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n + d (n) + d (d (n)) +... = 2021 where d (k) is max divisor of k with d<k

Source: 2021 Czech and Slovak Olympiad III A p4

June 6, 2021
number theorydivisordivides

Problem Statement

Find all natural numbers nn for which equality holds n+d(n)+d(d(n))+...=2021n + d (n) + d (d (n)) +... = 2021, where d(0)=d(1)=0d (0) = d (1) = 0 and for k>1k> 1, d(k) d (k) is the superdivisor of the number kk (i.e. its largest divisor of dd with property d<kd <k).
(Tomáš Bárta)
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