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CZ-SK-MO 2021 Problem 4

    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)

    Dataset text AI-MO AoPS. Quoted with source.

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