From 7623a7b99e8c219a87cbc506c1fa0d39e98ed77d Mon Sep 17 00:00:00 2001
From: filifa <filifa.breath652@8alias.com>
Date: Sat, 19 Apr 2025 00:44:17 -0400
Subject: [PATCH] add problem 34

---
 notebooks/problem0034.ipynb | 126 ++++++++++++++++++++++++++++++++++++
 1 file changed, 126 insertions(+)
 create mode 100644 notebooks/problem0034.ipynb

diff --git a/notebooks/problem0034.ipynb b/notebooks/problem0034.ipynb
new file mode 100644
index 0000000..37938c8
--- /dev/null
+++ b/notebooks/problem0034.ipynb
@@ -0,0 +1,126 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "eba5f5fb",
+   "metadata": {},
+   "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:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "06bc38a2",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "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",
+   "metadata": {},
+   "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)"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "SageMath 9.5",
+   "language": "sage",
+   "name": "sagemath"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.11.2"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}