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

1,1

Comments

For primes in this sequence see A146361.

Links

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

Example

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.

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 *)

Crossrefs

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

Sequence in context: A252131 A223823 A224555 * A067356 A230371 A056132

Adjacent sequences: A146336 A146337 A146338 * A146340 A146341 A146342

Keyword

nonn

Author

Artur Jasinski, Oct 30 2008

Extensions

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

More terms from Amiram Eldar, Mar 31 2020

Status

approved