An is equal to the sum of the factors of n divided by n. Thus `A12=(1+2+3+4+6+12)/12=28/12=2.bar(3)`

Another formulation is An is equal to the sum of the reciprocals of the factors of n. Here `A12=1/1+1/2+1/3+1/4+1/6+1/12=2.bar(3)`

The way to see this is to write `An=(f_1+f_2+...+f_k)/n=(f_1)/n+(f_2)/n+...(f_k)/n`

` ` Note that `(f_1)/n=1/f_k` and `(f_k)/n=1/(f_1)` . This works for all pairs of factors.

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**Given a number that is a multiple of 6. Then the factors include 1,2,3, and 6. Thus `An>=1/1+1/2+1/3+1/6=2` and the abundancy ratio is greater than or equal to 2 for all multiples of 6.**

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