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%I #28 Mar 31 2020 03:02:39
%S 73,97,233,277,349,353,613,821,877,1073,1181,1189,1277,1285,1313,1385,
%T 1613,1637,1693,1745,1865,2357,2581,2777,3233,3557,3989,4157,4469,
%U 4517,4553,4709,4889,4925,4933,5245,5261,5305,5597,6113,6205,6253,7213,7585,7837,8885
%N Numbers k such that the continued fraction of (1 + sqrt(k))/2 has period 9.
%C For primes in this sequence see A146354.
%C Superset of A146354. - _R. J. Mathar_, Nov 05 2008
%H Amiram Eldar, <a href="/A143577/b143577.txt">Table of n, a(n) for n = 1..10000</a>
%e 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 .
%p 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
%t Select[Range[1000], !IntegerQ @ Sqrt[#] && Length[ContinuedFraction[(1 + Sqrt[#])/2][[2]]] == 9 &] (* _Amiram Eldar_, Mar 19 2020 *)
%Y Cf. A000290, A078370, A146326-A146345, A146348-A146360.
%K nonn
%O 1,1
%A _Artur Jasinski_, Oct 30 2008
%E Extended by _R. J. Mathar_, Nov 05 2008
%E More terms from _Amiram Eldar_, Mar 19 2020