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A146327
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Numbers k such that the continued fraction of (1 + sqrt(k))/2 has period 2.
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3
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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
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OFFSET
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1,1
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COMMENTS
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For primes in this sequence see A056899, primes of the form k^2 + 2.
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LINKS
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EXAMPLE
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a(1) = 2 because continued fraction of (1 + sqrt(2))/2 = 1, 4, 1, 4, 1, 4, 1, ... has repeating part (1,4), period 2.
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MAPLE
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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
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MATHEMATICA
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Select[Range[1000], 2 == Length[ContinuedFraction[(1 + Sqrt[#])/2][[2]]] &]
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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