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 A146327 Numbers n such that continued fraction of (1 + sqrt(n))/2 has period 2. 2
 2, 3, 10, 11, 12, 15, 21, 26, 27, 30, 35, 45, 50, 51, 56, 63, 77, 82, 83, 84, 87, 90, 93, 99, 117, 122, 123, 132, 143, 165, 170, 171, 182, 195, 221, 226, 227, 228, 230, 231, 235, 237, 240, 245, 255, 285, 290, 291, 306, 323, 357, 362, 363, 380, 399, 437, 442, 443 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS For primes in this sequence see A056899, primes of the form n^2 + 2. LINKS EXAMPLE a(1) = 2 because continued fraction of (1 + sqrt(2))/2 = 1, 4, 1, 4, 1, 4, 1, ... has period (1,4) length 2. MAPLE A146326 := proc(n) if not issqr(n) then numtheory[cfrac]( (1+sqrt(n))/2, 'periodic', 'quotients') ; nops(%[2]) ; else 0 ; fi; end: isA146327 := proc(n) RETURN(A146326(n) = 2) ; end: for n from 2 to 450 do if isA146327(n) then printf("%d, ", n) ; fi; od: # R. J. Mathar, Sep 06 2009 MATHEMATICA Select[Range[1000], 2 == Length[ContinuedFraction[(1 + Sqrt[#])/2][[2]]] &] CROSSREFS Cf. A000290, A078370, A146326-A146345, A146348-A146360. Sequence in context: A007961 A212067 A060811 * A278742 A250174 A285622 Adjacent sequences:  A146324 A146325 A146326 * A146328 A146329 A146330 KEYWORD nonn AUTHOR Artur Jasinski, Oct 30 2008 EXTENSIONS 226, 227, 290, 291 added by R. J. Mathar, Sep 06 2009 STATUS approved

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Last modified August 20 20:50 EDT 2019. Contains 326155 sequences. (Running on oeis4.)