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 A146335 Numbers k such that continued fraction of (1 + sqrt(k))/2 has period 11. 4
 265, 541, 593, 661, 701, 857, 1061, 1109, 1217, 1237, 1709, 1733, 1949, 2333, 2509, 2557, 2957, 3125, 3229, 3677, 3701, 4181, 4373, 4685, 5081, 5237, 5309, 6133, 6425, 6445, 7013, 7025, 8185, 8545, 8693, 9305, 9533, 9553, 10333, 10525, 10853, 10961, 11125, 11141 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS For primes in this sequence see A146356. LINKS Amiram Eldar, Table of n, a(n) for n = 1..10000 EXAMPLE a(4) = 661 because continued fraction of (1+sqrt(661))/2 = 13, 2, 1, 4, 2, 8, 8, 2, 4, 1, 2, 25, 2, 1, 4, 2, 8, 8, 2, 4, 1, 2, 25, 2, 1, 4, 2, 8, 8, 2, 4, 1, 2, 25, 2, 1, 4, 2, 8, 8 ... has period (2, 1, 4, 2, 8, 8, 2, 4, 1, 2, 25) length 11. MAPLE A146326 := proc(n) if not issqr(n) then numtheory[cfrac]( (1+sqrt(n))/2, 'periodic', 'quotients') ; nops(%[2]) ; else 0 ; fi; end: isA146335 := proc(n) RETURN(A146326(n) = 11) ; end: for n from 2 to 2000 do if isA146335(n) then printf("%d, ", n) ; fi; od: # R. J. Mathar, Sep 06 2009 MATHEMATICA Select[Range[10^4], !IntegerQ @ Sqrt[#] && Length[ContinuedFraction[(1 + Sqrt[#])/2][[2]]] == 11 &] (* Amiram Eldar, Mar 31 2020 *) CROSSREFS Cf. A000290, A078370, A146326-A146345, A146348-A146360. Sequence in context: A345854 A051976 A255085 * A211714 A259741 A202416 Adjacent sequences: A146332 A146333 A146334 * A146336 A146337 A146338 KEYWORD nonn AUTHOR Artur Jasinski, Oct 30 2008 EXTENSIONS 916 removed by R. J. Mathar, Sep 06 2009 More terms from Amiram Eldar, Mar 31 2020 STATUS approved

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Last modified September 24 19:02 EDT 2023. Contains 365581 sequences. (Running on oeis4.)