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f(x)=⌊230.403243784−x2⌋×10−9f(x) = \lfloor 2^{30.403243784-x^2}\rfloor × 10^{-9}f(x)=⌊230.403243784−x2⌋×10−9 (⌊⋅⌋\lfloor \cdot \rfloor⌊⋅⌋は床関数)とし, 数列unu_nunを,u0=−1,un+1=f(un)u_0 = -1, u_{n+1} = f(u_n)u0=−1,un+1=f(un)と定義する.
n=1012n = 10^{12}n=1012でun+un+1u_n + u_{n+1}un+un+1を求めよ. 小数点以下を 9 桁で解答せよ.