A143577 Numbers k such that the continued fraction of (1 + sqrt(k))/2 has period 9.

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

Offset: 1

Author

Artur Jasinski, Oct 30 2008

Keywords

• nonn

Comments

For primes in this sequence see A146354.

Superset of A146354. - R. J. Mathar, Nov 05 2008

Examples

			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 .
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000290, A078370, A146326-A146345, A146348-A146360.

Programs

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 *)

Extensions

Extended by R. J. Mathar, Nov 05 2008

More terms from Amiram Eldar, Mar 19 2020