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    "# [Number Spiral Diagonals](https://projecteuler.net/problem=28)\n",
    "\n",
    "Whether you choose to program or just use pen and paper, there's a lot of different ways to tackle this problem. Here's one approach.\n",
    "\n",
    "If we have an $n \\times n$ spiral (note that $n$ must be odd!), it's pretty easy to get a formula for just the sum of the corners. The top right corner will always be $n^2$, and since it's an $n \\times n$ spiral (as mentioned one sentence ago), the top left corner will just be $n^2 - (n - 1)$. We can continue subtracting $n-1$ to get the values of the other corners. Adding these up, we get\n",
    "$$f(n) = n^2 + (n^2 - (n-1)) + (n^2 - 2(n-1)) + (n^2 - 3(n-1)) = 4n^2 - 6n + 6$$\n",
    "\n",
    "Of course, we want the sums of the *diagonals*, not just the outermost corners. We can think about this as getting the sum of the corners of the $n \\times n$ spiral, plus the sum of the corners of the $(n-2) \\times (n-2)$ spiral one layer deeper, and so on until we reach the 1 at the center.\n",
    "$$g(n) = f(n) + f(n-2) + f(n-4) + \\cdots + f(5) + f(3) + 1 = 1 + \\sum_{k=1}^{(n-1)/2} f(2k + 1) = 1 + \\sum_{k=1}^{(n-1)/2} (16k^2 + 4k + 4)$$\n",
    "\n",
    "Now we can apply [summation identities](https://en.wikipedia.org/wiki/Summation) (including the return of triangular numbers and square pyramidal numbers from [problem 6](https://projecteuler.net/problem=6)) to get a closed formula:\n",
    "$$g(n) = \\frac{2}{3}n^3 + \\frac{1}{2}n^2 + \\frac{4}{3}n - \\frac{3}{2}$$\n",
    "Plugging in 1001, we get our answer: $g(1001) = 669171001$.\n",
    "\n",
    "Side note: this practice of writing the natural numbers in a spiral, combined with marking the prime numbers, has been coined the [Ulam spiral](https://en.wikipedia.org/wiki/Ulam_spiral). Somewhat interestingly, lots of primes appear in vertical, horizontal, and diagonal lines when laid out this way.\n",
    "\n",
    "## Relevant sequences\n",
    "* Numbers on diagonals: [A200975](https://oeis.org/A200975)\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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