Solved 44 and 45.

ipython/EulerProblem045.ipynb

@@ -0,0 +1,128 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Triangular, pentagonal, and hexagonal (Euler Problem 45)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "collapsed": true
+   },
+   "source": [
+    "[https://projecteuler.net/problem=45](https://projecteuler.net/problem=45)\n",
+    "\n",
+    "Triangle, pentagonal, and hexagonal numbers are generated by the following formulae:\n",
+    "\n",
+    "Triangle\t \t$T_n=n(n+1)/2 = 1, 3, 6, 10, 15, ...$\n",
+    "\n",
+    "Pentagonal\t \t$P_n=n(3n−1)/2 = 1, 5, 12, 22, 35, ...$\n",
+    "\n",
+    "Hexagonal\t \t$H_n=n(2n−1) = 1, 6, 15, 28, 45, ...$\n",
+    "\n",
+    "It can be verified that T285 = P165 = H143 = 40755.\n",
+    "\n",
+    "Find the next triangle number that is also pentagonal and hexagonal."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "def T(n):\n",
+    "    return n * (n + 1) // 2\n",
+    "\n",
+    "assert(T(3) == 6)\n",
+    "\n",
+    "def P(n):\n",
+    "    return n * (3 * n - 1) // 2\n",
+    "\n",
+    "assert(P(3) == 12)\n",
+    "\n",
+    "def H(n):\n",
+    "    return n * (2 * n - 1)\n",
+    "\n",
+    "assert(H(3) == 15)"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "n_max = 100000\n",
+    "s = 0\n",
+    "\n",
+    "T_list = [T(n) for n in range(1, n_max + 1)]\n",
+    "P_set = {P(n) for n in range(1, n_max + 1)}\n",
+    "H_set = {H(n) for n in range(1, n_max + 1)}\n",
+    "\n",
+    "for n in range(1, n_max + 1):\n",
+    "    if T_list[n - 1] in P_set and T_list[n - 1] in H_set:\n",
+    "        t = T_list[n - 1]\n",
+    "        if t > 40755:\n",
+    "            s = t"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "1533776805\n"
+     ]
+    }
+   ],
+   "source": [
+    "print(s)\n",
+    "assert(s == 1533776805)"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "collapsed": true
+   },
+   "outputs": [],
+   "source": []
+  }
+ ],
+ "metadata": {
+  "completion_date": "Sat, 22 Dec 2018, 23:17",
+  "kernelspec": {
+   "display_name": "Python 3",
+   "language": "python3.6",
+   "name": "python3"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.6.5"
+  },
+  "tags": [
+   "triangular",
+   "pentagona",
+   "hexagonal"
+  ]
+ },
+ "nbformat": 4,
+ "nbformat_minor": 2
+}