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A146328 Numbers k such that the continued fraction of (1+sqrt(k))/2 has period 3. 2
17, 37, 61, 65, 101, 145, 185, 197, 257, 317, 325, 401, 461, 485, 557, 577, 677, 773, 785, 901, 985, 1025, 1129, 1157, 1297, 1429, 1445, 1601, 1765, 1877, 1901, 1937, 2117, 2285, 2305, 2501, 2705, 2873, 2917, 3077, 3137, 3281, 3293, 3341, 3365, 3601, 3845, 4045, 4097, 4357, 4597, 4625, 4901 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

COMMENTS

For primes in this sequence see A146348.

LINKS

Table of n, a(n) for n=1..53.

EXAMPLE

a(1) = 3 because continued fraction of (1+sqrt(17))/2 = 2, 1, 1, 3, 1, 1, 3, ... has period (1,1,3) length 3.

MAPLE

A146326 := proc(n) if not issqr(n) then numtheory[cfrac]( (1+sqrt(n))/2, 'periodic', 'quotients') ; nops(%[2]) ; else 0 ; fi; end: isA146328 := proc(n) RETURN(A146326(n) = 3) ; end: for n from 2 to 1801 do if isA146328(n) then printf("%d, ", n) ; fi; od: # R. J. Mathar, Sep 06 2009

MATHEMATICA

okQ[n_] := Module[{cf = ContinuedFraction[(1 + Sqrt[n])/2]}, Length[cf] > 1 && Length[cf[[2]]] == 3]; Select[Range[5000], okQ]

CROSSREFS

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

Sequence in context: A295338 A059425 A225077 * A161549 A269788 A146348

Adjacent sequences:  A146325 A146326 A146327 * A146329 A146330 A146331

KEYWORD

nonn

AUTHOR

Artur Jasinski, Oct 30 2008

EXTENSIONS

803 removed by R. J. Mathar, Sep 06 2009

Extended by T. D. Noe, Mar 09 2011

STATUS

approved

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Last modified July 21 22:02 EDT 2019. Contains 325210 sequences. (Running on oeis4.)