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   "source": [
    "# [Digit Factorials](https://projecteuler.net/problem=34)\n",
    "\n",
    "Numbers like 145 are called [factorions](https://en.wikipedia.org/wiki/Factorion).\n",
    "\n",
    "We can use similar logic as [problem 30](https://projecteuler.net/problem=30) to produce an upper bound for how many numbers to brute force. Let $f(n)$ be the sum of the factorials of the digits of $n$, and consider the largest seven digit number, 9999999. $f(9999999) = 7 \\times 9! = 2540160$. If we replaced any digits in 9999999, they would have to be smaller than 9, so the sum of the factorials of the digits would be smaller than 2540160. Therefore, for any $n \\leq 9999999$, $f(n) \\leq 2540160$, so for $n$ to *equal* $f(n)$, $n$ must be less than or equal to 2540160, as well.\n",
    "\n",
    "We can write a simple recursive function for calculating the sum of the factorials of the digits:"
   ]
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    "from functools import cache\n",
    "from math import factorial\n",
    "\n",
    "@cache\n",
    "def sfd(n):\n",
    "    q = n // 10\n",
    "    if q == 0:\n",
    "        return factorial(n)\n",
    "    \n",
    "    return factorial(n % 10) + sfd(q)"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "c6c47e17",
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   "source": [
    "Now we just iterate over all the numbers less than our upper bound."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "id": "b4642720",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "{145, 40585}"
      ]
     },
     "execution_count": 2,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "factorions = {n for n in range(3, 2540161) if sfd(n) == n}\n",
    "factorions"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "1257c102",
   "metadata": {},
   "source": [
    "Interestingly, there's only two factorions!"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "id": "caccb3bd",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "40730"
      ]
     },
     "execution_count": 3,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "sum(factorions)"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "b25666f7",
   "metadata": {},
   "source": [
    "## Relevant sequences\n",
    "* Factorions: [A014080](https://oeis.org/A014080)\n",
    "\n",
    "#### Copyright (C) 2025 filifa\n",
    "\n",
    "This work is licensed under the [Creative Commons Attribution-ShareAlike 4.0 International license](https://creativecommons.org/licenses/by-sa/4.0/) and the [BSD Zero Clause license](https://spdx.org/licenses/0BSD.html)."
   ]
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