Finding $\frac{1}{d_1}+\frac{1}{d_2}+\frac{1}{d_3}+...+\frac{1}{d_k}$

Question

If we assume that $d_1,d_2,d_3,...,d_k$ are the divisors for the positive integer $n$ except $1,n$ if $d_1+d_2+d_3+...+ d_k=72$ then how to find $$\frac{1}{d_1}+\frac{1}{d_2}+\frac{1}{d_3}+...+\frac{1}{d_k}$$

Answer 1

The divisor function satisfies the reflection formula, $$n^x\sigma_{-x}(n)=\sigma_{x}(n)$$

Since $$\sum_{d\mid n} d^x=\sum_{d\mid n} (n/d)^x$$

Answer 2

If you write: $$ \sum_{d \mid n} \frac{1}{d} = \frac{1}{n} \sum_{d \mid n} \frac{n}{d} = \frac{1}{n} \sum_{d \mid n} d = \frac{72}{n} $$ The divisor sum $\sum_{d \mid n} d = \sigma(n)$ is easily seen to satisfy $\sigma(n) \ge n + 1$ (those two are divisors always; they are the only ones if $n$ is prime, otherwise there are more). So you'd have to check up to $n = 71$. To find out what $n$ is so that $\sigma(n) = 72$, a fast trip to http://oeis.org/A000203 shows that the full list is 30, 46, 51, 55, 71.