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 A143577 Numbers k such that the continued fraction of (1 + sqrt(k))/2 has period 9. 3
 73, 97, 233, 277, 349, 353, 613, 821, 877, 1073, 1181, 1189, 1277, 1285, 1313, 1385, 1613, 1637, 1693, 1745, 1865, 2357, 2581, 2777, 3233, 3557, 3989, 4157, 4469, 4517, 4553, 4709, 4889, 4925, 4933, 5245, 5261, 5305, 5597, 6113, 6205, 6253, 7213, 7585, 7837, 8885 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS For primes in this sequence see A146354. Superset of A146354. - R. J. Mathar, Nov 05 2008 LINKS Amiram Eldar, Table of n, a(n) for n = 1..10000 EXAMPLE a(1) = 73 because continued fraction of (1+sqrt(73))/2 = 4, 1, 3, 2, 1, 1, 2, 3, 1, 7, 1, 3, 2, 1, 1, 2, 3, 1, 7, 1, 3, 2, 1, 1, 2, 3, 1, 7, 1, 3, ... has period (1, 3, 2, 1, 1, 2, 3, 1, 7) length 9 . MAPLE isA143577 := proc(k) local c; try c := numtheory[cfrac](1/2+sqrt(k)/2, 'periodic', 'quotients') ; if nops(c[2]) = 9 then RETURN(true) ; else RETURN(false) ; fi; catch: RETURN(false) ; end try; end: for k from 2 to 80000 do if isA143577(k) then printf("%d, ", k) ; fi; od: # R. J. Mathar, Nov 05 2008 MATHEMATICA Select[Range[1000], !IntegerQ @ Sqrt[#] && Length[ContinuedFraction[(1 + Sqrt[#])/2][[2]]] == 9 &]  (* Amiram Eldar, Mar 19 2020 *) CROSSREFS Cf. A000290, A078370, A146326-A146345, A146348-A146360. Sequence in context: A107008 A141375 A140621 * A146354 A050958 A139990 Adjacent sequences:  A143574 A143575 A143576 * A143578 A143579 A143580 KEYWORD nonn AUTHOR Artur Jasinski, Oct 30 2008 EXTENSIONS Extended by R. J. Mathar, Nov 05 2008 More terms from Amiram Eldar, Mar 19 2020 STATUS approved

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Last modified September 28 10:48 EDT 2021. Contains 347714 sequences. (Running on oeis4.)