A146335 Numbers k such that continued fraction of (1 + sqrt(k))/2 has period 11.

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

Offset: 1

Author

Artur Jasinski, Oct 30 2008

Keywords

• nonn

Comments

For primes in this sequence see A146356.

Examples

			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.
		

Links

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

Crossrefs

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

Programs

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 *)

Extensions

916 removed by R. J. Mathar, Sep 06 2009

More terms from Amiram Eldar, Mar 31 2020