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%I #13 Dec 01 2024 11:37:17
%S 2,21,321,7721,184729,185145,21110729,54929017,134099385,3331670169
%N The smallest number whose sum of proper divisors (aliquot parts) is the repunit number (10^n - 1)/9, denoted by R_n.
%e a(4) = 7721 because 7721 is the smallest number whose sum of proper divisors is R_4 = 1111.
%o (PARI) a(n) = my(k=1); while (sigma(k)-k != (10^n - 1)/9, k++); k; \\ _Michel Marcus_, Nov 16 2024
%Y Cf. A001065, A002275.
%K nonn,hard,more
%O 1,1
%A _Shyam Sunder Gupta_, Nov 16 2024