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Problem 435

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Polynomials of Fibonacci numbers

The Fibonacci numbers {f_n, n ≥ 0} are defined recursively as f_n = f_n-1 + f_n-2 with base cases f₀ = 0 and f₁ = 1.

Define the polynomials {F_n, n ≥ 0} as F_n(x) = ∑f_ixⁱ for 0 ≤ i ≤ n.

For example, F₇(x) = x + x² + 2x³ + 3x⁴ + 5x⁵ + 8x⁶ + 13x⁷, and F₇(11) = 268357683.

Let n = 10¹⁵. Find the sum [∑_0≤x≤100 F_n(x)] mod 1307674368000 (= 15!).

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斐波那契数多项式

斐波那契数{f_n, n ≥ 0}由递归式f_n = f_n-1 + f_n-2给出，初值是f₀ = 0和f₁ = 1。

定义多项式{F_n, n ≥ 0}为F_n(x) = ∑f_ixⁱ，其中0 ≤ i ≤ n。

例如，F₇(x) = x + x² + 2x³ + 3x⁴ + 5x⁵ + 8x⁶ + 13x⁷，而F₇(11) = 268357683.

取n = 10¹⁵。求[∑_0≤x≤100 F_n(x)] mod 1307674368000 (= 15!)。

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