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sequence with powers of primes, a_{j+1} = f (a_j)

Source: Austrian - Polish 1993 APMC

May 3, 2020
power of primeProductSumnumber theory

Problem Statement

Define f(n)=n+1f (n) = n + 1 if n=pk>1n = p^k > 1 is a power of a prime number, and f(n)=p1k1+...+prkrf (n) =p_1^{k_1}+... + p_r^{k_r} for natural numbers n=p1k1...prkrn = p_1^{k_1}... p_r^{k_r} (r>1,ki>0r > 1, k_i > 0). Given m>1m > 1, we construct the sequence a0=m,aj+1=f(aj)a_0 = m, a_{j+1} = f (a_j) for j≥0j \ge 0 and denote by g(m)g(m) the smallest term in this sequence. For each m>1m > 1, determine g(m)g(m).
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