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

Author

Artur Jasinski, Oct 30 2008

Keywords

• nonn

Comments

For primes in this sequence see A146359.

Examples

			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.
		

Links

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

Crossrefs

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

Programs

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 *)

Extensions

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