A060215 -id:A060215 - cp's OEIS frontend

A071296 a(n) is the least m such that a period of the continued fraction expansion of sqrt(m) is 1,1,1,...,1,1,1,Z and there are n ones in the period (Z is 2*floor(sqrt(m))). If no such m exists, a(n) = 0.

3, 0, 7, 13, 0, 58, 135, 0, 819, 2081, 0, 13834, 35955, 0, 244647, 639389, 0, 4374866, 11448871, 0, 78439683, 205337953, 0, 1407271538, 3684200835, 0, 25251313255, 66108441037, 0, 453111560266, 1186259960295, 0, 8130736409715, 21286537898177, 0

Offset: 1

Lekraj Beedassy, Jun 11 2002

nonn

			a(3) = 7 because sqrt(7)'s continued fraction is [2;1,1,1,4,...]; the period has 3 ones (and only one other number).

T. D. Noe, Table of n, a(n) for n = 1..500

Cf. A000045, A010342, A060215.

Table[If[Mod[n, 3] == 2, 0, x = (Fibonacci[n + 1] + 1)/2; x^2 + (Fibonacci[n - 1] + 2*x*Fibonacci[n])/Fibonacci[n + 1]], {n, 50}] (* T. D. Noe, Apr 07 2014 *)

from gmpy2 import fib2
def A071296(n):
    if n%3==2: return 0
    f, g = fib2(n)
    return int(f*(f + (g<<1) + 6) + g*(g + 2) + 5>>2) # Chai Wah Wu, Mar 21 2023

Let F(n) = n-th Fibonacci number (A000045). If n == 2 mod 3 then F(n+1) is even and there's no such m. Otherwise, let x = (F(n+1) + 1) / 2. Then a(n) = x^2 + (F(n-1) + 2*x*F(n))/F(n+1).

Edited by Don Reble, Jun 06 2003

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

Original entry on oeis.org

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

Greg Dresden, Aug 11 2019

nonn

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.

			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.

Cf. A060215, A071296.

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++
]

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)

cp's OEIS Frontend

A071296 a(n) is the least m such that a period of the continued fraction expansion of sqrt(m) is 1,1,1,...,1,1,1,Z and there are n ones in the period (Z is 2*floor(sqrt(m))). If no such m exists, a(n) = 0.

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