题目140：考察系数为二阶递归关系的无限多项式序列的值

欧拉计划

Modified Fibonacci golden nuggets

Consider the infinite polynomial series A_G(x) = xG₁ + x²G₂ + x³G₃ + ..., where G_k is the kth term of the second order recurrence relation G_k = G_k−1 + G_k−2, G₁ = 1 and G₂ = 4; that is, 1, 4, 5, 9, 14, 23, ... .

For this problem we shall be concerned with values of x for which A_G(x) is a positive integer.

The corresponding values of x for the first five natural numbers are shown below.

We shall call A_G(x) a golden nugget if x is rational, because they become increasingly rarer; for example, the 20th golden nugget is 211345365.

Find the sum of the first thirty golden nuggets.

题目：

考虑无限多项式序列 A_G(x) = xG₁ + x²G₂ + x³G₃ + ...，其中 G_k 是递归关系 G_k = G_k−1 + G_k−2, G₁ = 1, G₂ = 4 的第 k 项；也就是 1，4，5，9，14，23， ...

对于此题目，我们考虑使得 A_G(x) 为正整数的 x 值。

前五个自然数对应的 x 值如下表所示：

如果 x 是有理数，我们将 A_G(x) 称为一个金块，因为这样的数非常的少；例如，第 20 个金块是 211345365。

求前 30 个金块之和。

guosl

可以推得AG(x)=(x+3x^2)/(1-x-x^2)。
此后的解法可参考137题。
答案：5673835352990