A146338 Numbers k such that the continued fraction of (1 + sqrt(k))/2 has period 15.

193, 281, 481, 1417, 1861, 1933, 2089, 2141, 2197, 2437, 2741, 2837, 3037, 3065, 3121, 3413, 3589, 3625, 3785, 3925, 3977, 4001, 4637, 4777, 4877, 5317, 5821, 5941, 6025, 6641, 6653, 6749, 7673, 8117, 8177, 8345, 10069, 10273, 10457, 11197, 11281, 11549, 11821

Offset: 1

Author

Artur Jasinski, Oct 30 2008

Keywords

• nonn

Comments

For primes in this sequence see A146360.

Examples

			a(1) = 193 because continued fraction of (1+sqrt(193))/2 = 7, 2, 4, 6, 1, 2, 1, 1, 1, 1, 2, 1, 6, 4, 2, 13, 2, 4, 6, 1, 2, 1, 1, 1, 1, 2, 1, 6, 4, 2, 13, ... has period (2, 4, 6, 1, 2, 1, 1, 1, 1, 2, 1, 6, 4, 2, 13) length 15.
		

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:
isA146338 := proc(n) RETURN(A146326(n) = 15) ; end:
for n from 2 to 4000 do if isA146338(n) then printf("%d,\n",n) ; fi; od: # R. J. Mathar, Sep 06 2009

Mathematica

Select[Range[10^4], !IntegerQ @ Sqrt[#] && Length[ContinuedFraction[(1 + Sqrt[#])/2][[2]]] == 15 &]  (* Amiram Eldar, Mar 31 2020 *)

Extensions

Extended beyond 3 terms by R. J. Mathar, Sep 06 2009

More terms from Amiram Eldar, Mar 31 2020