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%I #21 Sep 28 2023 10:17:49
%S 15,77,99,240,354,18870,284481,302174,433197,440973,453086,2446619,
%T 5776855,9961111,17986255,19091527,28997648,37443680,40074848,
%U 47602448,67166528,129389763,141963648,146259296,152062688,202038871,203444576
%N Numbers k such that the sum of prime divisors of k is congruent to 2^k (mod k).
%C Prime factors counted with multiplicity. - _Harvey P. Dale_, Jul 25 2013
%H Giovanni Resta, <a href="/A051423/b051423.txt">Table of n, a(n) for n = 1..88</a> (terms < 10^11)
%e 15 = 3*5 and 2^15 = 3+5 (mod 15).
%t Select[Range[2500000],Total[Flatten[Table[#[[1]],{#[[2]]}]&/@ FactorInteger [#]]] ==PowerMod[2,#,#]&] (* _Harvey P. Dale_, Jul 25 2013 *)
%K nonn,easy,nice
%O 1,1
%A Joe K. Crump (joecr(AT)carolina.rr.com)
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