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Numbers m for which Sum_{i=1..k} (p(i)/(p(i)-1)) + Product_{i=1..k} (p(i)/(p(i)-1)) is an integer, where p(i) are the k prime factors of m (with multiplicity).
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%I #8 Feb 14 2021 18:34:32

%S 2,3,4,8,16,32,36,64,72,108,128,144,200,256,288,396,512,528,576,588,

%T 1024,1040,1152,1296,2000,2048,2304,2320,2400,2592,3888,4096,4160,

%U 4608,4752,4800,5184,5600,6552,7200,8192,8448,9216,9600,9936,10368,11316,12000

%N Numbers m for which Sum_{i=1..k} (p(i)/(p(i)-1)) + Product_{i=1..k} (p(i)/(p(i)-1)) is an integer, where p(i) are the k prime factors of m (with multiplicity).

%C Apart from 3 all terms are even.

%H Paolo P. Lava, <a href="/A224912/b224912.txt">Table of n, a(n) for n = 1..200</a>

%e Prime factors of 11316 are 2^2, 3, 23 and 41.

%e Sum_{i=1..5} (p(i)/(p(i)-1)) = 2*(2/(2-1)) + 3/(3-1) + 23/(23-1) + 41/(41-1) = 3331/440.

%e Sroduct_{i=1..5} (p(i)/(p(i)-1)) = 2*(2/(2-1)) * 3/(3-1) * 23/(23-1) * 41/(41-1) = 2829/440.

%e Their sum is an integer: 3331/440 + 2829/440 = 14.

%p with(numtheory);

%p A224912:=proc(i) local b,c,d,n,p;

%p for n from 2 to i do p:=ifactors(n)[2];

%p b:=add(op(2,d)*op(1,d)/(op(1,d)-1),d=p)+mul((op(1,d)/(op(1,d)-1))^op(2,d),d=p);

%p if trunc(b)=b then print(n); fi; od; end:

%p A224912(10^6);

%Y Cf. A199767, A198391.

%K nonn

%O 1,1

%A _Paolo P. Lava_, Apr 19 2013