%I #37 Jul 18 2021 03:04:11
%S 125,142,1005,118678,706862,18481615,122003411,30330043,5923078409,
%T 22110133333,120175787632,5971473681952
%N a(n) is the smallest number k such that the sum of the n-th powers of the digits of k equals the sum of the divisors of k other than 1 and k.
%e a(5) = 118678 because 1^5 + 1^5 + 8^5 + 6^5 + 7^5 + 8^5 = 90121, and sum of the divisors 1 < d < a(5) = sigma(118678) - 118678 - 1 = 90121.
%o (PARI) f(k, n) = my(d=digits(k)); sum(i=1, #d, d[i]^n);
%o a(n) = my(k=1); while(f(k, n) != sigma(k)-k-1, k++); k; \\ _Michel Marcus_, Sep 29 2018
%Y Cf. A070308 (n=2, "Canada perfect numbers").
%Y Cf. A202279 (n=3), A202147 (n=4), A202285 (n=5).
%K nonn,hard,base,more
%O 2,1
%A _Michel Lagneau_, Dec 16 2011
%E a(8)-a(9) added by _Amiram Eldar_, Sep 29 2018
%E a(10)-a(13) from _Giovanni Resta_, Oct 04 2018
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