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simplify sieving method

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filifa 2025-07-19 21:08:12 -04:00
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14
notebooks/problem0021.ipynb
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@ -68,7 +68,7 @@
"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 $\\sigma$. This is possible because the sum of divisors function is multiplicative."
"Since we need all the sums of divisors up to 10000, instead of factoring each number individually, we could sieve the values of $\\sigma$."
]
},
{
@ -82,18 +82,12 @@
" dsum = [1 for _ in range(0, limit)]\n",
" \n",
" for n in range(0, limit):\n",
" if n == 0 or n == 1 or dsum[n] != 1:\n",
" if n == 0 or n == 1:\n",
" yield dsum[n]\n",
" continue\n",
" \n",
" m = 1\n",
" while m * n < limit:\n",
" dsum[m*n] = dsum[m] + m*n\n",
" for k in range(2 * m * n, limit, m * n):\n",
" if k % (m*n^2) != 0:\n",
" dsum[k] *= dsum[m*n]\n",
" \n",
" m *= n\n",
" for k in range(n, limit, n):\n",
" dsum[k] += n\n",
" \n",
" yield dsum[n]"
]