A146340 Numbers k such that continued fraction of (1 + sqrt(k))/2 has period 17.

521, 617, 709, 1433, 1597, 2549, 2909, 2965, 3161, 3581, 3821, 4013, 4285, 4649, 5501, 5585, 5693, 5813, 6197, 6409, 7825, 7853, 8093, 8125, 8573, 8917, 9281, 9665, 9677, 9925, 10265, 10597, 10973, 11273, 12085, 12805, 13061, 13109, 13613, 13957, 14677

Offset: 1

Author

Artur Jasinski, Oct 30 2008

Keywords

• nonn

Comments

For primes in this sequence see A146362.

Examples

			a(1) = 521 because continued fraction of (1+sqrt(521))/2 = 11, 1, 10, 2, 5, 4, 2, 1, 1, 1, 1, 2, 4, 5, 2, 10, 1, 21, 1, 10, 2, 5, 4, 2, 1, 1, 1, 1, 2, 4, 5, 2, 10, 1, 21, 1, 10, 2, 5, ... has period (1, 10, 2, 5, 4, 2, 1, 1, 1, 1, 2, 4, 5, 2, 10, 1, 21) length 17.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..150 from Harvey P. Dale)

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: isA146340 := proc(n) RETURN(A146326(n) = 17) ; end: for n from 2 do if isA146340(n) then printf("%d,\n",n) ; fi; od: # R. J. Mathar, Sep 06 2009

Mathematica

cf17Q[n_]:=Module[{s=(1+Sqrt[n])/2},If[IntegerQ[s], 1,Length[ ContinuedFraction[ s][[2]]]]==17]; Select[Range[5000],cf17Q] (* Harvey P. Dale, Dec 20 2017 *)

Extensions

998 and 1006 removed, sequence extended by R. J. Mathar, Sep 06 2009

More terms from Harvey P. Dale, Dec 20 2017