A309666 a(n) is the least k such that the denominators of continued fraction convergents for sqrt(k) match the first n Fibonacci numbers.

2, 3, 7, 7, 13, 58, 58, 135, 819, 819, 2081, 13834, 13834, 35955, 244647, 244647, 639389, 4374866, 4374866, 11448871, 78439683, 78439683, 205337953, 1407271538, 1407271538, 3684200835, 25251313255, 25251313255, 66108441037, 453111560266, 453111560266, 1186259960295, 8130736409715, 8130736409715, 21286537898177

Offset: 1

Author

Greg Dresden, Aug 11 2019

Keywords

• nonn

Comments

Aside from the first term, this appears to be a subset of A060215.

Same as A071296 if you drop a(0) and replace each repeated pair x,x with 0,x (credit to Daniel Suteu for pointing this out).

These are also the least a(n) such that the continued fraction expansion for sqrt(a(n) - floor(a(n))) begins with (n-1) 1's.

Examples

			For n = 5 the convergents of sqrt(13) are 3/1, 4/1, 7/2, 11/3, 18/5, 119/33, ... and the first five denominators are 1, 1, 2, 3, 5, which match the first five Fibonacci numbers. Since 13 is the first number with this property, then a(5) = 13.
		

Crossrefs

Cf. A060215, A071296.

Programs

Mathematica

c = 1;
n = 2;
F = Table[Fibonacci[n], {n, 20}];
While[c <= 14,
If[! IntegerQ[Sqrt[n]]
   &&
   Denominator[Convergents[Sqrt[n], c]] == F[[1 ;; c]],
  Print[n, "  ", Denominator[Convergents[Sqrt[n], c]]];
  c++; n--];
n++
]

Formula

Conjectures from Colin Barker, Aug 26 2019: (Start)

G.f.: x*(2 + x + 4*x^2 - 42*x^3 - 15*x^4 - 39*x^5 + 100*x^6 + x^7 - 61*x^8 + 172*x^9 + 31*x^10 - 17*x^11 + 26*x^12 - 2*x^13 + x^14 - 2*x^15) / ((1 - x)*(1 + x)*(1 - 3*x + x^2)*(1 - x + x^2)*(1 - x - x^2)*(1 + x + 2*x^2 - x^3 + x^4)*(1 + 3*x + 8*x^2 + 3*x^3 + x^4)).

a(n) = a(n-1) + 21*a(n-3) - 21*a(n-4) - 50*a(n-6) + 50*a(n-7) - 86*a(n-9) + 86*a(n-10) - 13*a(n-12) + 13*a(n-13) + a(n-15) - a(n-16) for n>16.

(End)