A146328 Numbers k such that the continued fraction of (1 + sqrt(k))/2 has period 3.

17, 37, 61, 65, 101, 145, 185, 197, 257, 317, 325, 401, 461, 485, 557, 577, 677, 773, 785, 901, 985, 1025, 1129, 1157, 1297, 1429, 1445, 1601, 1765, 1877, 1901, 1937, 2117, 2285, 2305, 2501, 2705, 2873, 2917, 3077, 3137, 3281, 3293, 3341, 3365, 3601, 3845, 4045, 4097, 4357, 4597, 4625, 4901

Offset: 1

Author

Artur Jasinski, Oct 30 2008

Keywords

• nonn

Comments

For primes in this sequence see A146348.

Examples

			a(1) = 3 because continued fraction of (1+sqrt(17))/2 = 2, 1, 1, 3, 1, 1, 3, ... has period (1,1,3) length 3.
		

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: isA146328 := proc(n) RETURN(A146326(n) = 3) ; end: for n from 2 to 1801 do if isA146328(n) then printf("%d,",n) ; fi; od: # R. J. Mathar, Sep 06 2009

Mathematica

okQ[n_] := Module[{cf = ContinuedFraction[(1 + Sqrt[n])/2]}, Length[cf] > 1 && Length[cf[[2]]] == 3]; Select[Range[5000], okQ]

Extensions

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

Extended by T. D. Noe, Mar 09 2011