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VN-TST 1993 Problem 5

  • Day 2

Let an integer k>1k > 1 be given. For each integer n>1n > 1 , we put

f(n)=k⋅n⋅(1−1p1)⋅(1−1p2)⋯(1−1pr)f(n) = k \cdot n \cdot \left(1-\frac{1}{p_1}\right) \cdot \left(1-\frac{1}{p_2}\right) \cdots \left(1-\frac{1}{p_r}\right)

where p1,p2,…,prp_1, p_2, \ldots, p_r are all distinct prime divisors of nn . Find all values kk for which the sequence {xm}\{x_m\} defined by x0=ax_0 = a and xm+1=f(xm),m=0,1,2,3,…x_{m+1} = f(x_m), m = 0, 1, 2, 3, \ldots is bounded for all integers a>1a > 1 .

Dataset text AI-MO AoPS. Quoted with source.

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