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    "# [Amicable Numbers](https://projecteuler.net/problem=21)\n",
    "\n",
    "The sum of the proper divisors of a number is called the [aliquot sum](https://en.wikipedia.org/wiki/Aliquot_sum).\n",
    "\n",
    "SageMath [provides the `sigma` function](https://doc.sagemath.org/html/en/reference/rings_standard/sage/arith/misc.html#sage.arith.misc.Sigma), which can compute the sum of the divisors of $n$. The aliquot sum is then just `sigma(n) - n`.\n",
    "\n",
    "If a number is equal to its own aliquot sum, it's called a [perfect number](https://en.wikipedia.org/wiki/Perfect_number), and we exclude those numbers from our total."
   ]
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     "text": [
      "31626\n"
     ]
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   "source": [
    "aliquot_sum = lambda n: sigma(n) - n\n",
    "\n",
    "amicables = set()\n",
    "for a in range(2, 10000):\n",
    "    if a in amicables:\n",
    "        continue\n",
    "        \n",
    "    b = aliquot_sum(a)\n",
    "    if a == b:\n",
    "        continue\n",
    "    \n",
    "    if a == aliquot_sum(b):\n",
    "        amicables.update({a, b})\n",
    "\n",
    "print(sum(amicables))"
   ]
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   "source": [
    "Funny enough, there's only five pairs of amicable numbers below 10,000. If you looked up [amicable numbers](https://en.wikipedia.org/wiki/Amicable_numbers), you may have stumbled on all the numbers you need to answer the problem!\n",
    "\n",
    "## Sum of divisors\n",
    "Of course, you could implement your own [divisor sum function](https://en.wikipedia.org/wiki/Divisor_function). In [problem 12](https://projecteuler.net/problem=12), we implemented a divisor *counting* function, which is related.\n",
    "\n",
    "One important property of the divisor sum function $\\sigma(n)$ is that it is [multiplicative](https://en.wikipedia.org/wiki/Multiplicative_function). This means that if $a$ and $b$ are [coprime](https://en.wikipedia.org/wiki/Coprime_integers), $\\sigma(ab) = \\sigma(a)\\sigma(b)$. It follows that if $n = 2^{r_1}3^{r_2}5^{r_3}7^{r_4}\\cdots$, then\n",
    "$$\\sigma(n) = \\sigma(2^{r_1})\\sigma(3^{r_2})\\sigma(5^{r_3})\\sigma(7^{r_4})\\cdots$$\n",
    "\n",
    "Furthermore, for a prime $p$, $\\sigma(p^k) = 1 + p + p^2 + \\cdots + p^k$. This expression is actually a partial sum of a [geometric series](https://en.wikipedia.org/wiki/Geometric_series), which has a closed formula:\n",
    "$$1 + p + p^2 + \\cdots + p^k = \\frac{p^{k+1}-1}{p-1}$$\n",
    "\n",
    "Putting it all together, we can compute $\\sigma(n)$ as\n",
    "$$\\sigma(n) = \\frac{2^{r_1+1}-1}{2-1} \\cdot \\frac{3^{r_2+1}-1}{3-1} \\cdot \\frac{5^{r_3+1}-1}{5-1} \\cdot \\frac{7^{r_4+1}-1}{7-1} \\cdot \\cdots$$\n",
    "\n",
    "Therefore, if you have the number's factorization (see [problem 3](https://projecteuler.net/problem=3)), you can use it to compute the sum of its divisors.\n",
    "\n",
    "## Sieving the sums of divisors\n",
    "Since we need all the sums of divisors up to 10000, instead of factoring each number individually, we could sieve the values of $\\sigma$."
   ]
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   "source": [
    "def sum_of_divisors_sieve(limit):\n",
    "    dsum = [1 for _ in range(0, limit)]\n",
    "    \n",
    "    for n in range(0, limit):\n",
    "        if n == 0 or n == 1:\n",
    "            yield dsum[n]\n",
    "            continue\n",
    "            \n",
    "        for k in range(n, limit, n):\n",
    "            dsum[k] += n\n",
    "        \n",
    "        yield dsum[n]"
   ]
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   "source": [
    "Here's a check to make sure the sieve matches the values from SageMath."
   ]
  },
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   "cell_type": "code",
   "execution_count": 3,
   "id": "0539d1fc",
   "metadata": {},
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   "source": [
    "s = list(sum_of_divisors_sieve(10000))\n",
    "assert all(s[n] == sigma(n) for n in range(1, 10000))"
   ]
  },
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   "cell_type": "markdown",
   "id": "a847f754",
   "metadata": {},
   "source": [
    "## Relevent sequences\n",
    "* Amicable numbers: [A063990](https://oeis.org/A063990)"
   ]
  }
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