A146337 Numbers k such that continued fraction of (1 + sqrt(k))/2 has period 14.

118, 154, 179, 201, 212, 244, 251, 262, 286, 292, 307, 340, 347, 388, 403, 418, 422, 430, 467, 471, 474, 494, 497, 500, 519, 543, 548, 566, 587, 594, 598, 670, 683, 687, 692, 698, 699, 703, 713, 722, 742, 745, 754, 831, 833, 847, 873, 879, 932, 939, 945

Offset

1,1

Comments

For primes in this sequence see A146359.

Links

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

Example

a(1) = 421 because continued fraction of (1+sqrt(421))/2 = 17, 5, 3, 1, 1, 1, 2, 26, 2, 1, 1, 1, 3, 5, 13, 5, 3, 1, 1, 1, 2, 26, 2, 1, 1, 1, 3, 5, 13, 5, 3, 1, 1, 1, 2, 26... has period (5, 3, 1, 1, 1, 2, 26, 2, 1, 1, 1, 3, 5, 13) length 14.

Maple

A := proc(n) option remember ; local c; try c := numtheory[cfrac](1/2+sqrt(n)/2, 'periodic', 'quotients') ; RETURN(nops(c[2]) ); catch: RETURN(-1) end try ; end: isA146337 := proc(n) if A(n) = 14 then RETURN(true); else RETURN(false); fi; end: for k from 1 do if isA146337(k) then printf("%d, ", k) ; fi; od: # R. J. Mathar, Nov 08 2008

Mathematica

cf14Q[n_]:=Module[{s=(1+Sqrt[n])/2}, !IntegerQ[s]&&Length[ ContinuedFraction[ s][[2]]] == 14]; Select[Range[1000], cf14Q] (* Harvey P. Dale, Oct 15 2015 *)

Crossrefs

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

Sequence in context: A356729 A095627 A066734 * A165462 A261464 A250793

Adjacent sequences: A146334 A146335 A146336 * A146338 A146339 A146340

Keyword

nonn

Author

Artur Jasinski, Oct 30 2008

Extensions

More terms from R. J. Mathar, Nov 08 2008

Status

approved