4

Problems(1)

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

6/6/2021

Find all natural numbers

n

for which equality holds

n + d (n) + d (d (n)) +... = 2021

, where

d (0) = d (1) = 0

and for

k> 1

d (k)

is the superdivisor of the number

k

(i.e. its largest divisor of

d

with property

d <k

).
(Tomáš Bárta)

number theorydivisordivides