List all the abundant integers up to 200

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- Jun 13th 2009, 08:38 AMmiz.perfect84Abundant Integer
List all the abundant integers up to 200

- Jun 13th 2009, 11:38 AMTheAbstractionist
For each $\displaystyle p$ there always exist integers $\displaystyle r$ such that $\displaystyle 2^{r+1}>p+1.$ Hence there exist integers $\displaystyle r$ such that

$\displaystyle \sigma(n)\ =\ \sigma(2^rp)$

$\displaystyle =\ \sigma(2^r)\sigma(p)$$\displaystyle =\ (1+2+2^2+\cdots+2^r)(p+1)$$\displaystyle =\ (2^{r+1}-1)(p+1)$$\displaystyle =\ 2^{r+1}p+2^{r+1}-p-1$$\displaystyle >\ 2^{r+1}p$$\displaystyle =\ 2n$