A146339 Numbers k such that the continued fraction of (1 + sqrt(k))/2 has period 16.

172, 191, 217, 232, 249, 310, 311, 329, 343, 344, 355, 369, 391, 393, 416, 428, 431, 446, 496, 513, 520, 524, 536, 537, 550, 559, 589, 647, 655, 679, 682, 686, 700, 704, 748, 760, 768, 775, 802, 816, 848, 851, 872, 927, 995, 996, 1036, 1058, 1079, 1080, 1120, 1136

Offset: 1

Author

Artur Jasinski, Oct 30 2008

Keywords

• nonn

Comments

For primes in this sequence see A146361.

Examples

			a(1) = 191 because continued fraction of (1+sqrt(191))/2 = 7, 2, 2, 3, 1, 1, 4, 1, 26, 1, 4, 1, 1, 3, 2, 2, 13, 2, 2, 3, 1, 1, 4, 1, 26, 1, 4, 1, 1, 3, 2, 2, 13, 2, 2, 3, 1, 1, 4, 1, 26... has period (2, 2, 3, 1, 1, 4, 1, 26, 1, 4, 1, 1, 3, 2, 2, 13) length 16.
		

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:
isA146339 := proc(n) RETURN(A146326(n) = 16) ; end:
for n from 2 to 1000 do if isA146339(n) then printf("%d,",n) ; fi; od: # R. J. Mathar, Sep 06 2009

Mathematica

Select[Range[1000], !IntegerQ @ Sqrt[#] && Length[ContinuedFraction[(1 + Sqrt[#])/2][[2]]] == 16 &]  (* Amiram Eldar, Mar 31 2020 *)

Extensions

311 inserted, sequence extended by R. J. Mathar, Sep 06 2009

More terms from Amiram Eldar, Mar 31 2020