euler/ipython/EulerProblem027.ipynb

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    "# Euler Problem 27\n",
    "\n",
    "Euler discovered the remarkable quadratic formula:\n",
    "\n",
    "$n^2 + n + 41$\n",
    "\n",
    "It turns out that the formula will produce 40 primes for the consecutive integer values 0≤n≤39. However, when $n=40$, $40^2 + 40 + 41 = 40(40+1)+41$ is divisible by $41$, and certainly when $n=41,41^2+41+41$ is clearly divisible by 41.\n",
    "\n",
    "The incredible formula $n^2−79n+1601$ was discovered, which produces 80 primes for the consecutive values 0≤n≤79. The product of the coefficients, $−79$ and $1601$, is $−126479$.\n",
    "\n",
    "Considering quadratics of the form:\n",
    "\n",
    "$n^2 + an +b$, where |a|<1000 and |b|≤1000\n",
    "\n",
    "where |n| is the modulus/absolute value of n e.g. |11|=11 and |−4|=4.\n",
    "\n",
    "Find the product of the coefficients, a and b, for the quadratic expression that produces the maximum number of primes for consecutive values of n, starting with n=0."
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    "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",
    "$s = a\\times b = (-2 * 31 + 1) * (31^2 - 31 + 41) = -61 \\times 971 = -59231$"
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       "-59231"
      ]
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   "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"
   ]
  }
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