Problem 137
Fibonacci golden nuggets
Consider the infinite polynomial series AF(x) = xF1 + x2F2 + x3F3 + …, where Fk is the kth term in the Fibonacci sequence: 1, 1, 2, 3, 5, 8, … ; that is, Fk = Fk−1 + Fk−2, F1 = 1 and F2 = 1.
For this problem we shall be interested in values of x for which AF(x) is a positive integer.
Surprisingly
| AF(1/2) | = | (1/2).1 + (1/2)2.1 + (1/2)3.2 + (1/2)4.3 + (1/2)5.5 + … |
| = | 1/2 + 1/4 + 2/8 + 3/16 + 5/32 + … | |
| = | 2 |
The corresponding values of x for the first five natural numbers are shown below.
| x | AF(x) |
|---|---|
| √2−1 | 1 |
| 1/2 | 2 |
| (√13−2)/3 | 3 |
| (√89−5)/8 | 4 |
| (√34−3)/5 | 5 |
We shall call AF(x) a golden nugget if x is rational, because they become increasingly rarer; for example, the 10th golden nugget is 74049690.
Find the 15th golden nugget.
斐波那契金块
考虑无穷级数AF(x) = xF1 + x2F2 + x3F3 + …, 其中Fk是斐波那契数列的第k项：1, 1, 2, 3, 5, 8, …；该数列由如下方式定义：Fk = Fk−1 + Fk−2，其中F1 = 1且F2 = 1。
在这个问题中，我们感兴趣的是那些使得AF(x)为正整数的x。
其中一个特别的解是：
| AF(1/2) | = | (1/2).1 + (1/2)2.1 + (1/2)3.2 + (1/2)4.3 + (1/2)5.5 + … |
| = | 1/2 + 1/4 + 2/8 + 3/16 + 5/32 + … | |
| = | 2 |
对应于前五个自然数的x如下所示。
| x | AF(x) |
|---|---|
| √2−1 | 1 |
| 1/2 | 2 |
| (√13−2)/3 | 3 |
| (√89−5)/8 | 4 |
| (√34−3)/5 | 5 |
当x是有理数时，我们称AF(x)是一个斐波那契金块，因为这样的数将会变得越来越稀少，例如，第10个斐波那契金块将是74049690。
求第15个斐波那契金块。