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  • 特殊二阶递推式的一个关于最大公因数的性质

    定义

    对于互质的(a, b),即(gcd(a, b) = 1),有以下递推式:

    [f_0 = 0, f_1 = 1, f_n = a imes f_{n - 1} + b imes f_{n - 2} (n ge 2) ]

    引理1:(gcd(a, b) = gcd(b, a mod b))

    证明略

    引理2:若(gcd(b, c) = 1),则(gcd(a imes c, b) = gcd(a, b))

    证明略

    定理1:(gcd(f_n, f_{n + 1}) = 1)

    证明:

    (n le 2)时,结论显然成立

    (n > 2)时,(gcd(f_n, f_{n + 1}) = gcd(f_n, a imes f_n + b imes f_{n - 1}) = gcd(f_n, b imes f_{n - 1}) = gcd(f_n, b))

    (g = gcd(f_n, b)),则有(g|f_n, g|b)

    因为(f_n = a imes f_{n - 1} + b imes f_{n - 2}, gcd(a, b) = 1),所以(g|f_{n_1}),则有(g | gcd(f_n, f_{n - 1}))

    由此可得(gcd(f_n, f_{n - 1}) = 1 Longrightarrow gcd(f_n, f_{n + 1}) = 1)

    数学归纳即可

    定理2:(f_n = f_{i + 1} imes f_{n - 1} + b imes f_{n - i - 1} imes f_{n - i - 1} (1 le i < n))

    数学归纳即可

    定理3:(gcd(f_n, f_m) = f_{gcd(n, m)})

    证明:

    不失一般性假设(n ge m)

    (n = m),结论显然成立

    (n = m + 1),根据定理1,有(gcd(f_n, f_m) = 1),且(f_{gcd(n, m)} = f_1 = 1),结论成立

    (n > m + 1),根据定理2,有(f_n = f_{n - m} imes f_{m + 1} + b imes f_{n - m - 1} imes f_m)

    (gcd(f_n, f_m) = gcd(f_{n - m} imes f_{m + 1} + b imes f_{n - m - 1} imes f_m, f_m) = gcd(f_{n - m} imes f_{m + 1}, f_m) = gcd(f_{n - m}, f_m))

    即可得(gcd(f_n, f_m) = gcd(f_m, f_{n mod m}))

    证毕

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  • 原文地址:https://www.cnblogs.com/tkandi/p/10414428.html
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