题目140：考察系数为二阶递归关系的无限多项式序列的值
Modified Fibonacci golden nuggetsConsider the infinite polynomial series AG(x) = xG1 + x2G2 + x3G3 + ..., where Gk is the kth term of the second order recurrence relation Gk = Gk−1 + Gk−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 AG(x) is a positive integer.
The corresponding values of x for the first five natural numbers are shown below.
We shall call AG(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 AG(x) = xG1 + x2G2 + x3G3 + ...，其中 Gk 是递归关系 Gk = Gk−1 + Gk−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 AG(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 AG(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。
本帖最后由 guosl 于 2022-9-18 23:09 编辑
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
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