Finished 27 in ipython. Very nice one to be honest. Fuck yeah.

ipython/EulerProblem027.ipynb

@@ -24,13 +24,91 @@
    ]
   },
   {
-   "cell_type": "code",
-   "execution_count": null,
+   "cell_type": "markdown",
    "metadata": {
     "collapsed": true
    },
-   "outputs": [],
-   "source": []
+   "source": [
+    "Okay, bruteforcing this complete thing is definitely hard. The interesting thing is that euler provided two examples. If we calculate the primes for both terms we see that there is a certain overlap. This indicates that there is a relation between the two. Sure enough if we put $n-40$ into the first term we get the following.\n",
+    "\n",
+    "$(n - 40)^2 + (n - 40) + 41 = n^2 - 80n + 1600 + n - 40 + 41 = n^2 - 79n +1601$\n",
+    "\n",
+    "Let's assume that all incredible formulas can be derived by inserting $(n - p)$ into the formula. Of course, what ever value we choose for p the resulting terms must not exceed the boundaries for a or b. We calculate the boundaries.\n",
+    "\n",
+    "$(n-p)^2 + (n-p) + 41 = n^2 + n(-2p + 1) + p^2 - p + 41$\n",
+    "\n",
+    "Where $(-2p + 1) = a$ and $p^2 -p + 41 = b$. We can now calulate the bounds for a:\n",
+    "\n",
+    "$-2p + 1 > -1000 \\rightarrow p < 500.5$\n",
+    "\n",
+    "$-2 p + 1 < 1000 \\rightarrow p > -499.5$\n",
+    "\n",
+    "And b:\n",
+    "\n",
+    "$p^2 - p + 41 <= 1000$\n",
+    "\n",
+    "$-30.472 < p < 31.472$\n",
+    "\n",
+    "$p^2 - p + 41 >= 1000$ True for $\\forall p \\in \\mathbb{N}$\n",
+    "\n",
+    "So now we only have to check for the values p in range(-30, 32). Alternatively, for the example $p = 40$ was used, maybe the next smaller value $p = 31$ yields the correct solution:\n",
+    "\n",
+    "$s = a\\times b = (-2 * 31 + 1) * (31^2 - 31 + 41) = -61 \\times 971 = -59231$"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "-59231"
+      ]
+     },
+     "execution_count": 1,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "from functools import lru_cache\n",
+    "\n",
+    "@lru_cache(maxsize=1000)\n",
+    "def is_prime(n):\n",
+    "    if n < 2:\n",
+    "        return False\n",
+    "    from math import sqrt\n",
+    "    for s in range(2, int(sqrt(n) + 1)):\n",
+    "        if n % s == 0:\n",
+    "            return False\n",
+    "        if s * s > n:\n",
+    "            return True\n",
+    "    return True\n",
+    "                   \n",
+    "assert(is_prime(41) == True)\n",
+    "\n",
+    "def get_quadratic(n, p):\n",
+    "    return n*n + n * (-2* p + 1) + p*p - p + 41\n",
+    "\n",
+    "n_max = 0\n",
+    "p_max = 0\n",
+    "\n",
+    "for p in range(-30, 32):\n",
+    "    for n in range(0, 10000):\n",
+    "        if not is_prime(get_quadratic(n, p)) and n > n_max:\n",
+    "            n_max = n\n",
+    "            p_max = p\n",
+    "            break\n",
+    "\n",
+    "p = p_max\n",
+    "s = (-2 * p + 1) * (p*p - p + 41)\n",
+    "assert(s == -59231)\n",
+    "s"
+   ]
   }
  ],
  "metadata": {
@@ -53,7 +131,10 @@
    "version": "3.5.4"
   },
   "tags": [
-   "quadratic primes"
+   "quadratic",
+   "primes",
+   "formula",
+   "brain"
   ]
  },
  "nbformat": 4,