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Problem 137


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。