Amicable numbers

Let d(n) be defined as the sum of proper divisors of n (numbers less than n which divide evenly into n).
If d(a) = b and d(b) = a, where a ≠ b, then a and b are an amicable pair and each of a and b are called amicable numbers.

For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110; therefore d(220) = 284. The proper divisors of 284 are 1, 2, 4, 71 and 142; so d(284) = 220.

Evaluate the sum of all the amicable numbers under 10000.

Solution

p21.py
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#!/usr/bin/python
# -*- coding: utf-8 -*-
def sum_proper_factors(n):
(result, sqrt) = (1, n ** 0.5)
(start, step) = n % 2 == 1 and (3, 2) or (2, 1)
for i in range(start, int(sqrt) + 1, step):
if n % i == 0:
result += i + n / i
if sqrt == int(sqrt):
result -= sqrt
return result
def main():
result = 0
for i in range(1, 10000):
sum1 = sum_proper_factors(i)
if sum1 > i:
if i == sum_proper_factors(sum1):
result += i + sum1
print result
if __name__ == '__main__':
import time
startTime = time.time()
main()
print time.time() - startTime
p21.go
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package main
import (
"fmt"
"math"
"time"
)
func sum_proper_factors(n int) int {
sum, sqrt := 1, math.Sqrt(float64(n))
start, step := 2, 1
if n%2 == 1 {
start, step = 3, 2
}
for i := start; i <= int(sqrt); i += step {
if n%i == 0 {
sum += i + n/i
}
}
if sqrt == float64(int(sqrt)) {
sum -= int(sqrt)
}
return sum
}
func main() {
result, startTime := 0, time.Now()
for i := 1; i < 10000; i++ {
iSum := sum_proper_factors(i)
if iSum > i {
if i == sum_proper_factors(iSum) {
result += i + iSum
}
}
}
fmt.Println(result, time.Now().Sub(startTime))
}

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