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
Keywords
Examples
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).
Links
- T. D. Noe, Table of n, a(n) for n = 1..500
Programs
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Mathematica
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 *) -
Python
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
Formula
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).
Extensions
Edited by Don Reble, Jun 06 2003