294, 303, 308, 309, 326, 327, 330, 333, 341, 346
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OFFSET

1,1


COMMENTS

For primes in this sequence see A146351.


LINKS



EXAMPLE

a(2) = 19 because continued fraction of (1+sqrt(19))/2 = 2, 1, 2, 8, 2, 1, 3, 1, 2, 8, 2, 1, 3, 1, 2, 8, 2, 1, 3, 1, 2, 8, 2, 1, 3, 1, 2, 8, 2, 1 ... has period (1, 2, 8, 2, 1, 3) length 6.


MAPLE

A146326 := proc(n) if not issqr(n) then numtheory[cfrac]( (1+sqrt(n))/2, 'periodic', 'quotients') ; nops(%[2]) ; else 0 ; fi; end: isA146331 := proc(n) RETURN(A146326(n) = 6) ; end: for n from 2 to 380 do if isA146331(n) then printf("%d, ", n) ; fi; od: # R. J. Mathar, Sep 06 2009


MATHEMATICA

cf6Q[n_]:=Module[{c=(1+Sqrt[n])/2}, !IntegerQ[c]&&Length[ContinuedFraction[ c][[2]]]==6]; Select[Range[400], cf6Q] (* Harvey P. Dale, May 30 2012 *)


CROSSREFS



KEYWORD

nonn


AUTHOR



EXTENSIONS



STATUS

approved



