Editing 410: Math Paper
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:We need first to define a divisor function over the integers, written σ(n) if you're so inclined. To get it first we get all the integers that divide into n. So for 3, it's 1 and 3. For 4, it's 1, 2, and 4, and for 5 it's only 1 and 5. | :We need first to define a divisor function over the integers, written σ(n) if you're so inclined. To get it first we get all the integers that divide into n. So for 3, it's 1 and 3. For 4, it's 1, 2, and 4, and for 5 it's only 1 and 5. | ||
− | :Now sum them to get σ(n). So σ(3) = 1 + 3 = 4, or σ(4) = 1 + 2 + 4 = | + | :Now sum them to get σ(n). So σ(3) = 1 + 3 = 4, or σ(4) = 1 + 2 + 4 = 6, and so on. |
:For each of these n, there is something called a characteristic ratio. Now that's just the divisors function over the integer itself: σ(n)/n. (This is the formula shown at the top of Cueball's slide). So the characteristic ratio where n = 6 is σ(6)/6 = 12/6 = 2. | :For each of these n, there is something called a characteristic ratio. Now that's just the divisors function over the integer itself: σ(n)/n. (This is the formula shown at the top of Cueball's slide). So the characteristic ratio where n = 6 is σ(6)/6 = 12/6 = 2. |